Resolução Física -Sears e Zemansky Vol 4 - Young e Freedman 12ªed

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33-1 
THE NATURE AND PROPAGATION OF LIGHT 
 33.1. IDENTIFY: For reflection, r aθ θ= . 
SET UP: The desired path of the ray is sketched in Figure 33.1. 
EXECUTE: 14.0 cmtan
11.5 cm
φ = , so 50.6φ = ° . 90 39.4rθ φ= − =° ° and 39.4r aθ θ= = ° . 
EVALUATE: The angle of incidence is measured from the normal to the surface. 
 
Figure 33.1 
 33.2 IDENTIFY: For reflection, r aθ θ= . 
SET UP: The angles of incidence and reflection at each reflection are shown in Figure 33.2. For the rays to be 
perpendicular when they cross, 90α = ° . 
EXECUTE: (a) 90θ φ+ = ° and 90β φ+ = ° , so β θ= . 90
2
α β+ = ° and 180 2α θ= −° . 
(b) 1 12 2(180 ) (180 90 ) 45θ α= − = − =° ° ° ° . 
EVALUATE: As 0θ → ° , 180α → ° . This corresponds to the incident and reflected rays traveling in nearly the 
same direction. As 90θ → ° , 0α → ° . This corresponds to the incident and reflected rays traveling in nearly 
opposite directions. 
 
Figure 33.2 
33
33-2 Chapter 33 
 33.3. IDENTIFY and SET UP: Use Eqs.(33.1) and (33.5) to calculate v and .λ 
EXECUTE: (a) 
8
82.998 10 m/s so 2.04 10 m/s
1.47
c cn v
v n
×
= = = = × 
(b) 0 650 nm 442 nm
1.47n
λλ = = = 
EVALUATE: Light is slower in the liquid than in vacuum. By ,v f λ= when v is smaller, λ is smaller. 
 33.4. IDENTIFY: In air, 0c f λ= . In glass, 0n
λλ = . 
SET UP: 83.00 10 m/sc = × 
EXECUTE: (a) 
8
0 14
3.00 10 m/s 517 nm
5.80 10 Hz
c
f
λ ×= = =
×
 
(b) 0 517 nm 340 nm
1.52n
λλ = = = 
EVALUATE: In glass the light travels slower than in vacuum and the wavelength is smaller. 
 33.5. IDENTIFY: cn
v
= . 0
n
λλ = , where 0λ is the wavelength in vacuum. 
SET UP: 83.00 10 m/sc = × . n for air is only slightly larger than unity. 
EXECUTE: (a) 
8
8
3.00 10 m/s 1.54.
1.94 10 m/s
cn
v
×
= = =
×
 
(b) 7 70 (1.54)(3.55 10 m) 5.47 10 m.nλ λ
− −= = × = × 
EVALUATE: In quartz the speed is lower and the wavelength is smaller than in air. 
 33.6. IDENTIFY: 0
n
λλ = . 
SET UP: From Table 33.1, water 1.333n = and benzene 1.501n = . 
EXECUTE: (a) water water benzene benzene 0n nλ λ λ= = . waterbenzene water
benzene
1.333(438 nm) 389 nm
1.501
n
n
λ λ
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
. 
(b) 0 water water (438 nm)(1.333) 584 nmnλ λ= = = 
EVALUATE: λ is smallest in benzene, since n is largest for benzene. 
 33.7. IDENTIFY: Apply Eqs.(33.2) and (33.4) to calculate and .r bθ θ The angles in these equations are measured with 
respect to the normal, not the surface. 
(a) SET UP: The incident, reflected and refracted rays are shown in Figure 33.7. 
 
EXECUTE: 42.5r aθ θ= = ° 
The reflected ray makes an 
angle of 90.0 47.5rθ° − = ° 
with the surface of the glass. 
Figure 33.7 
(b) sin sin ,a a b bn nθ θ= where the angles are measured from the normal to the interface. 
sin (1.00)(sin 42.5 )sin 0.4070
1.66
a a
b
b
n
n
θθ °= = = 
24.0bθ = ° 
The refracted ray makes an angle of 90.0 66.0bθ° − = ° with the surface of the glass. 
EVALUATE: The light is bent toward the normal when the light enters the material of larger refractive index. 
 33.8. IDENTIFY: Use the distance and time to find the speed of light in the plastic. cn
v
= . 
SET UP: 83.00 10 m/sc = × 
EXECUTE: 89
2.50 m 2.17 10 m s
11.5 10 s
dv
t −
= = = ×
×
. 
8
8
3.00 10 m/s 1.38
2.17 10 m/s
cn
v
×
= = =
×
. 
EVALUATE: In air light travels this same distance in 8
2.50 m 8.3 ns
3.00 10 m/s
=
×
. 
The Nature and Propagation of Light 33-3 
 33.9. IDENTIFY and SET UP: Use Snell�s law to find the index of refraction of the plastic and then use Eq.(33.1) to 
calculate the speed v of light in the plastic. 
EXECUTE: sin sina a b bn nθ θ= 
sin sin 62.71.00 1.194
sin sin 48.1
a
b a
b
n n θ
θ
⎛ ⎞ °⎛ ⎞= = =⎜ ⎟ ⎜ ⎟°⎝ ⎠⎝ ⎠
 
8 8 so (3.00 10 m/s) /1.194 2.51 10 m/sc cn v
v n
= = = × = × 
EVALUATE: Light is slower in plastic than in air. When the light goes from air into the plastic it is bent toward 
the normal. 
33.10. IDENTIFY: Apply Snell�s law at both interfaces. 
SET UP: The path of the ray is sketched in Figure 33.10. Table 33.1 gives 1.329n = for the methanol. 
EXECUTE: (a) At the air-glass interface glass(1.00)sin 41.3 sinn α° = . At the glass-methanol interface 
glass sin (1.329)sinn α θ= . Combining these two equations gives sin 41.3 1.329sinθ° = and 29.8θ = °. 
(b) The same figures applies as for part (a), except 20.2θ = ° . (1.00)sin 41.3 sin 20.2n° = ° and 1.91n = . 
EVALUATE: The angle α is 25.2°. The index of refraction of methanol is less than that of the glass and the ray is 
bent away from the normal at the glass → methanol interface. The unknown liquid has an index of refraction 
greater than that of the glass, so the ray is bent toward the normal at the glass → liquid interface. 
 
Figure 33.10 
33.11. IDENTIFY: Apply Snell�s law to each refraction. 
SET UP: Let the light initially be in the material with refractive index an and let the final slab have refractive 
index bn . In part (a) let the middle slab have refractive index 1n . 
EXECUTE: (a) 1st interface: 1 1sin sina an nθ θ= . 2
nd interface: 1 1sin sinb bn nθ θ= . Combining the two equations 
gives sin sina a b bn nθ θ= . This is the equation that would apply if the middle slab were absent. 
(b) For N slabs, 1 1sin sina an nθ θ= , 1 1 2 2sin sinn nθ θ= , �, 2 2sin sinN N b bn nθ θ− − = . Combining all these equations 
gives sin sina a b bn nθ θ= . 
EVALUATE: The final direction of travel depends on the angle of incidence in the first slab and the refractive 
indices of the first and last slabs. 
33.12. IDENTIFY: Apply Snell's law to the refraction at each interface. 
SET UP: air 1.00n = . water 1.333n = . 
EXECUTE: (a) airwater air
water
1.00arcsin sin arcsin sin35.0 25.5 .
1.333
n
n
θ θ
⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 
EVALUATE: (b) This calculation has no dependence on the glass because we can omit that step in the 
air air glass glass wate waterchain: sin sin sin .rn n nθ θ θ= = 
33.13. IDENTIFY: When a wave passes from one material into another, the number of waves per second that cross the 
boundary is the same on both sides of the boundary, so the frequency does not change. The wavelength and speed 
of the wave, however, do change. 
SET UP: In a material having index of refraction n, the wavelength is 0 ,
n
λλ = where λ0 is the wavelength in 
vacuum, and the speed is .c
n
 
EXECUTE: (a) The frequency is the same, so it is still f. The wavelength becomes 0 ,
n
λλ = so λ0 = nλ. The speed 
is ,cv
n
= so c = nv. 
(b) The frequency is still f. The wavelength becomes 0 n n
n n n
λ λλ λ⎛ ⎞′ = = = ⎜ ⎟′ ′ ′⎝ ⎠
 and the speed becomes 
c nv nv v
n n n
⎛ ⎞′ = = = ⎜ ⎟′ ′ ′⎝ ⎠
 
EVALUATE: These results give the speed and wavelength in a new medium in terms of the original medium 
without referring them to the values in vacuum (or air). 
33-4 Chapter 33 
33.14. IDENTIFY: Apply the law of reflection. 
SET UP: The mirror in its original position and after being rotated by an angle θ are shown in Figure 33.14. α is 
the angle through which the reflected ray rotates when the mirror rotates. The two angles labeled φ are equal and 
the two angles labeled φ′are equal because of the law of reflection. The two angles labeled θ are equal because the 
lines forming one angle are perpendicular to the lines forming the other angle. 
EXECUTE: From the diagram, 2 2 2( )α φ φ φ φ′ ′= − = − and θ φ φ′= − . 2α θ= , as was to be shown. 
EVALUATE: This result is independent of the initial angle of incidence. 
 
Figure 33.14 
33.15. IDENTIFY: Apply sin sina a b bn nθ θ= . 
SET UP: The light refracts from the liquid into the glass, so 1.70an = , 62.0aθ = °. 1.58bn = . 
EXECUTE: 1.70sin sin sin 62.0 0.950
1.58
a
b a
b
n
n
θ θ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
° and 71.8bθ = ° . 
EVALUATE: The ray refracts into a material of smaller n, so it is bent away from the normal. 
33.16. IDENTIFY: Apply Snell's law. 
SET UP: aθ and bθ are measured relative to the normal to the surface of the interface. 60.0 15.0 45.0aθ = − =° ° ° . 
EXECUTE: 1.33arcsin sin arcsin sin 45.0 38.2
1.52
a
ba
b
n
n
θ θ⎛ ⎞ ⎛ ⎞= = ° =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
°. But this is the angle from the normal to the surface, 
so the angle from the vertical is an additional 15° because of the tilt of the surface. Therefore, the angle is 53.2°. 
EVALUATE: Compared to Example 33.1, aθ is shifted by 15° but the shift in bθ is only 53.2 49.3 3.9− =° ° ° . 
33.17. IDENTIFY: The critical angle for total internal reflection is aθ that gives 90bθ = ° in Snell's law. 
SET UP: In Figure 33.17 the angle of incidence aθ is related to angle θ by 90aθ θ+ = ° . 
EXECUTE: (a) Calculate aθ that gives 90bθ = °. 1.60an = , 1.00bn = so sin sina a b bn nθ θ= gives 
(1.60)sin (1.00)sin90aθ = °. 
1.00sin
1.60a
θ = and 38.7aθ = ° . 90 51.3aθ θ= =°− °. 
(b) 1.60an = , 1.333bn = . (1.60)sin (1.333)sin90aθ = °. 
1.333sin
1.60a
θ = and 56.4aθ = ° . 90 33.6aθ θ= =°− °. 
EVALUATE: The critical angle increases when the ratio a
b
n
n
 increases. 
 
Figure 33.17 
The Nature and Propagation of Light 33-5 
33.18. IDENTIFY: Since the refractive index of the glass is greater than that of air or water, total internal reflection will 
occur at the cube surface if the angle of incidence is greater than or equal to the critical angle. 
SET UP: At the critical angle θc, Snell�s law gives nglass sin θc = nair sin 90° and likewise for water. 
EXECUTE: (a) At the critical angle θc , nglass sin θc = nair sin 90°. 1.53 sin θc = (1.00)(1) andθc = 40.8°. 
(b) Using the same procedure as in part (a), we have 1.53 sin θc = 1.333 sin 90° and θc = 60.6°. 
EVALUATE: Since the refractive index of water is closer to the refractive index of glass than the refractive index 
of air is, the critical angle for glass-to-water is greater than for glass-to-air. 
33.19. IDENTIFY: Use the critical angle to find the index of refraction of the liquid. 
SET UP: Total internal reflection requires that the light be incident on the material with the larger n, in this case 
the liquid. Apply sin sina a b bn nθ θ= with a = liquid and b = air, so liqan n= and 1.0.bn = 
EXECUTE: crit liq crit when 90 , so sin (1.0)sin90a b nθ θ θ θ= = ° = ° 
liq
crit
1 1 1.48.
sin sin 42.5
n
θ
= = =
°
 
(a) sin sina a b bn nθ θ= (a = liquid, b = air) 
sin (1.48)sin35.0sin 0.8489 and 58.1
1.0
a a
b b
b
n
n
θθ θ°= = = = ° 
(b) Now sin sina a b bn nθ θ= with a = air, b = liquid 
sin (1.0)sin35.0sin 0.3876 and 22.8
1.48
a a
b b
b
n
n
θθ θ°= = = = ° 
EVALUATE: For light traveling liquid → air the light is bent away from the normal. For light traveling air → 
liquid the light is bent toward the normal. 
33.20. IDENTIFY: The largest angle of incidence for which any light refracts into the air is the critical angle for 
water air→ . 
SET UP: Figure 33.20 shows a ray incident at the critical angle and therefore at the edge of the ring of light. The 
radius of this circle is r and 10.0 md = is the distance from the ring to the surface of the water. 
EXECUTE: From the figure, crittanr d θ= . critθ is calculated from sin sina a b bn nθ θ= with 1.333an = , critaθ θ= , 
1.00bn = and 90bθ = °. crit
(1.00)sin90sin
1.333
θ = ° and crit 48.6θ = ° . (10.0 m) tan 48.6 11.3 mr = =° . 
2 2 2(11.3 m) 401 mA rπ π= = = . 
EVALUATE: When the incident angle in the water is larger than the critical angle, no light refracts into the air. 
 
Figure 33.20 
33.21. IDENTIFY and SET UP: For glass → water, crit 48.7 .θ = ° Apply Snell�s law with critaθ θ= to calculate the index 
of refraction an of the glass. 
EXECUTE: crit
crit
1.333sin sin90 , so 1.77
sin sin 48.7
b
a b a
nn n nθ
θ
= ° = = =
°
 
EVALUATE: For total internal reflection to occur the light must be incident in the material of larger refractive 
index. Our results give glass water ,n n> in agreement with this. 
33-6 Chapter 33 
33.22. IDENTIFY: If no light refracts out of the glass at the glass to air interface, then the incident angle at that interface 
is crit .θ 
SET UP: The ray has an angle of incidence of 0° at the first surface of the glass, so enters the glass without being 
bent, as shown in Figure 33.22. The figure shows that crit 90α θ+ = °. 
EXECUTE: (a) For the glass-air interface crit ,aθ θ= 1.52,an = 1.00bn = and 90 .bθ = ° sin sina a b bn nθ θ= gives 
crit
(1.00)(sin90 )sin
1.52
θ = ° and crit 41.1θ = °. crit90 48.9α θ= − =° °. 
(b) Now the second interface is glass water→ and 1.333bn = . sin sina a b bn nθ θ= gives crit
(1.333)(sin90 )sin
1.52
θ = ° 
and crit 61.3θ = °. crit90 28.7α θ= − =° ° . 
EVALUATE: The critical angle increases when the air is replaced by water and rays are bent as they refract out of 
the glass. 
 
Figure 33.22 
33.23. IDENTIFY: Apply sin sina a b bn nθ θ= . 
SET UP: The light is in diamond and encounters an interface with air, so 2.42an = and 1.00bn = . The largest 
aθ is when 90bθ = ° . 
EXECUTE: (2.42)sin (1.00)sin90aθ = ° . 
1sin
2.42a
θ = and 24.4aθ = ° . 
EVALUATE: Diamond has an usually large refractive index, and this results in a small critical angle. 
33.24. IDENTIFY: Snell's law is sin sina a b bn nθ θ= . 
cv
n
= . 
SET UP: aira = , glassb = . 
EXECUTE: (a) red: sin (1.00)sin57.0 1.36
sin sin38.1
a a
b
b
nn θ
θ
= = =
°
°
. violet: (1.00)sin57.0 1.40
sin36.7b
n = =°
°
. 
(b) red: 
8
83.00 10 m/s 2.21 10 m/s
1.36
cv
n
×
= = = × ; violet: 
8
83.00 10 m/s 2.14 10 m/s
1.40
cv
n
×
= = = × . 
EVALUATE: n is larger for the violet light and therefore this light is bent more toward the normal, and the violet 
light has a smaller speed in the glass than the red light. 
33.25. IDENTIFY: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 12 and the 
transmitted light is polarized along the axis of the polarizer. When polarized light of intensity maxI is incident on a 
polarizer, the transmitted intensity is 2max cosI I φ= , where φ is the angle between the polarization direction of the 
incident light and the axis of the filter. 
SET UP: For the second polarizer 60φ = ° . For the third polarizer, 90 60 30φ = − =° ° ° . 
EXECUTE: (a) At point A the intensity is 0 / 2I and the light is polarized along the vertical direction. At point B 
the intensity is 20 0( / 2)(cos60 ) 0.125I I=° , and the light is polarized along the axis of the second polarizer. At 
point C the intensity is 20 0(0.125 )(cos30 ) 0.0938I I=° . 
(b) Now for the last filter 90φ = ° and 0I = . 
EVALUATE: Adding the middle filter increases the transmitted intensity. 
33.26. IDENTIFY: Apply Snell's law. 
SET UP: The incident, reflected and refracted rays are shown in Figure 33.26. 
EXECUTE: From the figure, 37.0bθ = ° and 
sin sin 531.33 1.77.
sin sin 37
a
b a
b
n n θ
θ
°
= = =
°
 
The Nature and Propagation of Light 33-7 
EVALUATE: The refractive index of b is greater than that of a, and the ray is bent toward the normal when it 
refracts. 
 
Figure 33.26 
33.27. IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence equals 
the polarizing angle, so p 54.5 .θ = ° Use Eq.(33.8) to calculate the refractive index of the glass. Then use Snell�s 
law to calculate the angle of refraction. 
EXECUTE: (a) p glass air ptan gives tan (1.00) tan54.5 1.40.b
a
n n n
n
θ θ= = = ° = 
(b) sin sina a b bn nθ θ= 
sin (1.00)sin54.5sin 0.5815 and 35.5
1.40
a a
b b
b
n
n
θθ θ°= = = = ° 
EVALUATE: 
 
Note: 180.0 and .r b r aφ θ θ θ θ= ° − − = Thus 
180.0 54.5 35.5 90.0 ;φ = ° − ° − ° = ° the 
reflected ray and the refracted ray are 
perpendicular to each other. This agrees 
with Fig.33.28. 
Figure 33.27 
33.28. IDENTIFY: Set 0 /10I I= , where I is the intensity of light passed by the second polarizer. 
SET UP: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 12 and the 
transmitted light is polarized along the axis of the polarizer. When polarized light of intensity maxI is incident on a 
polarizer, the transmitted intensity is 2max cosI I φ= , where φ is the angle between the polarization direction of the 
incidentlight and the axis of the filter. 
EXECUTE: (a) After the first filter 0
2
II = and the light is polarized along the vertical direction. After the second 
filter we want 0 ,
10
II = so 20 0 (cos )
10 2
I I φ⎛ ⎞= ⎜ ⎟
⎝ ⎠
. cos 2 /10φ = and 63.4φ = ° . 
(b) Now the first filter passes the full intensity 0I of the incident light. For the second filter 
20
0 (cos )10
I I φ= . 
cos 1/10φ = and 71.6φ = ° . 
EVALUATE: When the incident light is polarized along the axis of the first filter, φ must be larger to achieve the 
same overall reduction in intensity than when the incident light is unpolarized. 
33.29. IDENTIFY: From Malus�s law, the intensity of the emerging light is proportional to the square of the cosine of the 
angle between the polarizing axes of the two filters. 
SET UP: If the angle between the two axes is θ, the intensity of the emerging light is I = Imax cos2θ. 
EXECUTE: At angle θ, I = Imax cos2θ, and at the new angle α, 
1
2
I = Imax cos2α. Taking the ratio of the intensities 
gives 
2 1
max 2
2
max
cos
cos
I I
I I
α
θ
= , which gives us coscos
2
θα = . Solving for α yields cosarccos
2
θα ⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 
EVALUATE: Careful! This result is not cos2θ. 
33-8 Chapter 33 
33.30. IDENTIFY: The reflected light is completely polarized when the angle of incidence equals the polarizing angle 
pθ , where ptan b
a
n
n
θ = . 
SET UP: 1.66bn = . 
EXECUTE: (a) 1.00an = . p
1.66tan
1.00
θ = and p 58.9θ = ° . 
(b) 1.333an = . p
1.66tan
1.333
θ = and p 51.2θ = ° . 
EVALUATE: The polarizing angle depends on the refractive indicies of both materials at the interface. 
33.31. IDENTIFY: When unpolarized light of intensity 0I is incident on a polarizing filter, the transmitted light has 
intensity 1 02 I and is polarized along the filter axis. When polarized light of intensity 0I is incident on a polarizing 
filter the transmitted light has intensity 20 cosI φ . 
SET UP: For the second filter, 62.0 25.0 37.0φ = − =° ° ° . 
EXECUTE: After the first filter the intensity is 21 02 10.0 W mI = and the light is polarized along the axis of the 
first filter. The intensity after the second filter is 20cosI I φ= , where 
2
0 10.0 W mI = and 37.0φ = ° . This 
gives 26.38 W m .I = 
EVALUATE: The transmitted intensity depends on the angle between the axes of the two filters. 
33.32. IDENTIFY: After passing through the first filter the light is linearly polarized along the filter axis. After the 
second filter, 2max (cos )I I φ= , where φ is the angle between the axes of the two filters. 
SET UP: The maximum amount of light is transmitted when 0φ = . 
EXECUTE: (a) 20 0(cos22.5 ) 0.854I I I= =° 
(b) 20 0(cos45.0 ) 0.500I I I= =° 
(c) 20 0(cos67.5 ) 0.146I I I= =° 
EVALUATE: As φ increases toward 90° the axes of the two filters are closer to being perpendicular to each other 
and the transmitted intensity decreases. 
33.33. IDENTIFY and SET UP: Apply Eq.(33.7) to polarizers #2 and #3. The light incident on the first polarizer is 
unpolarized, so the transmitted light has half the intensity of the incident light, and the transmitted light is 
polarized. 
(a) EXECUTE: The axes of the three filters are shown in Figure 33.33a. 
 
2
max cosI I φ= 
Figure 33.33a 
After the first filter the intensity is 11 02I I= and the light is linearly polarized along the axis of the first polarizer. 
After the second filter the intensity is 2 212 1 0 02cos ( )(cos45.0 ) 0.250I I I Iφ= = ° = and the light is linearly polarized 
along the axis of the second polarizer. After the third filter the intensity is 2 23 2 0cos 0.250 (cos45.0 )I I Iφ= = ° = 
00.125I and the light is linearly polarized along the axis of the third polarizer. 
(b) The axes of the remaining two filters are shown in Figure 33.33b. 
 
After the first filter the intensity is 11 02I I= and the 
light is linearly polarized along the axis of the first 
polarizer. 
 
Figure 33.33b 
After the next filter the intensity is 2 213 1 02cos ( )(cos90.0 ) 0.I I Iφ= = ° = No light is passed. 
EVALUATE: Light is transmitted through all three filters, but no light is transmitted if the middle polarizer is removed. 
The Nature and Propagation of Light 33-9 
33.34. IDENTIFY: Use the transmitted intensity when all three polairzers are present to solve for the incident intensity 
0I . Then repeat the calculation with only the first and third polarizers. 
SET UP: For unpolarized light incident on a filter, 1 02I I= and the light is linearly polarized along the filter axis. 
For polarized light incident on a filter, 2max (cos )I I φ= , where maxI is the intensity of the incident light, and the 
emerging light is linearly polarized along the filter axis. 
EXECUTE: With all three polarizers, if the incident intensity is 0I the transmitted intensity is 
2 21
0 02( )(cos23.0 ) (cos[62.0 23.0 ]) 0.256I I I= − =° ° ° . 
2
2
0
75.0 W/cm 293 W/cm
0.256 0.256
II = = = . With only the first 
and third polarizers, 2 2 21 0 02( )(cos62.0 ) 0.110 (0.110)(293 W/cm ) 32.2 W/cmI I I= = = =° . 
EVALUATE: The transmitted intensity is greater when all three filters are present. 
33.35. IDENTIFY: The shorter the wavelength of light, the more it is scattered. The intensity is inversely proportional to 
the fourth power of the wavelength. 
SET UP: The intensity of the scattered light is proportional to 1/λ4, we can write it as I = (constant)/ λ4. 
EXECUTE: (a) Since I is proportional to 1/λ4, we have I = (constant)/ λ4. Taking the ratio of the intensity of the 
red light to that of the green light gives 
4 44
R R G
4
G R
(constant) / 520 nm
(constant) / 665 nm
I
I
λ λ
λ λ
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 = 0.374, so IR = 0.374I. 
(b) Following the same procedure as in part (a) gives 
4 4
V G
V
520 nm
420 nm
I
I
λ
λ
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 = 2.35, so IV = 2.35I. 
EVALUATE: In the scattered light, the intensity of the short-wavelength violet light is about 7 times as great as 
that of the red light, so this scattered light will have a blue-violet color. 
33.36. IDENTIFY: As the wave front reaches the sharp object, every point on the front will act as a source of secondary 
wavelets. 
SET UP: Consider a wave front that is just about to go past the corner. Follow it along and draw the successive 
wave fronts. 
EXECUTE: The path of the wavefront is drawn in Figure 33.36. 
EVALUATE: The wave fronts clearly bend around the sharp point, just as water waves bend around a rock and 
light waves bend around the edge of a slit. 
 
Figure 33.36 
33.37. IDENTIFY: Reflection reverses the sign of the component of light velocity perpendicular to the reflecting surface 
but leaves the other components unchanged. 
SET UP: Consider three mirrors, M1 in the (x,y)-plane, M2 in the (y,z)-plane, and M3 in the (x,z)-plane. 
EXECUTE: A light ray reflecting from M1 changes the sign of the z-component of the velocity, reflecting from M2 
changes the x-component, and from M3 changes the y-component. Thus the velocity, and hence also the path, of the 
light beam flips by 180° 
EVALUATE: Example 33.3 discusses some uses of corner reflectors. 
33.38. IDENTIFY: The light travels slower in the jelly than in the air and hence will take longer to travel the length of the 
tube when it is filled with jelly than when it contains just air. 
SET UP: The definition of the index of refraction is n = c/v, where v is the speed of light in the jelly. 
EXECUTE: First get the length L of the tube using air. In the air, we have L = ct = (3.00 × 108 m/s)(8.72 ns) = 2.616 m. 
The speed in the jelly is v = L
t
 = (2.616 m)/(8.72 ns + 2.04 ns) = 2.431 × 108 m/s. n = c
v
 = 
(3.00 × 108 m/s)/(2.431 × 108 m/s) = 1.23 
EVALUATE: A high-speed timer would be needed to measure times as short as a few nanoseconds. 
33-10 Chapter 33 
33.39. IDENTIFY and SET UP: Apply Snell's law at each interface. 
EXECUTE: (a) 1 1 2 2sin sinn nθ θ= and 2 2 3 3sin sinn nθ θ= , so 1 1 3 3sin sinn nθ θ= and 3 1 1 3sin ( sin ) /n nθ θ= . 
(b) 3 3 2 2sin sinn nθ θ= and 2 2 1 1sin sinnnθ θ= , so 1 1 3 3sin sinn nθ θ= and the light makes the same angle with respect 
to the normal in the material that has refractive index 1n as it did in part (a). 
(c) For reflection, .r aθ θ= These angles are still equal if rθ becomes the incident angle; reflected rays are also reversible. 
EVALUATE: Both the refracted and reflected rays are reversible, in the sense that if the direction of the light is 
reversed then each of these rays follow the path of the incident ray. 
33.40. IDENTIFY: Use the change in transit time to find the speed v of light in the slab, and then apply cn
v
= and 0
n
λλ = . 
SET UP: It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab is inserted into the beam. 
EXECUTE: 0.840 m 0.840 m 0.840 m( 1) 4.2 ns.
/
n
c n c c
− = − = We can now solve for the index of refraction: 
9 8(4.2 10 s) (3.00 10 m s) 1 2.50.
0.840 m
n
−× ×
= + = The wavelength inside of the glass is 
490 nm 196 nm
2.50
λ = = . 
EVALUATE: Light travels slower in the slab than in air and the wavelength is shorter. 
33.41. IDENTIFY: The angle of incidence at A is to be the critical angle. Apply Snell�s law at the air to glass refraction at 
the top of the block. 
SET UP: The ray is sketched in Figure 33.41. 
EXECUTE: For glass air→ at point A, Snell's law gives crit(1.38)sin (1.00)sin90θ = °and crit 46.4θ = ° . 
crit90 43.6bθ θ= − =° ° . Snell's law applied to the refraction from air to glass at the top of the block gives 
(1.00)sin (1.38)sin(43.6 )aθ = ° and 72.1aθ = °. 
EVALUATE: If aθ is larger than 72.1°then the angle of incidence at point A is less than the initial critical angle and 
total internal reflection doesn�t occur. 
 
Figure 33.41 
33.42. IDENTIFY: As the light crosses the glass-air interface along AB, it is refracted and obeys Snell�s law. 
SET UP: Snell�s law is na sin θa = nb sin θb and n = 1.000 for air. At point B the angle of the prism is 30.0° . 
EXECUTE: Apply Snell�s law at AB. The prism angle at A is 60.0°, so for the upper ray, the angle of incidence at 
AB is 60.0° + 12.0° = 72.0°. Using this value gives n1 sin 60.0° = sin 72.0° and n1 = 1.10. For the lower ray, the 
angle of incidence at AB is 60.0° + 12.0° + 8.50° = 80.5°, giving n2 sin 60.0° = sin 80.5° and n2 = 1.14. 
EVALUATE: The lower ray is deflected more than the upper ray because that wavelength has a slightly greater 
index of refraction than the upper ray. 
The Nature and Propagation of Light 33-11 
33.43. IDENTIFY: Circularly polarized light consists of the superposition of light polarized in two perpendicular 
directions, with a quarter-cycle ( 90° ) phase difference between the two polarization components. 
SET UP: A quarter-wave plate shifts the relative phase of the two perpendicular polarization components by 90° . 
EXECUTE: In the circularly polarized light the two perpendicular polarization components are 90° out of phase. 
The quarter-wave plate shifts the relative phase by 90± ° and then the two components are either in phase or 
180° out of phase. Either corresponds to linearly polarized light. 
EVALUATE: Either left circularly polarized light or right circularly polarized light is converted to linearly 
polarized light by the quarter-wave plate. 
33.44. IDENTIFY: Apply 0
n
λλ = . The number of wavelengths in a distance d of a material is d
λ
 where λ is the 
wavelength in the material. 
SET UP: The distance in glass is glass 0.00250 md = . The distance in air is 
air 0.0180 m 0.00250 m 0.0155 md = − = 
EXECUTE: number of wavelengths = number in air + number in glass. 
glass 4air
7 7
0.0155 m 0.00250 mnumber of wavelengths (1.40) 3.52 10
5.40 10 m 5.40 10 m
dd n
λ λ − −
= + = + = ×
× ×
. 
EVALUATE: Without the glass plate the number of wavelengths between the source and screen is 
4
3
0.0180 m 3.33 10
5.40 10 m−
= ×
×
. The wavelength is shorter in the glass so there are more wavelengths in a distance in 
glass than there are in the same distance in air. 
33.45. IDENTIFY: Find the critical angle for glass → air. Light incident at this critical angle is reflected back to the 
edge of the halo. 
SET UP: The ray incident at the critical angle is sketched in Figure 33.45. 
 
Figure 33.45 
EXECUTE: From the distances given in the sketch, crit crit
2.67 mmtan 0.8613; 40.7 .
3.10 mm
θ θ= = = ° 
Apply Snell�s law to the total internal reflection to find the refractive index of the glass: sin sina a b bn nθ θ= 
glass critsin 1.00sin90n θ = ° 
glass
crit
1 1 1.53
sin sin 40.7
n
θ
= = =
°
 
EVALUATE: Light incident on the back surface is also totally reflected if it is incident at angles greater than crit .θ 
If it is incident at less than critθ it refracts into the air and does not reflect back to the emulsion. 
33.46. IDENTIFY: Apply Snell's law to the refraction of the light as it passes from water into air. 
SET UP: 1.5 marctan 51
1.2 ma
θ
⎛ ⎞
= = °⎜ ⎟
⎝ ⎠
. 1.00an = . 1.333bn = . 
EXECUTE: 1.00arcsin sin arcsin sin51 36 .
1.333
a
b a
b
n
n
θ θ
⎛ ⎞ ⎛ ⎞= = ° = °⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 Therefore, the distance along the bottom of the 
pool from directly below where the light enters to where it hits the bottom is (4.0 m)tan (4.0 m)tan36bx θ= = ° = 
2.9 m. total 1.5 m 1.5 m 2.9 m 4.4 m.x x= + = + = 
EVALUATE: The light ray from the flashlight is bent toward the normal when it refracts into the water. 
33-12 Chapter 33 
33.47. IDENTIFY: Use Snell�s law to determine the effect of the liquid on the direction of travel of the light as it enters 
the liquid. 
SET UP: Use geometry to find the angles of incidence and refraction. Before the liquid is poured in the ray along 
your line of sight has the path shown in Figure 33.47a. 
 
8.0 cmtan 0.500
16.0 cma
θ = = 
26.57aθ = ° 
Figure 33.47a 
After the liquid is poured in, aθ is the same and the refracted ray passes through the center of the bottom of the 
glass, as shown in Figure 33.47b. 
 
4.0 cmtan 0.250
16.0 cmb
θ = = 
14.04bθ = ° 
Figure 33.47b 
EXECUTE: Use Snell�s law to find ,bn the refractive index of the liquid: 
sin sina a b bn nθ θ= 
sin (1.00)(sin 26.57 ) 1.84
sin sin14.04
a a
b
b
nn θ
θ
°
= = =
°
 
EVALUATE: When the light goes from air to liquid (larger refractive index) it is bent toward the normal. 
33.48. IDENTIFY: Apply Snell�s law to each refraction and apply the law of reflection at the mirrored bottom. 
SET UP: The path of the ray is sketched in Figure 33.48. The problem asks us to calculate bθ ′. 
EXECUTE: Apply Snell's law to the air liquid→ refraction. (1.00)sin(42.5 ) (1.63)sin bθ=° and 24.5bθ = ° . 
bθ φ= and aφ θ ′= , so 24.5a bθ θ′ = = °. Snell's law applied to the liquid air→ refraction gives 
(1.63)sin(24.5 ) (1.00)sin bθ ′=° and 42.5bθ ′ = ° . 
The Nature and Propagation of Light 33-13 
EVALUATE: The light emerges from the liquid at the same angle from the normal as it entered the liquid. 
 
Figure 33.48 
33.49. IDENTIFY: Apply Snell�s law to the water → ice and ice → air interfaces. 
(a) SET UP: Consider the ray shown in Figure 33.49. 
 
We want to find the incident angle aθ at 
the water-ice interface that causes the 
incident angle at the ice-air interface to be 
the critical angle. 
Figure 33.49 
EXECUTE: ice-air interface: ice critsin 1.0sin90n θ = ° 
ice crit crit
ice
1sin 1.0 so sinn
n
θ θ= = 
But from the diagram we see that crit
ice
1, so sin .b b n
θ θ θ= = 
water-ice interface: w icesin sina bn nθ θ= 
But 
ice
1sin so sin 1.0.b w ann
θ θ= = 
w
1 1sin 0.7502 and 48.6 .
1.333a an
θ θ= = = = ° 
(b) EVALUATE: The angle calculated in part (a) is the critical angle for a water-air interface; the answer would be 
the same if the ice layer wasn�t there! 
33.50. IDENTIFY: The incident angle at the prism water→ interface is to be the critical angle. 
SET UP: The path of the ray is sketched in Figure 33.50. The ray enters the prism at normal incidence so is not 
bent. For water, water 1.333n = . 
EXECUTE: From the figure, crit 45θ = ° . sin sina a b bn nθ θ= gives glass sin 45 (1.333)sin90n=° ° . 
glass
1.333 1.89
sin 45
n = =
°
. 
33-14 Chapter 33 
EVALUATE: For total internal reflection the ray must be incident in the material of greater refractive index. 
glass watern n> , so that is the case here. 
 
Figure 33.50 
33.51. IDENTIFY: Apply Snell�s law to the refraction of each ray as it emerges from the glass. The angle of incidence 
equals the angle 25.0 .A = ° 
SET UP: The paths of the two rays are sketched in Figure 33.51. 
 
Figure 33.51 
EXECUTE: sin sina a b bn nθ θ= 
glas sin 25.0 1.00sin bn θ° = 
glasssin sin 25.0b nθ = ° 
sin 1.66sin 25.0 0.7015bθ = ° = 
44.55bθ = ° 
90.0 45.45bβ θ= ° − = ° 
Then 90.0 90.0 25.0 45.45 19.55 .Aδ β= ° − − = ° − ° − ° = ° The angle between the two rays is 2 39.1 .δ = ° 
EVALUATE: The light is incident normally on the front face of the prism so the light is not bent as it enters the prism. 
33.52. IDENTIFY: The ray shown in the figure that accompanies the problem is to be incident at the critical angle. 
SET UP: 90bθ = ° . The incident angle for the ray in the figure is 60° . 
EXECUTE: sin sina a b bn nθ θ= gives 
sin 1.62sin 60 1.40.
sin sin90
a a
b
b
nn θ
θ
⎛ ⎞ °⎛ ⎞= = =⎜ ⎟ ⎜ ⎟°⎝ ⎠⎝ ⎠
 
EVALUATE: Total internal reflection occurs only when the light is incident in the material of the greater 
refractive index. 
33.53. IDENTIFY: No light enters the gas because total internal reflection must have occurred at the water-gas interface. 
SET UP: At the minimum value of S, the light strikes the water-gas interface at the critical angle. We apply 
Snell�s law, na sinθa = nb sinθb, at that surface. 
EXECUTE: (a) In the water, θ = S
R
 = (1.09 m)/(1.10 m) = 0.991 rad = 56.77°. This is the critical angle. So, using 
the refractive index for water from Table 33.1, we get n = (1.333) sin 56.77° = 1.12 
(b) (i) The laser beam stays in the water all the time, so 
t = 2R/v = 2R/
water
c
n
⎛ ⎞
⎜ ⎟
⎝ ⎠
 = waterDn
c
 = (2.20 m)(1.333)/(3.00 × 108 m/s) = 9.78 ns 
The Nature and Propagation of Light 33-15 
(ii) The beam is in the water half the time and in the gas the other half of the time. 
tgas = gas
Rn
c
 = (1.10 m)(1.12)/(3.00 × 108 m/s) = 4.09 ns 
The total time is 4.09 ns + (9.78 ns)/2 = 8.98 ns 
EVALUATE: The gas must be under considerable pressure to have a refractive index as high as 1.12. 
33.54. IDENTIFY: No light enters the water because total internal reflection must have occurred at the glass-water surface. 
SET UP: A little geometry tells us that θ is the angle of incidence at the glass-water face in the water. Also, θ = 
59.2° must be the critical angle at that surface, so the angle of refraction is 90.0°. Snell�s law, na sin θa = nb sin θb, 
applies at that glass-water surface, and the index of refraction is defined as n = .c
v
 
EXECUTE: Snell�s law at the glass-water surface gives n sin 59.2° = (1.333)(1.00), which gives n = 1.55. v = c
n
 = 
(3.00 × 108 m/s)/1.55 = 1.93 × 108 m/s. 
EVALUATE: Notice that θ is not the angle of incidence at the reflector, but it is the angle of incidence at the glass-
water surface. 
33.55. (a) IDENTIFY: Apply Snell�s law to the refraction of the light as it enters the atmosphere. 
SET UP: The path of a ray from the sun is sketched in Figure 33.55. 
 
a bδ θ θ= − 
From the diagram sin b
R
R h
θ =
+
 
 arcsin b
R
R h
θ ⎛ ⎞= ⎜ ⎟+⎝ ⎠
 
Figure 33.55 
EXECUTE: Apply Snell�s law to the refraction that occurs at the top of the atmosphere: sin sina a b bn nθ θ= 
(a = vacuum of space, refractive, index 1.0; b = atmosphere, refractive index n) 
sin sin so arcsina b a
R nRn n
R h R h
θ θ θ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
 
arcsin arcsina b
nR R
R h R h
δ θ θ ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
 
(b) 
6
6 3
6.38 10 m 0.99688
6.38 10 m 20 10 m
R
R h
×
= =
+ × + ×
 
1.0003(0.99688) 0.99718nR
R h
= =
+
 
arcsin 85.47b
R
R h
θ ⎛ ⎞= = °⎜ ⎟+⎝ ⎠
 
arcsin 85.70b
nR
R h
θ ⎛ ⎞= °⎜ ⎟+⎝ ⎠
 
85.70 85.47 0.23a bδ θ θ= − = ° − ° = ° 
EVALUATE: The calculated δ is about the same as the angular radius of the sun. 
33.56. IDENTIFY and SET UP: Follow the steps specified in the problem. 
EXECUTE: (a) The distance traveled by the light ray is the sum of the two diagonal segments: 
2 2 1/2 2 2 1/2
1 2( ) (( ) ) .d x y l x y= + + − + Then the time taken to travel that distance is 
2 2 1/ 2 2 2 1/ 2
1 2( ) (( ) ) .d x y l x yt
c c
+ + − +
= = 
(b) Taking the derivative with respect to x of the time and setting it to zero yields 
( ) ( )1 2 1 22 2 1 2 2 2 2 2 1 2 2 21 2 1 21 1( ) ( ) ( ) ( ) ( ) 0dt d dtx y l x y x x y l x l x ydx c dt dx c
−−⎡ ⎤ ⎡ ⎤= + + − + = + − − − + =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
. This gives 
2 2 2 2
1 2
( )
( )
x l x
x y l x y
−
=
+ − +
, 1 2sin sinθ θ= and 1 2.θ θ= 
EVALUATE: For any other path between points 1 and 2, that includes a point on the reflective surface, the 
distance traveled and therefore the travel time is greater than for this path. 
33-16 Chapter 33 
33.57. IDENTIFY and SET UP: Find the distance that the ray travels in each medium. The travel time in each medium is 
the distance divided by the speed in that medium. 
(a) EXECUTE: The light travels a distance 2 21h x+ in traveling from point A to the interface. Along this path 
the speed of the light is 1,v so the time it takes to travel this distance is 
2 2
1
1
1
.
h x
t
v
+
= The light travels a 
distance 2 22 ( )h l x+ − in traveling from the interface to point B. Along this path the speed of the light is 2 ,v 
so the time it takes to travel this distance is 
2 2
2
2
2
( )
.
h l x
t
v
+ −
= The total time to go from A to B is 
2 2 2 2
1 2
1 2
1 2
( )
.
h x h l x
t t t
v v
+ + +
= + = + 
(b) 2 2 1/ 2 2 2 1/ 21 2
1 2
1 1 1 1( ) (2 ) ( ( ) ) 2( )( 1) 0
2 2
dt h x x h l x l x
dx v v
− −⎛ ⎞ ⎛ ⎞= + + + − − − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
2 2 2 2
1 1 2 2 ( )
x l x
v h x v h l x
−
=
+ + −
 
Multiplying both sides by c gives 
2 2 2 2
1 21 2 ( )
c x c l x
v vh x h l x
−
=
+ + −
 
1 2
1 2
 and c cn n
v v
= = (Eq.33.1) 
From Fig.33.55 in the textbook, 1 22 2 2 2
1 2
sin and sin .
( )
x l x
h x h l x
θ θ −= =
+ + −
 
So 1 1 2 2sin sin ,n nθ θ= which is Snell�s law. 
EVALUATE: Snell�s law is a result of a change in speed when light goes from one material to another. 
33.58. IDENTIFY: Apply Snell's law to each refraction. 
SET UP: Refer to the angles and distances defined in the figure that accompanies the problem. 
EXECUTE: (a) For light in air incident on a parallel-faced plate, Snell�s Law yields: 
sin sin sin sin sin sin .a b b a a a a an n n nθ θ θ θ θ θ θ θ′ ′ ′ ′ ′ ′= = = ⇒ = ⇒ = 
(b) Adding more plates just adds extra steps in the middle of the above equation that always cancel out. The 
requirement of parallel faces ensures that the angle n nθ θ′ = and the chain of equations can continue. 
(c) The lateral displacement of the beam can be calculated using geometry: 
sin( )sin( ) and .
cos cos
a b
a b
b b
t td L L d θ θθ θ
θ θ
′−′= − = ⇒ =
′ ′
 
(d) sin sin 66.0arcsin arcsin 30.5
1.80
a
b
n
n
θθ °⎛ ⎞ ⎛ ⎞′ = = = °⎜ ⎟ ⎜ ⎟′⎝ ⎠ ⎝ ⎠
 and (2.40 cm)sin(66.0 30.5 ) 1.62 cm.
cos30.5
d ° − °= =
°
 
EVALUATE: The lateral displacement in part (d) is large, of the same order as the thickness of the plate. 
33.59. IDENTIFY: Apply Snell�s law to each refraction and apply the law of reflection to each reflection. 
SET UP: The paths of rays A and B are sketched in Figure 33.59. Let θ be the angle of incidence for the 
combined ray. 
EXECUTE: For ray A its final direction of travel is at an angle θ with respect to the normal, by the law of 
reflection. Let the final direction of travel for ray B be at angleφ with respect to the normal. At the upper surface, 
Snell�s law gives 1 2sin sinn nθ α= . The lower surface reflects ray B at angle α . Ray B returns to the upper 
surface of the film at an angle of incidence α . Snell�s law applied to the refraction as ray B leaves the film gives 
2 1sin sinn nα φ= . Combining the two equations gives 1 1sin sinn nθ φ= and θ φ= ; the two rays are parallel after 
they emerge from the film.The Nature and Propagation of Light 33-17 
EVALUATE: Ray B is bent toward the normal as it enters the film and away from the normal as it refracts out of 
the film. 
 
Figure 33.59 
33.60. IDENTIFY: Apply Snell's law and the results of Problem 33.58. 
SET UP: From Figure 33.58 in the textbook, r 1.61n = for red light and v 1.66n = for violet. In the notation of 
Problem 33.58, t is the thickness of the glass plate and the lateral displacement is d. We want the difference in d for 
the two colors of light to be 1.0 mm. 70.0aθ = ° . For red light, sin sina a b bn nθ θ ′= gives 
(1.00)sin 70.0sin
1.61b
θ ′ = ° 
and 35.71bθ ′ = ° . For violet light, 
(1.00)sin 70.0sin
1.66b
θ ′ = ° and 34.48bθ ′ = ° . 
EXECUTE: (a) n decreases with increasing λ , so n is smaller for red than for blue. So beam a is the red one. 
(b) Problem 33.58 says sin( )
cos
a b
b
d t θ θ
θ
′−
=
′
. For red light, r
sin(70 35.71 ) 0.6938
cos35.71
d t t−= =° °
°
and for violet light, 
v
sin(70 35.48 ) 0.7048
cos35.48
d t t−= =° °
°
. v r 1.0 mmd d− = gives 
0.10 cm 9.1 cm
0.7048 0.6958
t = =
−
. 
EVALUATE: Our calculation shown that the violet light has greater lateral displacement and this is ray b. 
33.61. IDENTIFY: Apply Snell's law to the two refractions of the ray. 
SET UP: Refer to the figure that accompanies the problem. 
EXECUTE: (a) sin sina a b bn nθ θ= gives sin sin 2a b
Anθ = . But 
2a
Aθ α= + , so 2sin sin sin .
2 2 2
A A Anαα +⎛ ⎞+ = =⎜ ⎟
⎝ ⎠
 
At each face of the prism the deviation is α , so 2α δ= and sin sin .
2 2
A Anδ+ = 
(b) From part (a), 2arcsin sin
2
An Aδ ⎛ ⎞= −⎜ ⎟
⎝ ⎠
. 60.02arcsin (1.52)sin 60.0 38.9 .
2
δ °⎛ ⎞= − ° = °⎜ ⎟
⎝ ⎠
 
(c) If two colors have different indices of refraction for the glass, then the deflection angles for them will differ: 
red
violet
60.02arcsin (1.61)sin 60.0 47.2
2
60.02arcsin (1.66)sin 60.0 52.2 52.2 47.2 5.0
2
δ
δ δ
°⎛ ⎞= − ° = °⎜ ⎟
⎝ ⎠
°⎛ ⎞= − ° = ° ⇒ Δ = ° − ° = °⎜ ⎟
⎝ ⎠
 
EVALUATE: The violet light has a greater refractive index and therefore the angle of deviation is greater for the 
violet light. 
33.62. IDENTIFY: The reflected light is totally polarized when light strikes a surface at Brewster�s angle. 
SET UP: At the plastic wall, Brewster�s angle obeys the equation tan θp = nb /na, and Snell�s law, 
na sinθa = nb sinθb, applies at the air-water surface. 
EXECUTE: To be totally polarized, the reflected sunlight must have struck the wall at Brewster�s angle. tan θp = 
nb /na = (1.61)/(1.00) and θp = 58.15° 
This is the angle of incidence at the wall. A little geometry tells us that the angle of incidence at the water surface 
is 90.00° � 58.15° = 31.85°. Applying Snell�s law at the water surface gives 
(1.00) sin31.85° = 1.333 sin θ and θ = 23.3° 
EVALUATE: We have two different principles involved here: Reflection at Brewster�s angle at the wall and 
Snell�s law at the water surface. 
33-18 Chapter 33 
33.63. IDENTIFY and SET UP: The polarizer passes 12 of the intensity of the unpolarized component, independent of .φ 
Out of the intensity pI of the polarized component the polarizer passes intensity 
2
p cos ( ),I φ θ− where φ θ− is the 
angle between the plane of polarization and the axis of the polarizer. 
(a) Use the angle where the transmitted intensity is maximum or minimum to find .θ See 
Figure 33.63. 
 
Figure 33.63 
EXECUTE: The total transmitted intensity is 21 0 p2 cos ( ).I I I φ θ= + − This is maximum when θ φ= and from the 
table of data this occurs for φ between 30 and 40 ,° ° say at 35 and 35 .θ° = ° Alternatively, the total transmitted 
intensity is minimum when 90φ θ− = ° and from the data this occurs for 125 .φ = ° Thus, 
90 125 90 35 ,θ φ= − ° = ° − ° = ° in agreement with the above. 
(b) IDENTIFY and SET UP: 21 0 p2 cos ( )I I I φ θ= + − 
Use data at two values of φ to determine the two constants 0 pand .I I Use data where the pI term is large 
( 30 )φ = ° and where it is small ( 130 )φ = ° to have the greatest sensitivity to both 0 pand :I I 
EXECUTE: 2 21 0 p230 gives 24.8 W/m cos (30 35 )I Iφ = ° = + ° − ° 
2
0 p24.8 W/m 0.500 0.9924I I= + 
2 21
0 p2130 gives 5.2 W/m cos (130 35 )I Iφ = ° = + ° − ° 
2
0 p5.2 W/m 0.500 0.0076I I= + 
Subtracting the second equation from the first gives 2 2p p19.6 W/m 0.9848 and 19.9 W/m .I I= = And then 
2 2 2
0 2(5.2 W/m 0.0076(19.9 W/m )) 10.1 W/m .I = − = 
EVALUATE: Now that we have 0 p, and I I θ we can verify that 
21
0 p2 cos ( )I I I φ θ= + − describes that data in the 
table. 
33.64. IDENTIFY: The number of wavelengths in a distance D of material is /D λ , where λ is the wavelength of the 
light in the material. 
SET UP: The condition for a quarter-wave plate is 
1 2
1
4
D D
λ λ
= + , where we have assumed 1 2n n> so 2 1λ λ> . 
EXECUTE: (a) 1 2
0 0
1
4
n D n D
λ λ
= + and 0
1 2
.
4( )
D
n n
λ
=
−
 
(b)
7
70
1 2
5 89 10 m 6.14 10 m.
4( ) 4(1.875 1.635)
.D
n n
λ − −×= = = ×
− −
 
EVALUATE: The thickness of the quarter-wave plate in part (b) is 614 nm, which is of the same order as the 
wavelength in vacuum of the light. 
33.65. IDENTIFY: Follow the steps specified in the problem. 
SET UP: cos( ) sin sin cos cosα β α β α β− = + . sin( ) sin cos cos sinα β α β α β− = − . 
EXECUTE: (a) Multiplying Eq.(1) by sin β and Eq.(2) by sinα yields: 
(1): sin sin cos sin cos sin sinx t t
a
β ω α β ω α β= − and (2): sin sin cos sin cos sin siny t t
a
α ω β α ω β α= − . 
Subtracting yields: sin sin sin (cos sin cos sin ).x y t
a
β α ω α β β α− = − 
The Nature and Propagation of Light 33-19 
(b) Multiplying Eq. (1) by cos β and Eq. (2) by cosα yields: 
(1) : cos sin cos cos cos sin cosx t t
a
β ω α β ω α β= − and (2) : cos sin cos cos cos sin cosy t t
a
α ω β α ω β α= − . 
Subtracting yields: cos cos cos (sin cos sin cos ).x y t
a
β α ω α β β α− = − − 
(c) Squaring and adding the results of parts (a) and (b) yields: 
2 2 2 2( sin sin ) ( cos cos ) (sin cos sin cos )x y x y aβ α β α α β β α− + − = − 
(d) Expanding the left-hand side, we have: 
2 2 2 2 2 2
2 2 2 2
(sin cos ) (sin cos ) 2 (sin sin cos cos )
2 (sin sin cos cos ) 2 cos( ).
x y xy
x y xy x y xy
β β α α α β α β
α β α β α β
+ + + − +
= + − + = + − −
 
The right-hand side can be rewritten: 2 2 2 2(sin cos sin cos ) sin ( ).a aα β β α α β− = − Therefore, 
2 2 2 22 cos( ) sin ( ).x y xy aα β α β+ − − = − Or, 2 2 2 22 cos sin , where .x y xy aδ δ δ α β+ − = = − 
EVALUATE: (e) 2 2 20 : 2 ( ) 0 ,x y xy x y x yδ = + − = − = ⇒ = which is a straight diagonal line 
2
2 2: 2 , which is an ellipse
4 2
ax y xyπδ = + − = 
2 2 2: ,which is a circle.
2
x y aπδ = + = This pattern repeats for the remaining phase differences. 
33.66. IDENTIFY: Apply Snell's law to each refraction. 
SET UP: Refer to the figure that accompanies the problem. 
EXECUTE: (a) By the symmetry of the triangles, , and .A B C B B Ab a a r a bθ θ θ θ θ θ= = = = Therefore, 
sin sin sin sin .C C A A C Ab a b a b an nθ θ θ θ θ θ= = = = = 
(b) The total angular deflection of the ray is 2 2 4 .A A B C C A Aa b a b a a bθ θ π θ θ θ θ θ πΔ = − + − + − = − + 
(c) From Snell�s Law, sin 1sin arcsin sinA A A Aa b b an n
θ θ θ θ⎛ ⎞= ⇒ = ⎜ ⎟
⎝ ⎠
. 
12 4 2 4arcsin sin .A A A Aa b a an
θ θ π θ θ π⎛ ⎞Δ = − + = − +⎜ ⎟
⎝ ⎠
 
(d) 1
2
1
2
1 4 cos0 2 4 arcsin sin 0 2
sin1
A
aA A
a a
d d
d d n n
n
θθ
θ θ θ
Δ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = − ⇒ = − ⋅⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠ −
. 
2 2
1 1
2 2
sin 16cos4 1
n n
θ θ⎛ ⎞ ⎛ ⎞
− =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
. 
2 2 2
1 14cos 1 cosnθ θ= − + . 
2 2
13cos 1nθ = − . 
2 2
1
1cos ( 1).
3
nθ = − 
(e) For violet: 2 21
1 1arccos ( 1) arccos (1.342 1) 58.89
3 3
nθ
⎛ ⎞ ⎛ ⎞
= − = − = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
. 
violet violet139.2 40.8 .θΔ = ° ⇒ = ° 
For red: 2 21
1 1arccos ( 1) arccos (1.330 1) 59.58
3 3
nθ
⎛ ⎞ ⎛ ⎞
= − = − = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
. red red137.5 42.5 .θΔ = ° ⇒ = ° 
EVALUATE: The angles we have calculated agree with the values given in Figure 37.20d in the textbook. 1θ is 
larger for red than for violet,so red in the rainbow is higher above the horizon. 
33.67. IDENTIFY: Follow similar steps to Challenge Problem 33.66. 
SET UP: Refer to Figure 33.20e in the textbook. 
EXECUTE: The total angular deflection of the ray is 
2 2 2 6 2 ,A A A A A A A Aa b b b a b a bθ θ π θ π θ θ θ θ θ πΔ = − + − + − + − = − + where we have used the fact from the previous 
problem that all the internal angles are equal and the two external equals are equal. Also using the Snell�s Law 
relationship, we have: 1arcsin sinA Ab an
θ θ⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 12 6 2 2 6arcsin sin 2 .A A A Aa b a an
θ θ π θ θ π⎛ ⎞Δ = − + = − +⎜ ⎟
⎝ ⎠
 
33-20 Chapter 33 
(b) 2
2
2
1 6 cos0 2 6 arcsin sin 0 2 .
sin1
A
aA A
a a
d d
d d n n
n
θθ
θ θ θ
Δ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = − ⇒ = −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠ −
. 
2
2 2 2 22
2 22
sin1 ( 1 cos ) 9cosn n
n
θ θ θ
⎛ ⎞
− = − + =⎜ ⎟
⎝ ⎠
. 2 22
1cos ( 1)
8
nθ = − . 
(c) For violet, 2 22
1 1arccos ( 1) arccos (1.342 1) 71.55
8 8
nθ
⎛ ⎞ ⎛ ⎞
= − = − = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
. violet 233.2Δ = ° and violet 53.2 .θ = ° 
For red, 2 22
1 1arccos ( 1) arccos (1.330 1) 71.94 .
8 8
nθ
⎛ ⎞ ⎛ ⎞
= − = − = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 red 230.1Δ = ° and red 50.1 .θ = ° 
EVALUATE: The angles we calculated agree with those given in Figure 37.20e in the textbook. The color that 
appears higher above the horizon is violet. The colors appear in reverse order in a secondary rainbow compared to 
a primary rainbow. 
 
34-1 
GEOMETRIC OPTICS 
 34.1. IDENTIFY and SET UP: Plane mirror: s s′= − (Eq.34.1) and / / 1m y y s s′ ′= = − = + (Eq.34.2). We are given s 
and y and are asked to find and .s y′ ′ 
EXECUTE: The object and image are shown in Figure 34.1. 
 
39.2 cms s′ = − = − 
( 1)(4.85 cm)y m y′ = = + 
4.85 cmy′ = 
 
Figure 34.1 
The image is 39.2 cm to the right of the mirror and is 4.85 cm tall. 
EVALUATE: For a plane mirror the image is always the same distance behind the mirror as the object is in front 
of the mirror. The image always has the same height as the object. 
 34.2. IDENTIFY: Similar triangles say tree tree
mirror mirror
h d
h d
= . 
SET UP: mirror 0.350 m,d = mirror 0.0400 mh = and tree 28.0 m 0.350 m.d = + 
EXECUTE: treetree mirror
mirror
28.0 m 0.350 m0.040 m 3.24 m.
0.350 m
dh h
d
+
= = = 
EVALUATE: The image of the tree formed by the mirror is 28.0 m behind the mirror and is 3.24 m tall. 
 34.3. IDENTIFY: Apply the law of reflection. 
SET UP: If up is the +y-direction and right is the +x-direction, then the object is at 0 0( , )x y− − and 2P′ is at 
0 0( , )x y− . 
EXECUTE: Mirror 1 flips the y-values, so the image is at 0 0( , )x y which is 3.P′ 
EVALUATE: Mirror 2 uses 1P′ as an object and forms an image at 3P′ . 
 34.4. IDENTIFY: / 2f R= 
SET UP: For a concave mirror 0.R > 
EXECUTE: (a) 34.0 cm 17.0 cm
2 2
Rf = = = 
EVALUATE: (b) The image formation by the mirror is determined by the law of reflection and that is unaffected 
by the medium in which the light is traveling. The focal length remains 17.0 cm. 
 34.5. IDENTIFY and SET UP: Use Eq.(34.6) to calculate s′ and use Eq.(34.7) to calculate .y′ The image is real if s′ is 
positive and is erect if 0.m > Concave means R and f are positive, 22.0 cm; / 2 11.0 cm.R f R= + = = + 
EXECUTE: (a) 
 
Three principal rays, 
numbered as in Sect. 34.2, 
are shown in Figure 34.5. 
The principal ray diagram 
shows that the image is 
real, inverted, and 
enlarged. 
 
Figure 34.5 
34
34-2 Chapter 34 
(b) 1 1 1
s s f
+ =
′
 
1 1 1 (16.5 cm)(11.0 cm) so 33.0 cm
16.5 cm 11.0 cm
s f sfs
s f s sf s f
− ′= − = = = = +
′ − −
 
0s′ > so real image, 33.0 cm to left of mirror vertex 
33.0 cm 2.00
16.5 cm
sm
s
′
= − = − = − (m < 0 means inverted image) 2.00(0.600 cm) 1.20 cmy m y′ = = = 
EVALUATE: The image is 33.0 cm to the left of the mirror vertex. It is real, inverted, and is 1.20 cm tall 
(enlarged). The calculation agrees with the image characterization from the principal ray diagram. A concave 
mirror used alone always forms a real, inverted image if s > f and the image is enlarged if f < s < 2f. 
 34.6. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 and 
sm
s
′
= − . 
SET UP: For a convex mirror, 0R < . 22.0 cmR = − and 11.0 cm
2
Rf = = − . 
EXECUTE: (a) The principal-ray diagram is sketched in Figure 34.6. 
(b) 1 1 1
s s f
+ =
′
. (16.5 cm)( 11.0 cm) 6.6 cm
16.5 cm ( 11.0 cm)
sfs
s f
−′ = = = −
− − −
. 
6.6 cm 0.400
16.5 cm
sm
s
′ −
= − = − = + . 
(0.400)(0.600 cm) 0.240 cmy m y′ = = = . The image is 6.6 cm to the right of the mirror. It is 0.240 cm tall. 
0s′ < , so the image is virtual. 0m > , so the image is erect. 
EVALUATE: The calculated image properties agree with the image characterization from the principal-ray diagram. 
 
Figure 34.6 
 34.7. IDENTIFY: 1 1 1
s s f
+ =
′
. sm
s
′
= − . 
y
m
y
′
= . Find m and calculate y′ . 
SET UP: 1.75 mf = + . 
EXECUTE: s f! so 1.75 ms f′ = = . 
11
10
1.75 m 3.14 10 .
5.58 10 m
sm
s
−′= − = − = − ×
×
11 6 4(3.14 10 )(6.794 10 m) 2.13 10 m 0.213 mmy m y − −′ = = × × = × = . 
EVALUATE: The image is real and is 1.75 m in front of the mirror. 
 34.8. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 and sm
s
′
= − . 
SET UP: The mirror surface is convex so 3.00 cmR = − . 24.0 cm 3.00 cm 21.0 cms = − = . 
EXECUTE: 1.50 cm
2
Rf = = − . 1 1 1
s s f
+ =
′
. (21.0 cm)( 1.50 cm) 1.40 cm
21.0 cm ( 1.50 cm)
sfs
s f
−′ = = = −
− − −
. The image is 
1.40 cm behind the surface so it is 3.00 cm 1.40 cm 1.60 cm− = from the center of the ornament, on the same side 
as the object. 1.40 cm 0.0667
21.0 cm
sm
s
′ −
= − = − = + . (0.0667)(3.80 mm) 0.253 mmy m y′ = = = . 
EVALUATE: The image is virtual, upright and smaller than the object. 
 34.9. IDENTIFY: The shell behaves as a spherical mirror. 
SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 
1 1 1
s s f
+ =
′
, and its magnification is given by sm
s
′
= − . 
Geometric Optics 34-3 
EXECUTE: 1 1 1 1 2 1 18.0 cm
18.0 cm 6.00 cm
s
s s f s
+ = ⇒ = − ⇒ =
′ − −
 from the vertex. 
6.00 cm 1 1 (1.5 cm) = 0.50 cm
18.0 cm 3 3
sm y
s
′ − ′= − = − = ⇒ = . The image is 0.50 cm tall, erect, and virtual. 
EVALUATE: Since the magnification is less than one, the image is smaller than the object. 
34.10. IDENTIFY: The bottom surface of the bowl behaves as a spherical convex mirror. 
SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 
1 1 1
s s f
+ =
′
, and its magnification is given by sm
s
′
= − . 
EXECUTE: 1 1 1 1 2 1 15 cm
35 cm 90 cm
s
s s f s
− ′+ = ⇒ = − ⇒ = −
′ ′
 behind bowl. 
15 cm 0.167 (0.167)(2.0 cm) 0.33 cm
90 cm
sm y
s
′
′= − = = ⇒ = = . The image is 0.33 cm tall, erect, and virtual. 
EVALUATE: Since the magnification is less than one, the image is smaller than the object. 
34.11. IDENTIFY: We are dealing with a spherical mirror. 
SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 
1 1 1
s s f
+ =
′
, and its magnification is given by sm
s
′
= − . 
EXECUTE: (a) 1 1 1 1 1 1 . Also .s f sf s fs m
s s f s f s fs s f s f s
′− ′+ = ⇒ = − = ⇒ = = − =
′ ′ − −
 
(b) The graph is given in Figure 34.11a. 
(c) 0 for , 0.s s f s′ > > < 
(d) 0 for 0 .s s f′ < < < 
(e) The image is at negative infinity, �behind� the mirror. 
(f ) At the focal point, s = f. 
(g) The image is at the mirror, 0s′ = . 
(h) The graph is given in Figure 34.11b. 
(i) Erect and larger if 0 < s < f. 
(j) Inverted if .s f> 
(k) The image is smaller if 2 or 0.s f s> < 
(l) As the object is moved closer and closer to the focal point, the magnification increases to infinite values. 
EVALUATE: As the object crosses the focal point, both the image distance and the magnification undergo 
discontinuities. 
 
Figure 34.11 
34-4 Chapter 34 
34.12. IDENTIFY: sfs
s f
′ =
−
 and fm
f s
=
−
. 
SET UP: With f f= − , 
s f
s
s f
′ = −
+
 and 
f
m
s f
=
+
. 
EXECUTE: The graphs are given in Figure 34.12. 
(a) 0 for 0.s f s′> − < < 
(b) 0 for s and 0.s f s′ < < − < 
(c) If the object is at infinity, the image is at the outward going focal point. 
(d) If the object is next to the mirror, then the image is also at the mirror 
(e) The image is erect (magnification greater than zero) for .s f> − 
(f) The image is inverted (magnification less than zero) for .s f< − 
(g) The image is larger than the object (magnification greater than one) for 2 0.f s− < < 
(h) The image is smaller than the object (magnification less than one) for 0 and 2 .s s f> < − 
EVALUATE: For a real image ( 0)s > , the image formed by a convex mirror is always virtual and smaller than the 
object. 
 
Figure 34.12 
34.13. IDENTIFY: 1 1 1
s s f
+ =
′
 and y sm
y s
′ ′
= = − . 
SET UP: 2.00m = + and 1.25 cms = . An erect image must be virtual. 
EXECUTE: (a) sfs
s f
′ =
−
 and fm
s f
= −
−
. For a concave mirror, m can be larger than 1.00. For a convex mirror, 
f f= − so 
f
m
s f
= +
+
 and m is always less than 1.00. The mirror must be concave ( 0)f > . 
(b) 1 s s
f ss
′ +
=
′
. ssf
s s
′
=
′+
. 2.00sm
s
′
= − = + and 2.00s s′ = − . ( 2.00 ) 2.00 2.50 cm
2.00
s sf s
s s
−
= = + = +
−
. 
2 5.00 cmR f= = + . 
(c) The principal ray diagram is drawn in Figure 34.13. 
Geometric Optics 34-5 
EVALUATE: The principal-ray diagram agrees with the description from the equations. 
 
Figure 34.13 
34.14. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 and sm
s
′
= − . 
SET UP: For a concave mirror, 0R > . 32.0 cmR = and 16.0 cm
2
Rf = = . 
EXECUTE: (a) 1 1 1
s s f
+ =
′
. (12.0 cm)(16.0 cm) 48.0 cm
12.0 cm 16.0 cm
sfs
s f
′ = = = −
− −
. 48.0 cm 4.00
12.0 cm
sm
s
′ −
= − = − = + . 
(b) 48.0 cms′ = − , so the image is 48.0 cm to the right of the mirror. 0s′ < so the image is virtual. 
(c) The principal-ray diagram is sketched in Figure 34.14. The rules for principal rays apply only to paraxial rays. 
Principal ray 2, that travels to the mirror along a line that passes through the focus, makes a large angle with the 
optic axis and is not described well by the paraxial approximation. Therefore, principal ray 2 is not included in the 
sketch. 
EVALUATE: A concave mirror forms a virtual image whenever s f< . 
 
Figure 34.14 
34.15. IDENTIFY: Apply Eq.(34.11), with . R s′→∞ is the apparent depth. 
SET UP The image and object are shown in Figure 34.15. 
 
;a b b an n n n
s s R
−
+ =
′
 
R →∞ (flat surface), so 
0a bn n
s s
+ =
′
 
 
Figure 34.15 
EXECUTE: (1.00)(3.50 cm) 2.67 cm
1.309
b
a
n ss
n
′ = − = − = − 
The apparent depth is 2.67 cm. 
EVALUATE: When the light goes from ice to air (larger to smaller n), it is bent away from the normal and the 
virtual image is closer to the surface than the object is. 
34-6 Chapter 34 
34.16. IDENTIFY: The surface is flat so R →∞ and 0a bn n
s s
+ =
′
. 
SET UP: The light travels from the fish to the eye, so 1.333an = and 1.00bn = . When the fish is viewed, 
7.0 cms = . The fish is 20.0 cm 7.0 cm 13.0 cm− = above the mirror, so the image of the fish is 13.0 cm below the 
mirror and 20.0 cm 13.0 cm 33.0 cm+ = below the surface of the water. When the image is viewed, 33.0 cms = . 
EXECUTE: (a) 1.00 (7.0 cm) 5.25 cm
1.333
b
a
ns s
n
⎛ ⎞ ⎛ ⎞′ = − = − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
. The apparent depth is 5.25 cm. 
(b) 1.00 (33.0 cm) 24.8 cm
1.333
b
a
ns s
n
⎛ ⎞ ⎛ ⎞′ = − = − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
. The apparent depth of the image of the fish in the mirror is 24.8 cm. 
EVALUATE: In each case the apparent depth is less than the actual depth of what is being viewed. 
34.17. IDENTIFY: a b b an n n n
s s R
−
+ =
′
. a
b
n sm
n s
′
= − . Light comes from the fish to the person�s eye. 
SET UP: 14.0 cmR = − . 14.0 cms = + . 1.333an = (water). 1.00bn = (air). Figure 34.17 shows the object and the 
refracting surface. 
EXECUTE: (a) 1.333 1.00 1.00 1.333
14.0 cm 14.0 cms
−
+ =
′ −
. 14.0 cms′ = − . (1.333)( 14.0 cm) 1.33
(1.00)(14.0 cm)
m −= − = + . 
The fish�s image is 14.0 cm to the left of the bowl surface so is at the center of the bowl and the magnification is 1.33. 
(b) The focal point is at the image location when s →∞ . b b an n n
s R
−
=
′
. 1.00an = . 1.333bn = . 14.0 cmR = + . 
1.333 1.333 1.00
14.0 cms
−
=
′
. 56.0 cms′ = + . s′ is greater than the diameter of the bowl, so the surface facing the sunlight 
does not focus the sunlight to a point inside the bowl. The focal point is outside the bowl and there is no danger to the fish. 
EVALUATE: In part (b) the rays refract when they exit the bowl back into the air so the image we calculated is not 
the final image. 
 
Figure 34.17 
34.18. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
. 
SET UP: For a convex surface, 0R > . 3.00 cmR = + . 1.00an = , 1.60bn = . 
EXECUTE: (a) s →∞ . b b an n n
s R
−
=
′
. 1.60 ( 3.00 cm) 8.00 cm
1.60 1.00
b
b a
ns R
n n
⎛ ⎞ ⎛ ⎞′ = = + = +⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠
. The image is 
8.00 cm to the right of the vertex. 
(b) 12.0 cms = . 1.00 1.60 1.60 1.00
12.0 cm 3.00 cms
−
+ =
′
. 13.7 cms′ = + . The image is 13.7 cm to the right of the vertex. 
(c) 2.00 cms = . 1.00 1.60 1.60 1.00
2.00 cm 3.00 cms
−
+ =
′
. 5.33 cms′ = − . The image is 5.33 cm to the left of the vertex. 
EVALUATE: The image can be either real ( 0s′ > ) or virtual ( 0s′ < ), depending on the distance of the object 
from the refracting surface. 
34.19. IDENTIFY: The hemispherical glass surface forms an image by refraction. The location of this image depends on 
the curvature of the surface and the indices of refraction of the glass and oil. 
SET UP: The image and object distances are related to the indices of refraction and the radius of curvature by the 
equation a b b an n n n
s s R
−
+ =
′
. 
Geometric Optics 34-7 
EXECUTE: 1.45 1.60 0.15 0.395 cm
1.20 m 0.0300 m
a b b an n n n s
s s R s
−
+ = ⇒ + = ⇒ =
′
 
EVALUATE: The presence of the oil changes the location of the image. 
34.20. IDENTIFY: a b b an n n n
s s R
−
+ =
′
. a
b
n sm
n s
′
= − . 
SET UP: 4.00 cmR = + . 1.00an = . 1.60bn = . 24.0 cms = . 
EXECUTE: 1 1.60 1.60 1.00
24.0 cm 4.00 cms
−
+ =
′
. 14.8 cms′ = + . (1.00)(14.8 cm) 0.385
(1.60)(24.0 cm)
m = − = − . 
(0.385)(1.50 mm) 0.578 mmy m y′ = = = . The image is 14.8 cm to the right of the vertex and is 0.578 mm tall. 
0m < , so the image is inverted. 
EVALUATE: The image is real. 
34.21. IDENTIFY: Apply Eqs.(34.11) and (34.12). Calculate s′ and .y′ The image is erect if m > 0. 
SET UP: The object and refracting surface are shown in Figure 34.21. 
 
Figure 34.21 
EXECUTE: a b b an n n n
s s R
−
+ =
′
 
1.00 1.60 1.60 1.00
24.0 cm 4.00 cms
−
+ =
′ −
 
Multiplying by 24.0 cm gives 38.41.00 3.60
s
+ = −
′
 
38.4 cm 4.60
s
= −
′
 and 38.4 cm 8.35 cm
4.60
s′ = − = − 
Eq.(34.12): (1.00)( 8.35 cm) 0.217
(1.60)( 24.0 cm)
a
b
n sm
n s
′ −
= − = − = +
+
 
(0.217)(1.50 mm) 0.326 mmy m y′ = = = 
EVALUATE: The image is virtual ( 0)s′ < and is 8.35 cm to the left of the vertex. The image is erect ( 0)m > and 
is 0.326 mm tall. R is negative since the center of curvature of the surface is on the incoming side. 
34.22. IDENTIFY: The hemispherical glass surface forms an image by refraction. The location of this image depends on 
the curvature of the surface and the indices of refraction of the glass and liquid. 
SET UP: The image and object distances are related to the indices of refraction and the radius of curvature by the 
equation a b b an n n n
s s R
−
+ =
′
. 
EXECUTE: 1.60 1.60 1.24
14.0 cm 9.00 cm 4.00 cm
a b b a a a
a
n n n n n n n
s s R
− −
+ = ⇒ + = ⇒ =
′
. 
EVALUATE: The result is a reasonable refractive index for liquids. 
34.23. IDENTIFY: Use 
1 2
1 1 1( 1)n
f R R
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
 to calculate f. The apply 1 1 1
s s f
+ =
′
 and y sm
y s
′ ′
= = − . 
SET UP: 1R →∞ . 2 13.0 cmR = − . If the lens is reversed, 1 13.0 cmR = + and 2R →∞ . 
EXECUTE: (a) 1 1 1 0.70(0.70)
13.0 cm 13.0 cmf
⎛ ⎞= − =⎜ ⎟∞ −⎝ ⎠
and 18.6 cmf = . 1 1 1 s f
s f s sf
−
= − =
′
. 
(22.5 cm)(18.6 cm) 107cm
22.5 cm 18.6 cm
sfs
s f
′ = = =
− −
. 107 cm 4.76
22.5 cm
sm
s
′
= − = − = − . 
( 4.76)(3.75 mm) 17.8 mmy my′ = = − = − . The image is 107 cm to the right of the lens and is 17.8 mm tall. The 
image is real and inverted. 
(b) 1 1 1( 1)
13.0 cm
n
f
⎛ ⎞= − −⎜ ⎟∞⎝ ⎠
and 18.6 cmf = . The image is the same as in part (a). 
EVALUATE: Reversing a lens does not change the focal length of the lens. 
34-8 Chapter 34 
34.24. IDENTIFY: 1 1 1
s s f
+ =
′
. The sign of f determines whether the lens is converging or diverging. 
SET UP: 16.0 cms = . 12.0 cms′ = − . 
EXECUTE: (a) (16.0 cm)( 12.0 cm) 48.0 cm
16.0 cm ( 12.0 cm)
ssf
s s
′ −
= = = −
′+ + −
. 0f < and the lens is diverging. 
(b) 12.0 cm 0.750
16.0 cm
sm
s
′ −
= − = − = + . (0.750)(8.50 mm) 6.38 mmy m y′ = = = . 0m > and the image is erect. 
(c) The principal-ray diagram is sketched in Figure 34.24. 
EVALUATE: A diverging lens always forms an image that is virtual, erect and reduced in size. 
 
Figure 34.24 
34.25. IDENTIFY: The liquid behaves like a lens, so the lensmaker�s equation applies. 
SET UP: The lensmaker�s equation is 
1 2
1 1 1 1( 1)n
s s R R
⎛ ⎞
+ = − −⎜ ⎟′ ⎝ ⎠
, and the magnification of the lens is sm
s
′
= − . 
EXECUTE: (a) 
1 2
1 1 1 1 1 1 1 1( 1) (1.52 1)
24.0 cm 7.00 cm 4.00 cm
n
s s R R s
⎛ ⎞ ⎛ ⎞
+ = − − ⇒ + = − −⎜ ⎟ ⎜ ⎟′ ′ − −⎝ ⎠⎝ ⎠
 
71.2 cms′⇒ = , to the right of the lens. 
(b) 71.2 cm 2.97
24.0 cm
sm
s
′
= − = − = − 
EVALUATE: Since the magnification is negative, the image is inverted. 
34.26. IDENTIFY: Apply y sm
y s
′ ′
= = − to relate s′ and s and then use 1 1 1
s s f
+ =
′
. 
SET UP: Since the image is inverted, 0y′ < and 0m < . 
EXECUTE: 4.50 cm 1.406
3.20 cm
ym
y
′ −
= = = − . sm
s
′
= − gives 1.406s s′ = + . 1 1 1
s s f
+ =
′
 gives 
1 1 1
1.406 90.0 cms s
+ = and 154 cms = . (1.406)(154 cm) 217 cms′ = = . The object is 154 cm to the left of the 
lens. The image is 217 cm to the right of the lens and is real. 
EVALUATE: For a single lens an inverted image is always real. 
34.27. IDENTIFY: The thin-lens equation applies in this case. 
SET UP: The thin-lens equation is 1 1 1
s s f
+ =
′
, and the magnification is s ym
s y
′ ′
= − = . 
EXECUTE: 34.0 mm 12.0 cm4.25 2.82 cm
8.00 mm
y sm s
y s s
′ ′ −
= = = = − = − ⇒ = . The thin-lens equation gives 
1 1 1 3.69 cmf
s s f
+ = ⇒ =
′
. 
EVALUATE: Since the focal length is positive, this is a converging lens. The image distance is negative because 
the object is inside the focal point of the lens. 
34.28. IDENTIFY: Apply sm
s
′
= − to relate s and s′ . Then use 1 1 1
s s f
+ =
′
. 
SET UP: Since the image is to the right of the lens, 0s′ > . 6.00 ms s′ + = . 
EXECUTE: (a) 80.0s s′ = and 6.00 ms s′+ = gives 81.00 6.00 ms = and 0.0741 ms = . 5.93 ms′ = . 
Geometric Optics 34-9 
(b) The image is inverted since both the image and object are real ( 0, 0).s s′ > > 
(c) 1 1 1 1 1 0.0732 m,
0.0741 m 5.93 m
f
f s s
= + = + ⇒ =
′
and the lens is converging. 
EVALUATE: The object is close to the lens and the image is much farther from the lens. This is typical for slide 
projectors. 
34.29. IDENTIFY: Apply 
1 2
1 1 1( 1)n
f R R
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
. 
SET UP: For a distant object the image is at the focal point of the lens. Therefore, 1.87 cmf = . For the double-
convex lens, 1R R= + and 2R R= − , where 2.50 cmR = . 
EXECUTE: 1 1 1 2( 1)( 1) nn
f R R R
−⎛ ⎞= − − =⎜ ⎟−⎝ ⎠
. 2.50 cm1 1 1.67
2 2(1.87 cm)
Rn
f
= + = + = . 
EVALUATE: 0f > and the lens is converging. A double-convex lens is always converging. 
34.30. IDENTIFY and SET UP: Apply 
1 2
1 1 1( 1)n
f R R
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
 
EXECUTE: We have a converging lens if the focal length is positive, which requires 
1 2 1 2
1 1 1 1 1( 1) 0 0.n
f R R R R
⎛ ⎞ ⎛ ⎞
= − − > ⇒ − >⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 This can occur in one of three ways: 
(i) 1R and 2R both positive and 1 2R R< . 1 2(ii) 0, 0R R≥ ≤ (double convex and planoconvex). 
(iii) 1R and 2R both negative and 1 2R R> (meniscus). The three lenses in Figure 35.32a in the textbook fall into 
these categories. 
We have a diverging lens if the focal length is negative, which requires 
1 2 1 2
1 1 1 1 1( 1) 0 0.n
f R R R R
⎛ ⎞ ⎛ ⎞
= − − < ⇒ − <⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 This can occur in one of three ways: 
(i) 1R and 2R both positive and 1 2R R> (meniscus). (ii) 1R and 2R both negative and 2 1R R> . (iii) 1 20, 0R R≤ ≥ 
(planoconcave and double concave). The three lenses in Figure 34.32b in the textbook fall into these categories. 
EVALUATE: The converging lenses in Figure 34.32a are all thicker at the center than at the edges. The diverging 
lenses in Figure 34.32b are all thinner at the center than at the edges. 
34.31. IDENTIFY and SET UP: The equations 1 1 1
s s f
+ =
′
 and sm
s
′
= − apply to both thin lenses and spherical mirrors. 
EXECUTE: (a) The derivation of the equations in Exercise 34.11 is identical and one gets: 
1 1 1 1 1 1 , and also .s f sf s fs m
s s f s f s fs s f s f s
′− ′+ = ⇒ = − = ⇒ = = − =
′ ′ − −
 
(b) Again, one gets exactly the same equations for a converging lens rather than a concave mirror because the 
equations are identical. The difference lies in the interpretation of the results. For a lens, the outgoing side is not 
that on which the object lies, unlike for a mirror. So for an object on the left side of the lens, a positive image 
distance means that the image is on the right of the lens, and a negative image distance means that the image is on 
the left side of the lens. 
(c) Again, for Exercise 34.12, the change from a convex mirror to a diverging lens changes nothing in the 
exercises, except for the interpretation of the location of the images, as explained in part (b) above. 
EVALUATE: Concave mirrors and converging lenses both have 0f > . Convex mirrors and diverging lenses both 
have 0f < . 
34.32. IDENTIFY: Apply 1 1 1
s s f
+ =
′
and y sm
y s
′ ′
= = − . 
SET UP: 12.0 cmf = + and 17.0 cms′ = − . 
EXECUTE: 1 1 1 1 1 1 7.0 cm.
12.0 cm 17.0 cm
s
s s f s
+ = ⇒ = − ⇒ =
′ −
 
( 17.0) 0.800 cm2.4 0.34 cm,
7.2 2.4
s ym y
s m
′ ′−
= − = − = + ⇒ = = = +
+
so the object is 0.34 cm tall, erect, same side as the 
image. The principal-ray diagram is sketched in Figure 34.32. 
34-10 Chapter 34 
EVALUATE: When the object is inside the focal point, a converging lens forms a virtual, enlarged image. 
 
Figure 34.32 
34.33. IDENTIFY: Use Eq.(34.16) to calculate the object distance s. m calculated from Eq.(34.17) determines the size 
and orientation of the image. 
SET UP: 48.0 cm.f = − Virtual image 17.0 cm from lens so 17.0 cm.s′ = − 
EXECUTE: 1 1 1 1 1 1, so s f
s s f s f s sf
−
+ = = − =
′ ′
 
( 17.0 cm)( 48.0 cm) 26.3 cm
17.0 cm ( 48.0 cm)
s fs
s f
′ − −
= = = +
′ − − − −
 
17.0 cm 0.646
26.3 cm
sm
s
′ −
= − = − = +
+
 
8.00 mm so 12.4 mm
0.646
yym y
y m
′′
= = = = 
The principal-ray diagram is sketched in Figure 34.33. 
EVALUATE: Virtual image, real object (s > 0) so image and object are on same side of lens. 
m > 0 so image is erect with respect to the object. The height of the object is 12.4 mm. 
 
Figure 34.33 
34.34. IDENTIFY: Apply 1 1 1
s s f
+ =
′
. 
SET UP: The sign of f determines whether the lens is converging or diverging. 16.0 cms = . 36.0 cms′ = + . Use 
sm
s
′
= − to find the size and orientation of the image. 
EXECUTE: (a) (16.0 cm)(36.0 cm) 11.1 cm
16.0 cm 36.0 cm
ssf
s s
′
= = =
′+ +
. 0f > and the lens is converging. 
(b) 36.0 cm 2.25
16.0 cm
sm
s
′
= − = − = − . (2.25)(8.00 mm) 18.0 mmy m y′ = = = . 0m < so the image is inverted. 
(c) The principal-ray diagram is sketched in Figure 34.34. 
EVALUATE: The image is real so the lens must be converging. 
 
Figure 34.34 
Geometric Optics 34-11 
34.35. IDENTIFY: Apply 1 1 1
s s f
+ =
′
. 
SET UP: The image is to be formed on the film, so 20.4 cms′ = + . 
EXECUTE: 1 1 1 1 1 1 1020 cm 10.2 m.
20.4 cm 20.0 cm
s
s s f s
+ = ⇒ + = ⇒ = =
′
 
EVALUATE: The object distance is much greater than f, so the image is justoutside the focal point of the lens. 
34.36. IDENTIFY: Apply 1 1 1
s s f
+ =
′
and y sm
y s
′ ′
= = − . 
SET UP: 3.90 ms = . 0.085 mf = . 
EXECUTE: 1 1 1 1 1 1 0.0869 m.
3.90 m 0.085 m
s
s s f s
′+ = ⇒ + = ⇒ =
′ ′
 0.08691750 mm 39.0 mm,
3.90
sy y
s
′
′ = − = − = − so 
it will not fit on the 24-mm× 36-mm film. 
EVALUATE: The image is just outside the focal point and s f′ ≈ . To have 36 mmy′ = , so that the image will fit 
on the film, (0.085 m)(1.75 m) 4.1 m
0.036 m
s ys
y
′
= − ≈ − =
′ −
. The person would need to stand about 4.1 m from the lens. 
34.37. IDENTIFY: sm
s
′
= . 
SET UP: s f! , so s f′ ≈ . 
EXECUTE: (a) 428 mm 1.4 10 .
200,000 mm
s fm m
s s
−′= ≈ ⇒ = = × 
(b) 4105 mm 5.3 10 .
200,000 mm
s fm m
s s
−′= ≈ ⇒ = = × 
(c) 3300 mm 1.5 10 .
200,000 mm
s fm m
s s
−′= ≈ ⇒ = = × 
EVALUATE: The magnitude of the magnification increases when f increases. 
34.38. IDENTIFY: 
ysm
s y
′′
= = 
SET UP: s f! , so s f′ ≈ . 
EXECUTE: 3
5.00 m (70.7 m) 0.0372 m 37.2 mm.
9.50 10 m
s fy y y
s s
′
′ = ≈ = = =
×
 
EVALUATE: A very long focal length lens is needed to photograph a distant object. 
34.39. IDENTIFY and SET UP: Find the lateral magnification that results in this desired image size. Use Eq.(34.17) to 
relate m and s′ and Eq.(34.16) to relate s and s′ to f. 
EXECUTE: (a) We need 
3
424 10 m 1.5 10 .
160 m
m
−
−×= − = − × Alternatively, 
3
436 10 m 1.5 10 .
240 m
m
−
−×= − = − × 
 so s f s f′ ≈! 
Then 41.5 10s fm
s s
−′= − = − = − × and 4(1.5 10 )(600 m) 0.090 m 90 mm.f −= × = = 
A smaller f means a smaller s′ and a smaller m, so with f = 85 mm the object�s image nearly fills the picture area. 
(b) We need 
3
336 10 m 3.75 10 .
9.6 m
m
−
−×= − = − × Then, as in part (a), 33.75 10f
s
−= × and 
3(40.0 m)(3.75 10 ) 0.15 m 150 mm.f −= × = = Therefore use the 135 mm lens. 
EVALUATE: When s f! and , ( / ).s f y f y s′ ′≈ = − For the mobile home y/s is smaller so a larger f is needed. 
Note that m is very small; the image is much smaller than the object. 
34.40. IDENTIFY: Apply 1 1 1
s s f
+ =
′
to each lens. The image of the first lens serves as the object for the second lens. 
SET UP: For a distant object, s →∞ 
EXECUTE: (a) 1 1 1 12 cm.s s f′= ∞⇒ = = 
(b) 2 4.0 cm 12 cm 8 cm.s = − = − 
34-12 Chapter 34 
(c) 2
2
1 1 1 1 1 1 24 cm,
8 cm 12 cm
s
s s f s
′+ = ⇒ + = ⇒ =
′ ′− −
to the right. 
(d) 
11 1
12 cm.s s f′= ∞⇒ = = 2 8.0 cm 12 cm 4 cm.s = − = − 2
2
1 1 1 1 1 1 6 cm.
4 cm 12 cm
s
s s f s
′+ = ⇒ + = ⇒ =
′ ′− −
 
EVALUATE: In each case the image of the first lens serves as a virtual object for the second lens, and 2 0s < . 
34.41. IDENTIFY: The f-number of a lens is the ratio of its focal length to its diameter. To maintain the same exposure, 
the amount of light passing through the lens during the exposure must remain the same. 
SET UP: The f-number is f/D. 
EXECUTE: (a) 180.0 mm-number -number -number = /11
16.36 mm
ff f f f
D
= ⇒ = ⇒ . (The f-number is an integer.) 
(b) f/11 to f/2.8 is four steps of 2 in intensity, so one needs 1/16th the exposure. The exposure should be 1/480 s = 
32.1 10−× s = 2.1 ms. 
EVALUATE: When opening the lens from f/11 to f/2.8, the area increases by a factor of 16, so 16 times as much 
light is allowed in. Therefore the exposure time must be decreased by a factor of 1/16 to maintain the same 
exposure on the film or light receptors of a digital camera. 
34.42. IDENTIFY and SET UP: The square of the aperture diameter is proportional to the length of the exposure time 
required. 
EXECUTE: 
2
1 8 mm 1 s s
30 23.1 mm 250
⎛ ⎞⎛ ⎞ ⎛ ⎞≈⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
 
EVALUATE: An increase in the aperture diameter decreases the exposure time. 
34.43. IDENTIFY and SET UP: Apply 1 1 1
s s f
+ =
′
to calculate s′ . 
EXECUTE: (a) A real image is formed at the film, so the lens must be convex. 
(b) 1 1 1 1so and , with 50.0.0 mms f sfs f
s s f s sf s f
− ′+ = = = = +
′ ′ −
. For 45 cm 450 mm, s 56 mm.s ′= = = For 
, 50 mm.s s f′= ∞ = = The range of distances between the lens and film is 50 mm to 56 mm. 
EVALUATE: The lens is closer to the film when photographing more distant objects. 
34.44. IDENTIFY: The projector lens can be modeled as a thin lens. 
SET UP: The thin-lens equation is 1 1 1
s s f
+ =
′
, and the magnification of the lens is sm
s
′
= − . 
EXECUTE: (a) 1 1 1 1 1 1 147.5 mm
0.150 m 9.00 m
f
s s f f
+ = ⇒ = + ⇒ =
′
, so use a f = 148 mm lens. 
(b) 60 Area 1.44 m 2.16 msm m
s
′
= − ⇒ = ⇒ = × . 
EVALUATE: The lens must produce a real image to be viewed on the screen. Since the magnification comes out 
negative, the slides to be viewed must be placed upside down in the tray. 
34.45. (a) IDENTIFY: The purpose of the corrective lens is to take an object 25 cm from the eye and form a virtual 
image at the eye�s near point. Use Eq.(34.16) to solve for the image distance when the object distance is 25 cm. 
SET UP: 1 2.75
f
= + diopters means 1 m 0.3636 m
2.75
f = + = + (converging lens) 
36.36 cm; 25 cm; ?f s s′= = = 
EXECUTE: 1 1 1 so
s s f
+ =
′
 
(25 cm)(36.36 cm) 80.0 cm
25 cm 36.36 cm
sfs
s f
′ = = = −
− −
 
The eye�s near point is 80.0 cm from the eye. 
(b) IDENTIFY: The purpose of the corrective lens is to take an object at infinity and form a virtual image of it at 
the eye�s far point. Use Eq.(34.16) to solve for the image distance when the object is at infinity. 
SET UP: 1 1.30
f
= − diopters means 1 m 0.7692 m
1.30
f = − = − (diverging lens) 
76.92 cm; ; ?f s s′= − = ∞ = 
EXECUTE: 1 1 1
s s f
+ =
′
 and s =∞ says 1 1
s f
=
′
 and 76.9 cm.s f′ = = − The eye�s far point is 76.9 cm from the eye. 
Geometric Optics 34-13 
EVALUATE: In each case a virtual image is formed by the lens. The eye views this virtual image instead of the 
object. The object is at a distance where the eye can�t focus on it, but the virtual image is at a distance where the 
eye can focus. 
34.46. IDENTIFY: a b b an n n n
s s R
−
+ =
′
 
SET UP: 1.00an = , 1.40bn = . 40.0 cms = , 2.60 cms′ = . 
EXECUTE: 1 1.40 0.40
40.0 cm 2.60 cm R
+ = and 0.710 cmR = . 
EVALUATE: The cornea presents a convex surface to the object, so 0R > . 
34.47. IDENTIFY: In each case the lens forms a virtual image at a distance where the eye can focus. Power in diopters 
equals 1/ f , where f is in meters. 
SET UP: In part (a), 25 cms = and in part (b), s →∞ . 
EXECUTE: (a) 1 1 1 1 1 1power 2.33
0.25 m 0.600 mf s s f
= + = + ⇒ = = +
′ −
 diopters. 
(b) 1 1 1 1 1 1power 1.67
0.600 mf s s f
= + = + ⇒ = = −
′ ∞ −
 diopters. 
EVALUATE: A converging lens corrects the near vision and a diverging lens corrects the far vision. 
34.48. IDENTIFY: When the object is at the focal point, 25.0 cmM
f
= . In part (b), apply 1 1 1
s s f
+ =
′
 to calculate s for 
25.0 cms′ = − . 
SET UP: Our calculation assumes the near point is 25.0 cm from the eye. 
EXECUTE: (a) Angular magnification 25.0 cm 25.0 cm 4.17.
6.00 cm
M
f
= = = 
(b) 1 1 1 1 1 1 4.84 cm.
25.0 cm 6.00 cm
s
s s f s
+ = ⇒ + = ⇒ =
′ −
 
EVALUATE: In part (b), y
s
θ′ = , 
25.0 cm
yθ = and 25.0 cm 25.0 cm 5.17
4.84 cm
M
s
= = = . M is greater when the image 
is at the near point than when the image is at infinity. 
34.49. IDENTIFY: Use Eqs.(34.16) and (34.17) to calculate s and .y′ 
(a) SET UP: 8.00 cm; 25.0 cm; ?f s s′= = − = 
1 1 1 1 1 1, so s f
s s f s f s s f
′ −
+ = = − =
′ ′ ′
 
EXECUTE: ( 25.0 cm)( 8.00 cm) 6.06 cm
25.0 cm 8.00 cm
s fs
s f
′ − +
= = = +
′ − − −
 
(b) 25.0 cm 4.125
6.06 cm
sm
s
′ −
= − = − = + 
 so (4.125)(1.00 mm) 4.12 mm
y
m y m y
y
′
′= = = = 
EVALUATE: The lens allows the object to be much closer to the eye than the near point. The lense allows the eye 
to view an image at the near point rather than the object. 
34.50. IDENTIFY: For a thin lens, s y
s y
′ ′
− = , so y y
s s
′
=
′
, and the angular size of the image equals the angular size of the 
object. 
SET UP: The object has angular size y
f
θ = , with θ in radians. 
EXECUTE: 2.00mm 80.0 mm 8.00 cm.
0.025 rad
y yf
f
θ
θ
= ⇒ = = = = 
EVALUATE: If the insect is at the near point of a normal eye, its angular size is 2.00 mm 0.0080 rad
250 mm
= . 
34.51. IDENTIFY: The thin-lens equation applies to the magnifying lens. 
SET UP: The thin-lens equation is 1 1 1
s s f
+ =
′
. 
34-14 Chapter 34 
EXECUTE: The image is behind the lens, so 0s′ < . The thin-lens equation gives 
1 1 1 1 1 1 4.17 cm
5.00 cm 25.0 cm
s
s s f s
+ = ⇒ = − ⇒ =
′ −
, on the same side of the lens as the ant. 
EVALUATE: Since 0s′ < , the image will be erect. 
34.52. IDENTIFY: Apply Eq.(34.24). 
SET UP: 1 160 mm 5.0 mm 165 mms′ = + = 
EXECUTE: (a) 1
1 2
(250 mm) (250 mm)(165 mm) 317.
(5.00 mm)(26.0 mm)
sM
f f
′
= = = 
(b) The minimum separation is 40.10 mm 0.10 mm 3.15 10 mm.
317M
−= = × 
EVALUATE: The angular size of the image viewed by the eye when looking through the microscope is 317 times 
larger than if the object is viewed at the near-point of the unaided eye. 
34.53. (a) IDENTIFY and SET UP: 
 
Figure 34.53 
Final image is at ∞ so the object for the eyepiece is at its focal point. But the object for the eyepiece is the image 
of the objective so the image formed by the objective is 19.7 cm � 1.80 cm = 17.9 cm to the right of the lens. 
Apply Eq.(34.16) to the image formation by the objective, solve for the object distance s. 
0.800 cm; 17.9 cm; ?f s s′= = = 
1 1 1 1 1 1, so s f
s s f s f s s f
′ −
+ = = − =
′ ′ ′
 
EXECUTE: (17.9 cm)( 0.800 cm) 8.37 mm
17.9 cm 0.800 cm
s fs
s f
′ +
= = = +
′ − −
 
(b) SET UP: Use Eq.(34.17). 
EXECUTE: 1
17.9 cm 21.4
0.837 cm
sm
s
′
= − = − = − 
The linear magnification of the objective is 21.4. 
(c) SET UP: Use Eq.(34.23): 1 2M m M= 
EXECUTE: 2
2
25 cm 25 cm 13.9
1.80 cm
M
f
= = = 
1 2 ( 21.4)(13.9) 297M m M= = − = − 
EVALUATE: M is not accurately given by 1 1 2(25 cm) / 311,s f f′ = because the object is not quite at the focal point 
of the objective 1 1( 0.837 cm and 0.800 cm).s f= = 
34.54. IDENTIFY: Eq.(34.24) can be written 11 2 2
1
sM m M M
f
′
= = . 
SET UP: 1 1 120 mms f′ = + 
EXECUTE: 16 mm : 120 mm 16 mm 136 mm; 16 mmf s s′= = + = = . 1
136 mm 8.5
16 mm
sm
s
′
= = = . 
1
124 mm4 mm : 120 mm 4 mm 124 mm; 4 mm 31
4 mm
sf s s m
s
′
′= = + = = ⇒ = = = . 
1
122 mm1.9 mm : 120 mm 1.9 mm 122 mm; 1.9 mm 64
1.9 mm
sf s s m
s
′
′= = + = = ⇒ = = = . 
The eyepiece magnifies by either 5 or 10, so: 
(a) The maximum magnification occurs for the 1.9-mm objective and 10x eyepiece: 
1 e (64)(10) 640.M m M= = = 
(b) The minimum magnification occurs for the 16-mm objective and 5x eyepiece: 
1 e (8.5)(5) 43.M m M= = = 
EVALUATE: The smaller the focal length of the objective, the greater the overall magnification. 
Geometric Optics 34-15 
34.55. IDENTIFY: -number /f f D= 
SET UP: 1.02 mD = 
EXECUTE: 19.0 (19.0) (19.0)(1.02 m) 19.4 m.f f D
D
= ⇒ = = = 
EVALUATE: Camera lenses can also have an f-number of 19.0. For a camera lens, both the focal length and lens 
diameter are much smaller, but the f-number is a measure of their ratio. 
34.56. IDENTIFY: For a telescope, 1
2
fM
f
= − . 
SET UP: 2 9.0 cmf = . The distance between the two lenses equals 1 2f f+ . 
EXECUTE: 1 2 11.80 m 1.80 m 0.0900 m 1.71 mf f f+ = ⇒ = − = . 1
2
171 19.0.
9.00
fM
f
= − = − = − 
EVALUATE: For a telescope, 1 2f f! . 
34.57. (a) IDENTIFY and SET UP: Use Eq.(34.24), with 1 95.0 cmf = (objective) and 2 15.0 cmf = (eyepiece). 
EXECUTE: 1
2
95.0 cm 6.33
15.0 cm
fM
f
= − = − = − 
(b) IDENTIFY and SET UP: Use Eq.(34.17) to calculate .y′ 
SET UP: 33.00 10 ms = × 
1 95.0 cms f′ = = (since s is very large, s f′ ≈ ) 
EXECUTE: 43
0.950 m 3.167 10
3.00 10 m
sm
s
−′= − = − = − ×
×
 
4(3.167 10 )(60.0 m) 0.0190 m 1.90 cmy m y −′ = = × = = 
(c) IDENTIFY: Use Eq.(34.21) and the angular magnification M obtained in part (a) to calculate .θ′ The angular 
size θ of the image formed by the objective (object for the eyepiece) is its height divided by its distance from the 
objective. 
EXECUTE: The angular size of the object for the eyepiece is 0.0190 m 0.0200 rad.
0.950 m
θ = = 
(Note that this is also the angular size of the object for the objective: 3
60.0 m 0.0200 rad.
3.00 10 m
θ = =
×
 For a thin lens 
the object and image have the same angular size and the image of the objective is the object for the eyepiece.) 
M θ
θ
′
= (Eq.34.21) so the angular size of the image is (6.33)(0.0200 rad) 0.127 radMθ θ′ = = − = − (The minus 
sign shows that the final image is inverted.) 
EVALUATE: The lateral magnification of the objective is small; the image it forms is much smaller than the 
object. But the total angular magnification is larger than 1.00; the angular size of the final image viewed by the eye 
is 6.33 times larger than the angular size of the original object, as viewed by the unaided eye. 
34.58. IDENTIFY: The angle subtended by Saturn with the naked eye is the same as the angle subtended by the image of 
Saturn formed by the objective lens (see Fig. 34.53 in the textbook). 
SET UP: The angle subtended by Saturn is 
1
diameter of Saturn
distance to Saturn
y
f
θ
′
= = . 
EXECUTE: Putting in the numbers gives 5
1
1.7 mm 0.0017 m 9.4 10 rad 0.0054
18 m 18 m
y
f
θ −
′
= = = = × = ° 
EVALUATE: The angle subtended by the final image, formed by the eyepiece, would be much larger than 0.0054°. 
34.59. IDENTIFY: / 2f R= and 1
2
fM
f
= − . 
SET UP: For object and image both at infinity, 1 2f f+ equals the distance d between the two mirrors. 
2 1.10 cmf = . 1 1.30 mR = . 
EXECUTE: (a) 11 1 20.650 m 0.661 m.2
Rf d f f= = ⇒ = + = 
(b) 1
2
0.650 m 59.1.
0.011 m
fM
f
= = = 
EVALUATE: For a telescope, 1 2f f! . 
34-16 Chapter 34 
34.60. IDENTIFY: The primary mirror forms an image which then acts as the object for the secondary mirror. 
SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 
1 1 1
s s f
+ =
′
. 
EXECUTE: For the first image (formed by the primary mirror): 
1 1 1 1 1 1 2.5 m
2.5 m
s
s s f s
′+ = ⇒ = − ⇒ =
′ ′ ∞
. 
For the second image (formed by the secondary mirror), the distance between the two vertices is x. Assuming that 
the image formed by the primary mirror is to the right of the secondary mirror, the object distance is s = x � 2.5 m 
and the image distance is s′ = x + 0.15 m. Therefore we have 
1 1 1 1 1 1
2.5 m 0.15 m 1.5 ms s f x x
+ = ⇒ + =
′ − + −
 
The positive root of the quadratic equation gives x = 1.7 m, which is the distance between the vertices. 
EVALUATE: Some light is blocked by the secondary mirror, but usually not enough to make much difference. 
34.61. IDENTIFY and SET UP: For a plane mirror . dss s v
dt
′ = − = and ,dsv
dt
′
′ = so .v v′ = − 
EXECUTE: The velocities of the object and image relative to the mirror are equal in magnitude and opposite in 
direction. Thus both you and your image are receding from the mirror surface at 2.40 m/s, in opposite directions. 
Your image is therefore moving at 4.80 m/s relative to you. 
EVALUATE: The result derives from the fact that for a plane mirror the image is the same distance behind the 
mirror as the object is in front of the mirror. 
34.62. IDENTIFY: Apply the law of reflection. 
SET UP: The image of one mirror can serve as the object for the other mirror. 
EXECUTE: (a) There are three images formed, as shown in Figure 34.62a. 
(b) The paths of rays for each image are sketched in Figure 34.62b. 
EVALUATE: Our results agree with Figure 34.9 in the textbook. 
 
Figure 34.62 
34.63. IDENTIFY: Apply the law of reflection for rays from the feet to the eyes and from the top of the head to the eyes. 
SET UP: In Figure 34.63, ray 1 travels from the feet of the woman to her eyes and ray 2 travels from the top of 
her head to her eyes. The total height of the woman is h. 
EXECUTE: The two angles labeled 1θ are equal because of the law of reflection, as are the two angles labeled 
2θ . Since these angles are equal, thetwo distances labeled 1y are equal and the two distances labeled 2y are equal. 
The height of the woman is w 1 22 2h y y= + . As the drawing shows, the height of the mirror is m 1 2h y y= + . 
Comparing, we find that m w / 2h h= . The minimum height required is half the height of the woman. 
Geometric Optics 34-17 
EVALUATE: The height of the image is the same as the height of the woman, so the height of the image is twice 
the height of the mirror. 
 
Figure 34.63 
34.64. IDENTIFY: Apply 1 1 2
s s R
+ =
′
and sm
s
′
= − . 
SET UP: Since the image is projected onto the wall it is real and 0s′ > . sm
s
′
= − so m is negative and 
2.25m = − . The object, mirror and wall are sketched in Figure 34.64. The sketch shows that 400 cms s′ − = . 
EXECUTE: 2.25 sm
s
′
= − = − and 2.25s s′ = . 2.25 400 cms s s s′ − = − = and 320 cms = . 
400 cms′ = + 320 cm 720 cm= . The mirror should be 7.20 m from the wall. 1 1 2
s s R
+ =
′
. 1 1 2
320 cm 720 cm R
+ = . 
4.43 m.R = 
EVALUATE: The focal length of the mirror is / 2 222 cmf R= = . s f> , as it must if the image is to be real. 
 
Figure 34.64 
34.65. IDENTIFY: We are given the image distance, the image height, and the object height. Use Eq.(34.7) to calculate 
the object distance s. Then use Eq.(34.4) to calculate R. 
(a) SET UP: Image is to be formed on screen so is real image; 0.s′ > Mirror to screen distance is 8.00 m, so 
800 cm.s′ = + 0sm
s
′
= − < since both s and s′ are positive. 
34-18 Chapter 34 
EXECUTE: 36.0 m 60.0
0.600 cm
y
m
y
′
= = = and 60.0.m = − Then sm
s
′
= − gives 800 cm 13.3 cm.
60.0
ss
m
′
= − = − = +
−
 
(b) 1 1 2 ,
s s R
+ =
′
 so 2 s s
R ss
′+
=
′
 
(13.3 cm)(800 cm)2 2 26.2 cm
800 cm 13.3 cm
ssR
s s
′⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟′+ +⎝ ⎠ ⎝ ⎠
 
EVALUATE: R is calculated to be positive, which is correct for a concave mirror. Also, in part (a) s is calculated 
to be positive, as it should be for a real object. 
34.66. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 to calculate s′ and then use s ym
s y
′ ′
= − = to find the height of the image. 
SET UP: For a convex mirror, 0R < , so 18.0 cmR = − and 9.00 cm
2
Rf = = − . 
EXECUTE: (a) 1 1 1
s s f
+ =
′
. (1300 cm)( 9.00 cm) 8.94 cm
1300 cm ( 9.00 cm)
sfs
s f
−′ = = = −
− − −
. 38.94 cm 6.88 10
1300 cm
sm
s
−′ −= − = − = × . 
3(6.88 10 )(1.5 m) 0.0103 m 1.03 cmy m y −′ = = × = = . 
(b) The height of the image is much less than the height of the car, so the car appears to be farther away than its 
actual distance. 
EVALUATE: The image formed by a convex mirror is always virtual and smaller than the object. 
34.67. IDENTIFY: Apply 1 1 2
s s R
+ =
′
 and sm
s
′
= − . 
SET UP: 19.4 cmR = + . 
EXECUTE: (a) 1 1 2 1 1 2 46 cm,
8.0 cm 19.4 cm
s
s s R s
′+ = ⇒ + = ⇒ = −
′ ′
 so the image is virtual. 
(b) 46 5.8,
8.0
sm
s
′ −
= − = − = so the image is erect, and its height is (5.8) (5.8)(5.0 mm) 29 mm.y y′ = = = 
EVALUATE: (c) When the filament is 8 cm from the mirror, the image is virtual and cannot be projected onto a wall. 
34.68. IDENTIFY: Combine 1 1 2
s s R
+ =
′
 and sm
s
′
= − . 
SET UP: 2.50m = + . 0R > . 
EXECUTE: 2.50sm
s
′
= − = + . 2.50s s′ = − . 1 1 2
2.50s s R
+ =
−
. 0.600 2
s R
= and 0.300s R= . 
2.50s s′ = − = ( 2.50)(0.300 ) 0.750R R− = − . The object is a distance of 0.300R in front of the mirror and the image 
is a distance of 0.750R behind the mirror. 
EVALUATE: For a single mirror an erect image is always virtual. 
34.69. IDENTIFY and SET UP: Apply Eqs.(34.6) and (34.7). For a virtual object 0.s < The image is real if 0.s′ > 
EXECUTE: (a) Convex implies 0; 24.0 cm; / 2 12.0 cmR R f R< = − = = − 
1 1 1 ,
s s f
+ =
′
 so 1 1 1 s f
s f s sf
−
= − =
′
 
( 12.0 cm)
12.0 cm
sf ss
s f s
−′ = =
− +
 
s is negative, so write as 
(12.0 cm)
; .
12.0 cm
s
s s s
s
′= − = +
−
 Thus 0s′ > (real image) for 12.0 cm.s < Since s is negative 
this means 12.0 cm 0.s− < < A real image is formed if the virtual object is closer to the mirror than the focus. 
(b) ;sm
s
′
= − real image implies 0;s′ > virtual object implies 0.s < Thus 0m > and the image is erect. 
Geometric Optics 34-19 
(c) The principal-ray diagram is given in Figure 34.69. 
 
Figure 34.69 
EVALUATE: For a real object, only virtual images are formed by a convex mirror. The virtual object considered 
in this problem must have been produced by some other optical element, by another lens or mirror in addition to 
the convex one we considered. 
34.70. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
, with R →∞ since the surfaces are flat. 
SET UP: The image formed by the first interface serves as the object for the second interface. 
EXECUTE: For the water-benzene interface to get the apparent water depth: 
1.33 1.500 0 7.33 cm.
6.50 cm
a bn n s
s s s
′+ = ⇒ + = ⇒ = −
′ ′
 
For the benzene-air interface, to get the total apparent distance to the bottom: 
1.50 10 0 6.62 cm.
(7.33 cm 2.60 cm)
a bn n s
s s s
′+ = ⇒ + = ⇒ = −
′ ′+
 
EVALUATE: At the water-benzene interface the light refracts into material of greater refractive index and the 
overall effect is that the apparent depth is greater than the actual depth. 
34.71. IDENTIFY: The focal length is given by 
1 2
1 1 1( 1)n
f R R
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
. 
SET UP: 1 4.0 cmR = ± or 8.0 cm± . 2 8.0 cmR = ± or 4.0 cm± . The signs are determined by the location of the 
center of curvature for each surface. 
EXECUTE: 1 1 1(0.60)
4.00 cm 8.00 cmf
⎛ ⎞
= −⎜ ⎟± ±⎝ ⎠
, so 4.44 cm, 13.3 cm.f = ± ± The possible lens shapes are 
sketched in Figure 34.71. 
1 2 3 4 5 6
7 8
13.3 cm; 4.44 cm; 4.44 cm; 13.3 cm; 13.3 cm; 13.3 cm;
4.44 cm; 4.44 cm.
f f f f f f
f f
= + = + = = − = − = +
= − = −
 
EVALUATE: f is the same whether the light travels through the lens from right to left or left to right, so for the 
pairs (1,6), (4,5) and (7,8) the focal lengths are the same. 
 
Figure 34.71 
34.72. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 and the concept of principal rays. 
SET UP: 10.0 cms = . If extended backwards the ray comes from a point on the optic axis 18.0 cm from the lens 
and the ray is parallel to the optic axis after it passes through the lens. 
EXECUTE: (a) The ray is bent toward the optic axis by the lens so the lens is converging. 
(b) The ray is parallel to the optic axis after it passes through the lens so it comes from the focal point; 
18.0 cmf = . 
34-20 Chapter 34 
(c) The principal ray diagram is drawn in Figure 34.72. The diagram shows that the image is 22.5 cm to the left of 
the lens. 
(d) 1 1 1
s s f
+ =
′
 gives (10.0 cm)(18.0 cm) 22.5 cm
10.0 cm 18.0 cm
sfs
s f
′ = = = −
− −
. The calculated image position agrees with the 
principal ray diagram. 
EVALUATE: The image is virtual. A converging lens produces a virtual image when the object is inside the focal point. 
 
Figure 34.72 
34.73. IDENTIFY: Since the truck is moving toward the mirror, its image will also be moving toward the mirror. 
SET UP: The equation relating the object and image distances to the focal length of a spherical mirror is 
1 1 1
s s f
+ =
′
, where f = R/2. 
EXECUTE: Since the mirror is convex, f = R/2 = (�1.50 m)/2 = �0.75 m. Applying the equation for a spherical 
mirror gives 1 1 1 fss
s s f s f
′+ = ⇒ =
′ −
. 
Using the chain rule from calculus and the fact that v = ds/dt, we have 
2
2( )
ds ds ds fv v
dt ds dt s f
′ ′
′ = = =
−
 
Solving for v gives 
2 22.0 m ( 0.75 m)(1.5 m/s) 20.2 m/s
0.75 m
s fv v
f
⎛ ⎞− − −⎡ ⎤′= = =⎜ ⎟ ⎢ ⎥−⎣ ⎦⎝ ⎠
. 
This is the velocity of the truck relative to the mirror, so the truck is approaching the mirror at 20.2 m/s. You are 
traveling at 25 m/s, so the truck must be traveling at 25 m/s + 20.2 m/s = 45 m/s relative to the highway. 
EVALUATE: Even though the truck and car are moving at constant speed, the image of the truck is not moving at 
constant speed because its location depends on the distance from the mirror to the truck. 
34.74. IDENTIFY: In this context, the microscope just looks at an image or object.Apply 0a bn n
s s
+ =
′
to the image 
formed by refraction at the top surface of the second plate. In this calculation the object is the bottom surface of the 
second plate. 
SET UP: The thickness of the second plate is 2.50 mm 0.78 mm+ , and this is s. The image is 2.50 mm below 
the top surface, so 2.50 mms′ = − . 
EXECUTE: 1 2.50 mm 0.780 mm0 0 1.31.
2.50 mm
a bn n n sn
s s s s s
+
+ = ⇒ + = ⇒ = − = − =
′ ′ ′ −
 
EVALUATE: The object and image distances are measured from the front surface of the second plate, and the 
image is virtual. 
34.75. IDENTIFY and SET UP: In part (a) use 1 1 1
s s f
+ =
′
 to evaluate /ds ds′ . Compare to sm
s
′
= − . In part (b) use 
1 1 2
s s R
+ =
′
to find the location of the image of each face of the cube. 
EXECUTE: (a) 1 1 1
s s f
+ =
′
 and taking its derivative with respect to s we have 2 2
1 1 1 1 10 d ds
ds s s f s s ds
′⎛ ⎞
= + − = − −⎜ ⎟′ ′⎝ ⎠
 
and 
2
2
2
ds s m
ds s
′ ′
= − = − . But ds m
ds
′
′= , so 2.m m′ = − Images are always inverted longitudinally. 
(b) (i) Front face: 1 1 2 1 1 2 120.00 cm.
200.000 cm 150.000 cm
s
s s R s
′+ = ⇒ + = ⇒ =
′ ′
 
Geometric Optics 34-21 
Rear face: 1 1 2 1 1 2 119.96 cm.
200.100 cm 150.000 cm
s
s s R s
′+ = ⇒ + = ⇒ =
′ ′
 
(ii) 120.000 0.600
200.000
sm
s
′
= − = − = − . ( )22 0.600 0.360.m m′ = − = − − = − 
(iii) The two faces perpendicular to the axis (the front and rear faces): squares with side length 0.600 mm. The 
four faces parallel to the axis (the side faces): rectangles with sides of length 0.360 mm parallel to the axis and 
0.600 mm perpendicular to the axis. 
EVALUATE: Since the lateral and longitudinal magnifications have different values the image of the cube is not a cube. 
34.76. IDENTIFY: /m ds ds′ ′= and a
b
n sm
n s
′
= − . 
SET UP: Use a b b an n n n
s s R
−
+ =
′
 to evaluate /ds ds′ . 
EXECUTE: a b b an n n n
s s R
−
+ =
′
 and taking its derivative with respect to s we have 
2 20
a b b a a bd n n n n n n ds
ds s s R s s ds
′−⎛ ⎞= + − = − −⎜ ⎟′ ′⎝ ⎠
 and 
2 2 2
2
2 2 2 .
a a b b
b b a a
ds s n s n n nm
ds s n s n n n
⎛ ⎞′ ′ ′
= − = − = −⎜ ⎟
⎝ ⎠
 
But ds m
ds
′
′= , so 2 b
a
nm m
n
′ = − . 
EVALUATE: m′ is always negative. This means that images are always inverted longitudinally. 
34.77. IDENTIFY and SET UP: Rays that pass through the hole are undeflected. All other rays are blocked. sm
s
′
= − . 
EXECUTE: (a) The ray diagram is drawn in Figure 34.77. The ray shown is the only ray from the top of the object 
that reaches the film, so this ray passes through the top of the image. An inverted image is formed on the far side 
of the box, no matter how far this side is from the pinhole and no matter how far the object is from the pinhole. 
(b) 1.5 ms = . 20.0 cms′ = . 20.0 cm 0.133
150 cm
sm
s
′
= − = − = − . ( 0.133)(18 cm) 2.4 cmy my′ = = − = − . The image is 
2.4 cm tall. 
EVALUATE: A defect of this camera is that not much light energy passes through the small hole each second, so 
long exposure times are required. 
 
Figure 34.77 
34.78. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
 and a
b
n sm
n s
′
= − to each refraction. The overall magnification is 1 2m m m= . 
SET UP: For the first refraction, 6.0 cmR = + , 1.00an = and 1.60bn = . For the second refraction, 12.0 cmR = − , 
1.60an = and 1.00bn = . 
EXECUTE: (a) The image from the left end acts as the object for the right end of the rod. 
(b) 1 1.60 0.60 28.3 cm.
23.0 cm 6.0 cm
a b b an n n n s
s s R s
− ′+ = ⇒ + = ⇒ =
′ ′
 
So the second object distance is 2 40.0 cm 28.3 cm 11.7 cm.s = − = ( )( )1
28.3 0.769.
1.60 23.0
a
b
n sm
n s
′
= − = − = − 
(c) The object is real and inverted. 
(d) 
2 2 2
1.60 1 0.60 11.5 cm.
11.7 cm 12.0 cm
a b b an n n n s
s s R s
− − ′+ = ⇒ + = ⇒ = −
′ ′ −
 
( )( ) ( )( )2 1 2
1.60 11.5
1.57 0.769 1.57 1.21.
11.7
a
b
n sm m m m
n s
−′
= − = − = ⇒ = = − = − 
(e) The final image is virtual, and inverted. 
34-22 Chapter 34 
(f) ( )( )1.50 mm 1.21 1.82 mm.y′ = − = − 
EVALUATE: The first image is to the left of the second surface, so it serves as a real object for the second surface, 
with positive object distance. 
34.79. IDENTIFY: Apply Eqs.(34.11) and (34.12) to the refraction as the light enters the rod and as it leaves the rod. The 
image formed by the first surface serves as the object for the second surface. The total magnification is 
tot 1 2 ,m m m= where 1m and 2m are the magnifications for each surface. 
SET UP: The object and rod are shown in Figure 34.79. 
 
Figure 34.79 
(a) image formed by refraction at first surface (left end of rod): 
23.0 cm; 1.00; 1.60; 6.00 cma bs n n R= + = = = + 
a b b an n n n
s s R
−
+ =
′
 
EXECUTE: 1 1.60 1.60 1.00
23.0 cm 6.00 cms
−
+ =
′
 
1.60 1 1 23 10 13
10.0 cm 23.0 cm 230 cm 230 cms
−
= − = =
′
 
230 cm1.60 28.3 cm;
13
s ⎛ ⎞′ = = +⎜ ⎟
⎝ ⎠
 image is 28.3 cm to right of first vertex. 
This image serves as the object for the refraction at the second surface (right-hand end of rod). It is 
28.3 cm 25.0 cm 3.3 cm− = to the right of the second vertex. For the second surface 3.3 cms = − (virtual object). 
(b) EVALUATE: Object is on side of outgoing light, so is a virtual object. 
(c) SET UP: Image formed by refraction at second surface (right end of rod): 
3.3 cm; 1.60; 1.00; 12.0 cma bs n n R= − = = = − 
a b b an n n n
s s R
−
+ =
′
 
EXECUTE: 1.60 1.00 1.00 1.60
3.3 cm 12.0 cms
−
+ =
′− −
 
1.9 cm; 0s s′ ′= + > so image is 1.9 cm to right of vertex at right-hand end of rod. 
(d) 0s′ > so final image is real. 
Magnification for first surface: 
(1.60)( 28.3 cm) 0.769
(1.00)( 23.0 cm)
a
b
n sm
n s
′ +
= − = − = −
+
 
Magnification for second surface: 
(1.60)( 1.9 cm) 0.92
(1.00)( 3.3 cm)
a
b
n sm
n s
′ +
= − = − = +
−
 
The overall magnification is tot 1 2 ( 0.769)( 0.92) 0.71m m m= = − + = − tot 0m < so final image is inverted with respect 
to the original object. 
(e) tot ( 0.71)(1.50 mm) 1.06 mmy m y′ = = − = − 
The final image has a height of 1.06 mm. 
EVALUATE: The two refracting surfaces are not close together and Eq.(34.18) does not apply. 
34.80. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 and y sm
y s
′ ′
= = − . The type of lens determines the sign of f. The sign of 
s′ determines whether the image is real or virtual. 
SET UP: 8.00 cms = + . 3.00 cms′ = − . s′ is negative because the image is on the same side of the lens as the 
object. 
EXECUTE: (a) 1 s s
f ss
′+
=
′
 and (8.00 cm)( 3.00 cm) 4.80 cm
8.00 cm 3.00 cm
ssf
s s
′ −
= = = −
′+ −
. f is negative so the lens is 
diverging. 
Geometric Optics 34-23 
(b) 3.00 cm 0.375
8.00 cm
sm
s
′ −
= − = − = + . (0.375)(6.50 mm) 2.44 mmy my′ = = = . 0s′ < and the image is virtual. 
EVALUATE: A converging lens can also form a virtual image, if the object distance is less than the focal length. 
But in that case s s′ > and the image would be farther from the lens than the object is. 
34.81. IDENTIFY: 1 1 1
s s f
+ =
′
. The type of lens determines the sign of f. y sm
y s
′ ′
= = − . The sign of s′ depends on 
whether the image is real or virtual. 16.0 cms = . 
SET UP: 22.0 cms′ = − ; s′ is negative because the image is on the same side of the lens as the object. 
EXECUTE: (a) 1 s s
f ss
′+
=
′
 and (16.0 cm)( 22.0 cm) 58.7 cm
16.0 cm 22.0 cm
ssf
s s
′ −
= = = +
′+ −
. f is positive so the lens is converging. 
(b) 22.0 cm 1.38
16.0 cm
sm
s
′ −
= − = − = . (1.38)(3.25 mm) 4.48 mmy my′ = = = . 0s′ < and the image is virtual. 
EVALUATE: A converging lens forms a virtual image when the object is closer to the lens than the focal point. 
34.82. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
. Use the image distance when viewed from the flat end to determine the 
refractive index n of the rod. 
SET UP: When viewing from the flat end, an n= , 1.00bn = and R →∞ . When viewing from the curved end, 
an n= , 1.00bn = and 10.0 cmR = − . 
EXECUTE: 1 15.00 0 1.58.
15.0 cm 9.50 cm 9.50
a bn n n n
s s
+ = ⇒ + = ⇒ = =′ −
 When viewed from the curved end of the 
rod 1 1 1.58 1 0.58
15.0 cm 10.0 cm
a b b an n n n n n
s s R s s R s
− − −
+ = ⇒ + = ⇒ + =
′ ′ ′ −
 , and 21.1cms′ = − . The image is 21.1 cm 
within the rod from the curved end. 
EVALUATE: In each case the image is virtual and on the same side of the surface as the object. 
34.83. (a) IDENTIFY: Apply Snell�s law to the refraction of a ray at each side of the beam to find where these rays strike 
the table. 
SET UP: The path of a ray is sketched in Figure 34.83. 
 
Figure 34.83 
The width of the incident beam is exaggerated in the sketch, to make it easier to draw. Since the diameter of the 
beam is much less than the radius of the hemisphere, angles aθ and bθ are small. The diameter of the circle of 
light formed on the table is 2 .x Note the two right triangles containing the angles aθ and .bθ 
0.190 cmr = is the radius of the incident beam. 
12.0 cmR = is the radius of the glass hemisphere. 
EXECUTE: aθ and bθ small imply ; sin , sina b
r x xx x
R R R
θ θ
′
′≈ = = ≈ 
Snell�s law: sin sina a b bn nθ θ= 
Using the above expressions for sin aθ and sin bθ gives a b
r xn n
R R
= 
a bn r n x= so 
1.00(0.190 cm) 0.1267 cm
1.50
a
b
n rx
n
= = = 
The diameter of the circle on the table is 2 2(0.1267 cm) 0.253 cm.x = = 
(b) EVALUATE: R divides out of the expression; the result for the diameter of the spot is independent of the radius R 
of the hemisphere. It depends only on the diameter of the incident beam and the index of refraction of the glass. 
34-24 Chapter 34 
34.84. IDENTIFY and SET UP: Treating each of the goblet surfaces as spherical surfaces, we have to pass, from left to 
right, through four interfaces. Apply a b b an n n n
s s R
−
+ =
′
to each surface. The image formed by one surface serves as 
the object for the next surface. 
EXECUTE: (a) For the empty goblet: 
1
1
1 1.50 0.50 12 cm
4.00 cm
a b b an n n n s
s s R s
− ′+ = ⇒ + = ⇒ =
′ ′∞
. 
2 2
2
1.50 1 0.500.60 cm 12 cm 11.4 cm 64.6 cm.
11.4 cm 3.40 cm
s s
s
− ′= − = − ⇒ + = ⇒ = −
′−
3 3
3
1 1.50 0.5064.6 cm 6.80 cm 71.4 cm 9.31cm.
71.4 cm 3.40 cm
s s
s
′= + = ⇒ + = ⇒ = −
′ −
 
4 4
4
1.50 1 0.509.31cm 0.60 cm 9.91cm 37.9 cm.
9.91cm 4.00 cm
s s
s
− ′= + = ⇒ + = ⇒ = −
′ −
 The final image is 
37.9 cm 2(4.0 cm) 29.9 cm− = to the left of the goblet. 
(b) For the wine-filled goblet: 
1
1
1 1.50 0.50 12 cm
4.00 cm
a b b an n n n s
s s R s
− ′+ = ⇒ + = ⇒ =
′ ′∞
. 
2 2
2
1.50 1.37 0.130.60 cm 12 cm 11.4 cm 14.7 cm.
11.4 cm 3.40 cm
s s
s
− ′= − = − ⇒ + = ⇒ =
′−
 
3 3
3
1.37 1.50 0.136.80 cm 14.7 cm 7.9 cm 11.1cm.
7.9 cm 3.40 cm
s s
s
′= − = − ⇒ + = ⇒ =
′− −
 
4 4
4
1.50 1 0.500.60 cm 11.1cm 10.5 cm 3.73 cm
10.5 cm 4.00 cm
s s
s
− ′= − = − ⇒ + = ⇒ =
′− −
. The final image is 3.73 cm to the 
right of the goblet. 
EVALUATE: If the object for a surface is on the outgoing side of the light, then the object is virtual and the object 
distance is negative. 
34.85. IDENTIFY: The image formed by refraction at the surface of the eye is located by a b b an n n n
s s R
−
+ =
′
. 
SET UP: 1.00an = , 1.35bn = . 0R > . For a distant object, s ≈ ∞ and 
1 0
s
≈ . 
EXECUTE: (a) s ≈ ∞ and 2.5 cms′ = : 1.35 1.35 1.00
2.5 cm R
−
= and 0.648 cm 6.48 mmR = = . 
(b) 0.648 cmR = and 25 cms = : 1.00 1.35 1.35 1.00
25 cm 0.648s
−
+ =
′
. 1.35 0.500
s
=
′
 and 2.70 cm 27.0 mms′ = = . The 
image is formed behind the retina. 
(c) Calculate s′ for s ≈ ∞ and 0.50 cmR = : 1.35 1.35 1.00
0.50 cms
−
=
′
. 1.93 cm 19.3 mms′ = = . The image is formed in 
front of the retina. 
EVALUATE: The cornea alone cannot achieve focus of both close and distant objects. 
34.86. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
 and a
b
n sm
n s
′
= − to each surface. The overall magnification is 1 2m m m= . The 
image formed by the first surface is the object for the second surface. 
SET UP: For the first surface, 1.00an = , 1.60bn = and 15.0 cmR = + . For the second surface, 1.60an = , 
1.00bn = and R =→∞ . 
EXECUTE: (a) 1 1.60 0.60 36.9 cm.
12.0 cm 15.0 cm
a b b an n n n s
s s R s
− ′+ = ⇒ + = ⇒ = −
′ ′
 The object distance for the far end 
of the rod is 50.0 cm ( 36.9 cm) 86.9 cm.− − = The final image is 4.3 cm to the left of the vertex of the 
hemispherical surface. 1.60 1 0 54.3 cm.
86.9 cm
a b b an n n n s
s s R s
− ′+ = ⇒ + = ⇒ = −
′ ′
 
(b) The magnification is the product of the two magnifications: 
1 2 1 2
36.9 1.92, 1.00 1.92.
(1.60)(12.0)
a
b
n sm m m m m
n s
′ −
= − = − = = ⇒ = = 
EVALUATE: The final image is virtual, erect and larger than the object. 
Geometric Optics 34-25 
34.87. IDENTIFY: Apply Eq.(34.11) to the image formed by refraction at the front surface of the sphere. 
SET UP: Let gn be the index of refraction of the glass. The image formation is shown in Figure 34.87. 
 
s = ∞ 
2 ,s r′ = + where r is the 
radius of the sphere 
1.00, , a b gn n n R r= = = + 
 
Figure 34.87 
a b b an n n n
s s R
−
+ =
′
 
EXECUTE: 
1.001
2
g gn n
r r
−
+ =
∞
 
1 1; 
2 2
g g gn n n
r r r r r
= − = and 2.00gn = 
EVALUATE: The required refractive index of the glass does not depend on the radius of the sphere. 
34.88. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
to each surface. The image of the first surface is the object for the second 
surface. The relation between 1s′ and 2s involves the length d of the rod. 
SET UP: For the first surface, 1.00an = , 1.55bn = and 6.00 cmR = + . For the second surface, 1.55an = , 
1.00bn = and 6.00 cmR = − . 
EXECUTE: We have images formed from both ends. From the first surface: 
1 1.55 0.55 30.0 cm.
25.0 cm 6.00 cm
a b b an n n n s
s s R s
− ′+ = ⇒ + = ⇒ =
′ ′
 
This image becomes the object for the second end: 1.55 1 0.55
30.0 cm 65.0 cm 6.00 cm
a b b an n n n
s s R d
− −
+ = ⇒ + =
′ − −
. 
30.0 cm 20.3 cm 50.3 cm.d d− = ⇒ = 
EVALUATE: The final image is real. The first image is 20.3 cm to the right of the second surface and serves as a 
real object. 
34.89. IDENTIFY: The first lens forms an image which then acts as the object for the second lens. 
SET UP: The thin-lens equation is 1 1 1
s s f
+ =
′
 and the magnification is sm
s
′
= − . 
EXECUTE: (a) For the first lens: 1 1 1 1 1 1 3.75 cm
5.00 cm 15.0 cm
s
s s f s
′+ = ⇒ + = ⇒ = −
′ ′ −
, to the left of the lens 
(virtual image). 
(b) For the second lens, s = 12.0 cm + 3.75 cm = 15.75 cm. 
1 1 1 1 1 1 315 cm
15.75 cm 15.0 cm
s
s s f s
′+ = ⇒ + = ⇒ =
′ ′
, or 332 cm from the object. 
(c) The final image is real. 
(d) 1 2 total, 0.750, 20.0, 15.0 6.00 cm,
sm m m m y
s
′
′= − = = − = − ⇒ = − inverted. 
EVALUATE: Note that the total magnification is the product of the individual magnifications. 
34.90. IDENTIFY and SET UP: Use 
1 2
1 1 1( 1)n
f R R
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
 to calculate the focal length of the lenses. The image formed 
by the first lens serves at the object for the second lens. tot 1 2m m m= . 
1 1 1
s s f
+ =
′
 gives sfs
s f
′ =
−
. 
EXECUTE: (a) 1 1 1(0.60)
12.0 cm 28.0 cmf
⎛ ⎞= −⎜ ⎟
⎝ ⎠
 and 35.0 cmf = + . 
Lens 1: 1 35.0 cmf = + . 1 45.0 cms = + . 1 11
1 1
(45.0 cm)(35.0 cm) 158 cm
45.0 cm 35.0 cm
s fs
s f
′ = = = +
− −
. 
1
1
1
158 cm 3.51
45.0 cm
sm
s
′
= − = − = − . 1 1 1 (3.51)(5.00 mm) 17.6 mmy m y′ = = = . The image of the first lens is 158 cm 
to the right of lens 1 and is 17.6 mm tall. 
34-26 Chapter 34 
(b) The image of lens 1 is 315 cm 158 cm 157 cm− = to the left of lens 2. 2 35.0 cmf = + . 2 157 cms = + . 
2 2
2
2 2
(157 cm)(35.0 cm) 45.0 cm
157 cm 35.0 cm
s fs
s f
′ = = = +
− −
. 22
2
45.0 cm 0.287
157 cm
sm
s
′
= − = − = − . 
tot 1 2 ( 3.51)( 0.287) 1.00m m m= = − − = + . The final image is 45.0 cm to the right of lens 2. The final image is 5.00 
mm tall. tot 0m > . So the final image is erect. 
EVALUATE: The final image is real. It is erect because each lens produces an inversion of the image, and two 
inversions return the image to the orientation of the object. 
34.91. IDENTIFYand SET UP: Apply Eq.(34.16) for each lens position. The lens to screen distance in each case is the 
image distance. There are two unknowns, the original object distance x and the focal length f of the lens. But each 
lens position gives an equation, so there are two equations for these two unknowns. The object, lens and screen 
before and after the lens is moved are shown in Figure 34.91. 
 
; 30.0 cms x s′= = 
1 1 1
s s f
+ =
′
 
1 1 1
30.0 cmx f
+ = 
 
Figure 34.91 
4.00 cm; 22.0 cms x s′= + = 
1 1 1
s s f
+ =
′
 gives 1 1 1
4.00 cm 22.0 cmx f
+ =
+
 
EXECUTE: Equate these two expressions for 1/f: 
1 1 1 1
30.0 cm 4.00 cm 22.0 cmx x
+ = +
+
 
1 1 1 1
4.00 cm 22.0 cm 30.0 cmx x
− = −
+
 
4.00 cm 30.0 22.0
( 4.00 cm) 660 cm
x x
x x
+ − −
=
+
 and 4.00 cm 8
( 4.00 cm) 660 cmx x
=
+
 
2 2(4.00 cm) 330 cm 0x x+ − = and 1 ( 4.00 16.0 4(330)) cm
2
x = − ± + 
x must be positive so 1 ( 4.00 36.55) cm 16.28 cm
2
x = − + = 
Then 1 1 1
30.0 cmx f
+ = and 1 1 1
16.28 cm 30.0 cmf
= + 
10.55 cm,f = + which rounds to 10.6 cm. 0;f > the lens is converging. 
EVALUATE: We can check that 16.28 cms = and 10.55 cmf = gives 30.0 cms′ = and that 
(16.28 4.0) cm 20.28 cms = + = and 10.55 cmf = gives 22.0 cm.s′ = 
34.92. IDENTIFY and SET UP: Apply a b b an n n n
s s R
−
+ =
′
. 
EXECUTE: (a) and .a b b a a b b a a b b an n n n n n n n n n n n
s s R f R f R
− − −
+ = ⇒ + = + =
′ ′∞ ∞
 
 and a b a b b an n n n n n
f R f R
− −
= =
′
. Therefore, a bn n
f f
=
′
 and /a b
fn n
f
=
′
. 
(b) (1 )a b b a b b bn n n n n f n n f f
s s R sf s R
′− −
+ = ⇒ + =
′ ′ ′
. Therefore, (1 ) 1f f f f f f f
s s R R
′ ′ ′ ′− −
+ = = =
′
. 
EVALUATE: For a thin lens the first and second focal lengths are equal. 
34.93. (a) IDENTIFY: Use Eq.(34.6) to locate the image formed by each mirror. The image formed by the first mirror 
serves as the object for the 2nd mirror. 
Geometric Optics 34-27 
SET UP: The positions of the object and the two mirrors are shown in Figure 34.93a. 
 
0.360 mR = 
/ 2 0.180 mf R= = 
 
Figure 34.93a 
EXECUTE: Image formed by convex mirror (mirror #1): 
convex means 1 10.180 m; f s L x= − = − 
1 1
1
1 1
( )( 0.180 m) 0.600 m(0.180 m) 0
0.180 m 0.780 m
s f L x xs
s f L x x
− − −⎛ ⎞′ = = = − <⎜ ⎟− − + −⎝ ⎠
 
The image is 0.600 m(0.180 m)
0.780 m
x
x
−⎛ ⎞
⎜ ⎟−⎝ ⎠
 to the left of mirror #1 so is 
20.600 m 0.576 m (0.780 m)0.600 m (0.180 m)
0.780 m 0.780 m
x x
x x
− −⎛ ⎞+ =⎜ ⎟− −⎝ ⎠
 to the left of mirror #2. 
Image formed by concave mirror (mirror #2); 
concave implies 2 0.180 mf = + 
2
2
0.576 m (0.780 m)
0.780 m
xs
x
−
=
−
 
Rays return to the source implies 2 .s x′ = Using these expressions in 2 22
2 2
s fs
s f
′
=
′ −
 gives 
20.576 m (0.780 m) (0.180 m)
0.780 m 0.180 m
x x
x x
−
=
− −
 
2 20.600 (0.576 m) 0.10368 m 0x x− + = 
21 1(0.576 (0.576) 4(0.600)(0.10368) ) m (0.576 0.288) m
1.20 1.20
x = ± − = ± 
0.72 mx = (impossible; can�t have 0.600 m)x L> = or 0.24 m.x = 
(b) SET UP: Which mirror is #1 and which is #2 is now reversed form part (a). This is shown in Figure 34.93b. 
 
Figure 34.93b 
EXECUTE: Image formed by concave mirror (mirror #1): 
concave means 1 10.180 m; f s x= + = 
1 1
1
1 1
(0.180 m)
0.180 m
s f xs
s f x
′ = =
− −
 
The image is (0.180 m)
0.180 m
x
x −
 to the left of mirror #1, so 
2
2
(0.180 m) (0.420 m) 0.180 m0.600 m
0.180 m 0.180 m
x xs
x x
−
= − =
− −
 
Image formed by convex mirror (mirror #2): 
convex means 2 0.180 mf = − 
rays return to the source means 2 0.600 ms L x x′ = − = − 
1 1 1
s s f
+ =
′
 gives 
2
0.180 m 1 1
(0.420 m) 0.180 m 0.600 m 0.180 m
x
x x
−
+ = −
− −
 
2 2
0.180 m 0.780 m
(0.420 m) 0.180 m 0.180 m (0.180 m)
x x
x x
⎛ ⎞− −
= −⎜ ⎟− −⎝ ⎠
 
2 20.600 (0.576 m) 0.1036 m 0x x− + = 
This is the same quadratic equation as obtained in part (a), so again 0.24 m.x = 
34-28 Chapter 34 
EVALUATE: For 0.24 mx = the image is at the location of the source, both for rays that initially travel from the 
source toward the left and for rays that travel from the source toward the right. 
34.94. IDENTIFY: 1 1 1
s s f
+ =
′
 gives sfs
s f
′ =
−
, for both the mirror and the lens. 
SET UP: For the second image, the image formed by the mirror serves as the object for the lens. For the mirror, 
m 10.0 cmf = + . For the lens, 32.0 cmf = . The center of curvature of the mirror is m2 20.0 cmR f= = to the right 
of the mirror vertex. 
EXECUTE: (a) The principal-ray diagrams from the two images are sketched in Figures 34.94a-b. In Figure 
34.94b, only the image formed by the mirror is shown. This image is at the location of the candle so the principal 
ray diagram that shows the image formation when the image of the mirror serves as the object for the lens is 
analogous to that in Figure 34.94a and is not drawn. 
(b) Image formed by the light that passes directly through the lens: The candle is 85.0 cm to the left of the lens. 
(85.0 cm)(32.0 cm) 51.3 cm
85.0 cm 32.0 cm
sfs
s f
′ = = = +
− −
. 51.3 cm 0.604
85.0 cm
sm
s
′
= − = − = − . This image is 51.3 cm to the right of 
the lens. 0s′ > so the image is real. 0m < so the image is inverted. Image formed by the light that first reflects off 
the mirror: First consider the image formed by the mirror. The candle is 20.0 cm to the right of the mirror, so 
20.0 cms = + . (20.0 cm)(10.0 cm) 20.0 cm
20.0 cm 10.0 cm
sfs
s f
′ = = =
− −
. 11
1
20.0 cm 1.00
20.0 cm
sm
s
′
= − = − = − . The image formed by 
the mirror is at the location of the candle, so 2 85.0 cms = + and 2 51.3 cms ′ = . 2 0.604m = − . tot 1 2m m m= = 
( 1.00)( 0.604) 0.604− − = . The second image is 51.3 cm to the right of the lens. 2 0s ′ > , so the final image is real. 
tot 0m > , so the final image is erect. 
EVALUATE: The two images are at the same place. They are the same size. One is erect and one is inverted. 
 
Figure 34.94 
34.95. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
 to each case. 
SET UP: 20.0 cms = . 0R > . Use 9.12 cms′ = + to find R. For this calculation, 1.00an = and 1.55bn = . Then 
repeat the calculation with 1.33an = . 
EXECUTE: a b b an n n n
s s R
−
+ =
′
 gives 1.00 1.55 1.55 1.00
20.0 cm 9.12 cm R
−
+ = . 2.50 cmR = . 
Geometric Optics 34-29 
Then 1.33 1.55 1.55 1.33
20.0 cm 2.50 cms
−
+ =
′
gives 72.1 cms′ = − . The image is 72.1 cm to the left of the surface vertex. 
EVALUATE: With the rod in air the image is real and with the rod in water the image is virtual. 
34.96. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 to each lens. The image formed by the first lens serves as the object for the second 
lens. The focal length of the lens combination is defined by 
1 2
1 1 1
s s f
+ =
′
. In part (b) use 
1 2
1 1 1( 1)n
f R R
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
to 
calculate f for the meniscus lens and for the 4CCl , treated as a thin lens. 
SET UP: With two lenses of different focal length in contact, the image distance from the first lens becomes 
exactly minus the object distance for the second lens. 
EXECUTE: (a) 
1 1 1 1 1 1
1 1 1 1 1 1
s s f s f s
+ = ⇒ = −
′ ′
 and 
2 2 1 2 1 1 2 2
1 1 1 1 1 1 1 1 .
s s s s s f s f
⎛ ⎞
+ = + = − + =⎜ ⎟′ ′ ′ ′− ⎝ ⎠
 But overall for the lens 
system, 
1 2 2 1
1 1 1 1 1 1 .
s s f f f f
+ = ⇒ = +
′
 
(b) With carbon tetrachloride sitting in a meniscus lens, we have two lenses in contact. All we need in order to 
calculate the system�s focal length is calculate the individual focal lengths, and then use the formula from part (a). 
For the meniscus lens 1
m 1 2
1 1 1 1 1( ) (0.55) 0.061 cm
4.50 cm 9.00 cmb a
n n
f R R
−⎛ ⎞ ⎛ ⎞= − − = − =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 and m 16.4 cmf = . 
For the 14
w 1 2
1 1 1 1 1CCl : ( ) (0.46) 0.051 cm
9.00 cmb a
n n
f R R
−⎛ ⎞ ⎛ ⎞= − − = − =⎜ ⎟ ⎜ ⎟∞⎝ ⎠⎝ ⎠
 and w 19.6 cmf = . 
1
w m
1 1 1 0.112 cm
f f f
−= + = and 8.93 cmf = . 
EVALUATE: 1 2
1 2
f ff
f f
=
+
, so f for the combination is less than either 1f or 2f . 
34.97. IDENTIFY: Apply Eq.(34.11) with R →∞ to the refractionat each surface. For refraction at the first surface the 
point P serves as a virtual object. The image formed by the first refraction serves as the object for the second 
refraction. 
SET UP: The glass plate and the two points are shown in Figure 37.97. 
 
plane faces means R →∞ and 
0a bn n
s s
+ =
′
 
b
a
ns s
n
′ = − 
 
Figure 34.97 
EXECUTE: refraction at the first (left-hand) surface of the piece of glass: 
The rays converging toward point P constitute a virtual object for this surface, so 
14.4 cm.s = − 
1.00, 1.60.a bn n= = 
1.60 ( 14.4 cm) 23.0 cm
1.00
s′ = − − = + 
This image is 23.0 cm to the right of the first surface so is a distance 23.0 cm t− to the right of the second surface. 
This image serves as a virtual object for the second surface. 
refraction at the second (right-hand) surface of the piece of glass: 
The image is at P′ so 14.4 cm 0.30 cm 14.7 cm .s t t′ = + − = − (23.0 cm ); 1.60; 1.00a bs t n n= − − = = b
a
ns s
n
′ = − 
gives 1.0014.7 cm ( [23.0 cm ]).
1.60
t t⎛ ⎞− = − − −⎜ ⎟
⎝ ⎠
 14.7 cm 14.4 cm 0.625 .t t− = + − 
0.375 0.30 cmt = and 0.80 cmt = 
EVALUATE: The overall effect of the piece of glass is to diverge the rays and move their convergence point to the 
right. For a real object, refraction at a plane surface always produces a virtual image, but with a virtual object the 
image can be real. 
34-30 Chapter 34 
34.98. IDENTIFY: Apply the two equations 
1 1 1 2 2 2
anda b b a b c c bn n n n n n n n
s s R s s R
− −
+ = + =
′ ′
. 
SET UP: liqa cn n n= = , bn n= , and 1 2s s′ = − . 
EXECUTE: (a) liq liq liq liq
1 1 1 1 2 2
and
n n n n n nn n
s s R s s R
− −
+ = + =
′ ′ ′−
. liq
1 2 1 2
1 1 1 1 1 1 1( 1)n n
s s s s f R R
⎛ ⎞
+ = + = = − −⎜ ⎟′ ′ ′ ⎝ ⎠
. 
(b) Comparing the equations for focal length in and out of air we have: 
liq liq
liq
liq liq
( 1)
( 1) ( 1) .
n n n n
f n f n n f f f
n n n
⎛ ⎞ ⎡ ⎤− −
′ ′ ′− = − = ⇒ =⎜ ⎟ ⎢ ⎥⎜ ⎟ −⎢ ⎥⎝ ⎠ ⎣ ⎦
 
EVALUATE: When liq 1n = , f f′ = , as it should. 
34.99. IDENTIFY: Apply 1 1 1
s s f
+ =
′
. 
SET UP: The image formed by the converging lens is 30.0 cm from the converging lens, and becomes a virtual 
object for the diverging lens at a position 15.0 cm to the right of the diverging lens. The final image is projected 
15 cm 19.2 cm 34.2 cm+ = from the diverging lens. 
EXECUTE: 1 1 1 1 1 1 26.7 cm.
15.0 cm 34.2 cm
f
s s f f
+ = ⇒ + = ⇒ = −
′ −
 
EVALUATE: Our calculation yields a negative value of f, which should be the case for a diverging lens. 
34.100. IDENTIFY: The spherical mirror forms an image of the object. It forms another image when the image of the 
plane mirror serves as an object. 
SET UP: For the convex mirror 24.0 cmf = − . The image formed by the plane mirror is 10.0 cm to the right of 
the plane mirror, so is 20.0 cm 10.0 cm 30.0 cm+ = from the vertex of the spherical mirror. 
EXECUTE: The first image formed by the spherical mirror is the one where the light immediately strikes its 
surface, without bouncing from the plane mirror. 
1 1 1 1 1 1 7.06 cm,
10.0 cm 24.0 cm
s
s s f s
′+ = ⇒ + = ⇒ = −
′ ′ −
and the image height 
is 7.06 (0.250 cm) 0.177 cm.
10.0
sy y
s
′ −′ = − = − = 
The second image is of the plane mirror image is located 30.0 cm from the vertex of the spherical mirror. 
1 1 1 1 1 1 13.3 cm
30.0 cm 24.0 cm
s
s s f s
′+ = ⇒ + = ⇒ = −
′ ′ −
 and the image height is 
13.3 (0.250 cm) 0.111cm.
30.0
sy y
s
′ −′ = − = − = 
EVALUATE: Other images are formed by additional reflections from the two mirrors. 
34.101. IDENTIFY: In the sketch in Figure 34.101 the light travels upward from the object. Apply Eq.(34.11) with 
R →∞ to the refraction at each surface. The image formed by the first surface serves as the object for the second 
surface. 
SET UP: The locations of the object and the glass plate are shown in Figure 34.101. 
 
For a plane (flat) surface 
R →∞ so 0a bn n
s s
+ =
′
 
b
a
ns s
n
′ = − 
 
Figure 34.101 
EXECUTE: First refraction (air glass):→ 
1.00; 1.55; 6.00 cma bn n s= = = 
1.55 (6.00 cm) 9.30 cm
1.00
b
a
ns s
n
′ = − = − = − 
The image is 9.30 cm below the lower surface of the glass, so is 9.30 cm 3.50 cm 12.8 cm+ = below the upper 
surface. 
Geometric Optics 34-31 
Second refraction (glass air):→ 
1.55; 1.00; 12.8 cma bn n s= = = + 
1.00 (12.8 cm) 8.26 cm
1.55
b
a
ns s
n
′ = − = − = − 
The image of the page is 8.26 cm below the top surface of the glass plate and therefore 
9.50 cm 8.26 cm 1.24 cm− = above the page. 
EVALUATE: The image is virtual. If you view the object by looking down from above the plate, the image of the 
page that you see is closer to your eye than the page is. 
34.102. IDENTIFY: Light refracts at the front surface of the lens, refracts at the glass-water interface, reflects from the 
plane mirror and passes through the two interfaces again, now traveling in the opposite direction. 
SET UP: Use the focal length in air to find the radius of curvature R of the lens surfaces. 
EXECUTE: (a) 
1 2
1 1 1 1 2( 1) 0.52 41.6 cm.
40 cm
n R
f R R R
⎛ ⎞ ⎛ ⎞= − − ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 
At the air�lens interface: 
1
1 1.52 0.52
70.0 cm 41.6 cm
a b b an n n n
s s R s
−
+ = ⇒ + =
′ ′
 and 1 2851 cm and 851 cm.s s′ = − = 
At the lens�water interface:
2
1.52 1.33 0.187
851cm 41.6 cms
−
⇒ + =
′ −
and 2 491cms′ = . 
The mirror reflects the image back (since there is just 90 cm between the lens and mirror.) So, the position of the 
image is 401 cm to the left of the mirror, or 311 cm to the left of the lens. 
At the water�lens interface:
3
1.33 1.52 0.187
311cm 41.6 cms
⇒ + =
′−
 and 3 173 cms′ = + . 
At the lens�air interface:
4
1.52 1 0.52
173 cm 41.6 cms
−
⇒ + =
′− −
 and 4 47.0 cms′ = + , to the left of lens. 
1 1 2 2 3 3 4 4
1 2 3 4
1 1 2 2 3 3 4 4
851 491 173 47.0 1.06.
70 851 311 173
a a a a
b b b b
n s n s n s n sm m m m m
n s n s n s n s
⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞′ ′ ′ ′ − + +⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= = = = −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− − −⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
 
(Note all the indices of refraction cancel out.) 
(b) The image is real. 
(c) The image is inverted. 
(d) The final height is (1.06)(4.00 mm) 4.24 mm.y my′ = = = 
EVALUATE: The final image is real even though it is on the same side of the lens as the object! 
34.103. IDENTIFY: The camera lens can be modeled as a thin lens that forms an image on the film. 
SET UP: The thin-lens equation is 1 1 1
s s f
+ =
′
, and the magnification of the lens is sm
s
′
= − . 
EXECUTE: (a) 41 (0.0360 m) (7.50 10 )
4 (12.0 m)
s ym s s
s y
−′ ′ ′= − = = ⇒ = × , 
4 4
1 1 1 1 1 1 1 11 46.7 m
(7.50 10 ) 7.50 10 0.0350 m
s
s s s s s f− −
⎛ ⎞+ = + = + = = ⇒ =⎜ ⎟′ × ×⎝ ⎠
. 
(b) To just fill the frame, the magnification must be 33.00 10−× so: 
3
1 1 1 11 11.7 m
3.00 10 0.0350 m
s
s f−
⎛ ⎞+ = = ⇒ =⎜ ⎟×⎝ ⎠
. 
Since the boat is originally 46.7 m away, the distance you must move closer to the boat is 
46.7 m � 11.7 m = 35.0 m. 
EVALUATE: This result seems to imply that if you are 4 times as far, the image is ¼ as large on the film. 
However this result is only an approximation, and would not be true for very close distances. It is a better 
approximation for large distances. 
34.104. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 and sm
s
′
= − . 
SET UP: 18.0 cms s′+ = 
EXECUTE: (a) 1 1 1
18.0 cm 3.00 cms s
+ =
′ ′−
. 2 2( ) (18.0 cm) 54.0 cm 0s s′ ′− + = so 14.2 cm or 3.80 cms′ = . 
3.80 cm or 14.2 cms = , so the screen must either be 3.80 cm or 14.2 cm from the object. 
34-32 Chapter 34 
(b) 3.803.80 cm : 0.268.
14.2
ss m
s
′
= = − = − = − 14.214.2 cm : 3.74.
3.80
ss m
s
′
= = − = − = − 
EVALUATE: Since the image is projected onto the screen, the image is real and s′ is positive. We assumed this 
when we wrote the condition 18.0 cms s′+ = . 
34.105. IDENTIFY: Apply Eq.(34.16) to calculate the image distance for each lens. The image formed by the 1st lens 
serves as the object for the 2nd lens, and the image formed by the 2nd lens serves as the object for the 3rd lens. 
SET UP: The positionsof the object and lenses are shown in Figure 34.105. 
 
1 1 1
s s f
+ =
′
 
1 1 1 s f
s f s sf
−
= − =
′
 
sfs
s f
′ =
−
 
 
Figure 34.105 
EXECUTE: lens #1 
80.0 cm; 40.0 cms f= + = + 
( 80.0 cm)( 40.0 cm) 80.0 cm
80.0 cm 40.0 cm
sfs
s f
+ +′ = = = +
− + −
 
The image formed by the first lens is 80.0 cm to the right of the first lens, so it is 80.0 cm 52.0 cm 28.0 cm− = to 
the right of the second lens. 
lens #2 
28.0 cm; 40.0 cms f= − = + 
( 28.0 cm)( 40.0 cm) 16.47 cm
28.0 cm 40.0 cm
sfs
s f
− +′ = = = +
− − −
 
The image formed by the second lens is 16.47 cm to the right of the second lens, so it is 
52.0 cm 16.47 cm 35.53 cm− = to the left of the third lens. 
lens #3 
35.53 cm; 40.0 cms f= + = + 
( 35.53 cm)( 40.0 cm) 318 cm
+35.53 cm 40.0 cm
sfs
s f
+ +′ = = = −
− −
 
The final image is 318 cm to the left of the third lens, so it is 318 cm 52 cm 52 cm 80 cm 134 cm− − − = to the left 
of the object. 
EVALUATE: We used the separation between the lenses and the sign conventions for s and s′ to determine the 
object distances for the 2nd and 3rd lenses. The final image is virtual since the final s′ is negative. 
34.106. IDENTIFY: Apply 1 1 1
s s f
+ =
′
 and calculate s′ for each s. 
SET UP: 90 mmf = 
EXECUTE: 1 1 1 1 1 1 96.7 mm.
1300 mm 90 mm
s
s s f s
′+ = ⇒ + = ⇒ =
′ ′
 
1 1 1 1 1 1 91.3 mm.
6500 mm 90 mm
96.7 mm 91.3 mm 5.4 mm toward the film
s
s s f s
s
′+ = ⇒ + = ⇒ =
′ ′
′⇒ Δ = − =
 
EVALUATE: sfs
s f
′ =
−
. For 0f > and s f> , s′ decreases as s increases. 
34.107. IDENTIFY and SET UP: The generalization of Eq.(34.22) is near point ,M
f
= so near point .f
M
= 
EXECUTE: (a) age 10, near point 7 cm= 
7 cm 3.5 cm
2.0
f = = 
(b) age 30, near point 14 cm= 
14 cm 7.0 cm
2.0
f = = 
Geometric Optics 34-33 
(c) age 60, near point 200 cm= 
200 cm 100 cm
2.0
f = = 
(d) 3.5 cmf = (from part (a)) and near point 200 cm= (for 60-year-old) 
200 cm 57
3.5 cm
M = = 
(e) EVALUATE: No. The reason 3.5 cmf = gives a larger M for a 60-year-old than for a 10-year-old is that the 
eye of the older person can�t focus on as close an object as the younger person can. The unaided eye of the 60-
year-old must view a much smaller angular size, and that is why the same f gives a much larger M. The angular 
size of the image depends only on f and is the same for the two ages. 
34.108. IDENTIFY: Use 1 1 1
s s f
+ =
′
 to calculate s that gives 25 cms′ = − . M θ
θ
′
= . 
SET UP: Let the height of the object be y , so y
s
θ′ = and 
25 cm
yθ = . 
EXECUTE: (a) 1 1 1 1 1 1 (25 cm) .
25 cm 25 cm
fs
s s f s f f
+ = ⇒ + = ⇒ =
′ − +
 
(b) ( 25 cm) ( 25 cm)arctan arctan .
(25 cm) (25 cm)
y y f y f
s f f
θ
⎛ ⎞+ +⎛ ⎞′ = = ≈⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
(c) ( 25 cm) 1 25 cm .
(25 cm) / 25 cm
y f fM
f y f
θ
θ
′ + +
= = = 
(d) If 10 cm 25 cm10 cm 3.5.
10 cm
f M += ⇒ = = This is 1.4 times greater than the magnification obtained if the image 
if formed at infinity 25 cm( 2.5).M
f∞
= = 
EVALUATE: (e) Having the first image form just within the focal length puts one in the situation described above, 
where it acts as a source that yields an enlarged virtual image. If the first image fell just outside the second focal 
point, then the image would be real and diminished. 
34.109. IDENTIFY: Apply 1 1 1
s s f
+ =
′
. The near point is at infinity, so that is where the image must be formed for any 
objects that are close. 
SET UP: The power in diopters equals 1
f
, with f in meters. 
EXECUTE: 1 1 1 1 1 1 4.17
24 cm 0.24 mf s s
= + = + = =
′ −∞
diopters. 
EVALUATE: To focus on closer objects, the power must be increased. 
34.110. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
. 
SET UP: 1.00an = , 1.40bn = . 
EXECUTE: 1 1.40 0.40 2.77 cm.
36.0 cm 0.75 cm
s
s
′+ = ⇒ =
′
 
EVALUATE: This distance is greater than the normal eye, which has a cornea vertex to retina distance of about 
2.6 cm. 
34.111. IDENTIFY: Use similar triangles in Figure 34.63 in the textbook and Eq.(34.16) to derive the expressions called 
for in the problem. 
(a) SET UP: The effect of the converging lens on the ray bundle is sketched in Figure 34.111. 
 
EXECUTE: From similar 
triangles in Figure 34.111a, 
0 0
1 1
r r
f f d
′
=
−
 
 
Figure 34.111a 
34-34 Chapter 34 
Thus 10 0
1
,f dr r
f
⎛ ⎞−′ = ⎜ ⎟
⎝ ⎠
 as was to be shown. 
(b) SET UP: The image at the focal point of the first lens, a distance 1f to the right of the first lens, serves as the 
object for the second lens. The image is a distance 1f d− to the right of the second lens, so 2 1 1( ) .s f d d f= − − = − 
EXECUTE: 2 2 1 22
2 2 1 2
( )s f d f fs
s f d f f
−′ = =
− − −
 
2 0f < so 2 2f f= − and 
1 2
2
2 1
( )
,
f d f
s
f f d
−
′ =
− +
 as was to be shown. 
(c) SET UP: The effect of the diverging lens on the ray bundle is sketched in Figure 34.111b. 
 
EXECUTE: From similar 
triangles in the sketch, 0 0
2
r r
f s
′
=
′
 
Thus 0
0 2
r f
r s
=
′ ′
 
Figure 34.111b 
From the results of part (a), 0 1
0 1
.r f
r f d
=
′ −
 Combining the two results gives 1
1 2
f f
f d s
=
′−
 
( )
1 2 1 1 21
2
1 2 1 1 2 1
( )
,
( )
f d f f f fff s
f d f f d f d f f d
−⎛ ⎞
′= = =⎜ ⎟− − + − − +⎝ ⎠
 as was to be shown. 
(d) SET UP: Put the numerical values into the expression derived in part (c). 
EXECUTE: 1 2
2 1
f f
f
f f d
=
− +
 
2
1 2
216 cm12.0 cm, 18.0 cm, so 
6.0 cm
f f f
d
= = =
+
 
0d = gives 36.0 cm;f = maximum f 
4.0 cmd = gives 21.6 cm;f = minimum f 
30.0 cmf = says 
2216 cm30.0 cm
6.0 cm d
=
+
 
6.0 cm 7.2 cmd+ = and 1.2 cmd = 
EVALUATE: Changing d produces a range of effective focal lengths. The effective focal length can be both 
smaller and larger than 1 2 .f f+ 
34.112. IDENTIFY: M θ
θ
′
= . 1 2
1 2
, andy y
f s
θ θ
′ ′
′= =
′
. This gives 2 1
2 1
.y fM
s y
′
=
′ ′
. 
SET UP: Since the image formed by the objective is used as the object for the eyepiece, 1 2y y′ = . 
EXECUTE: 2 1 2 1 2 1 1
2 2 2 2 2 2 2
. . . .y f y f s f fM
s y y s s s s
′ ′ ′
= = = =
′ ′ ′
 Therefore, 12
48.0 cm 1.33 cm,
36
fs
M
= = = and this is just 
outside the eyepiece focal point. 
Now the distance from the mirror vertex to the lens is 1 2 49.3 cm,f s+ = and so 
1
2
2 2 2
1 1 1 1 1 12.3 cm.
1.20 cm 1.33 cm
s
s s f
−
⎛ ⎞
′+ = ⇒ = − =⎜ ⎟′ ⎝ ⎠
 Thus we have a final image which is real and 12.3 cm from 
the eyepiece. (Take care to carry plenty of figures in the calculation because two close numbers are subtracted.) 
EVALUATE: Eq.(34.25) gives 40M = , somewhat larger than M for this telescope. 
34.113. IDENTIFY and SET UP: The image formed by the objective is the object for the eyepiece. The total lateral 
magnification is tot 1 2 1. 8.00 mmm m m f= = (objective); 2 7.50 cmf = (eyepiece) 
Geometric Optics 34-35 
(a) The locations of the object, lenses and screen are shown in Figure 34.113. 
 
Figure 34.113 
EXECUTE: 1Find the object distance for the objective:s 
1 1 118.0 cm, 0.800 cm, ?s f s′ = + = = 
1 1 1
1 1 1 ,
s s f
+ =
′
 so 1 1
1 1 1 1 1
1 1 1 s f
s f s s f
′ −
= − =
′ ′
 
1 1
1
1 1
(18.0 cm)(0.800 cm) 0.8372 cm
18.0 cm 0.800 cm
s fs
s f
′
= = =
′ − −
 
Find the object distance 2s for the eyepiece: 
2 2 2200 cm, 7.50 cm, ?s f s′ = + = = 
2 2 2
1 1 1
s s f
+ =
′
 
2 2
2
2 2
(200 cm)(7.50 cm) 7.792 cm
200 cm 7.50 cm
s fs
s f
′
= = =
′ − −
 
Now we calculate the magnification for each lens: 
1
1
1
18.0 cm 21.50
0.8372 cm
sm
s
′
= − = − = − 
2
2
2
200 cm 25.67
 7.792 cm
sm
s
′
= − = − = − 
tot 1 2 ( 21.50)( 25.67) 552.m m m= = − − = 
(b) From the sketch we can see that the distance between the two lenses is 1 2 18.0 cm 7.792 cm 25.8 cm.s s′ + = + = 
EVALUATE: The microscope is not being used in the conventional way; it merely serves as a two-lens system. In 
particular, the final image formed by the eyepiece in the problem is real, not virtual as is the case normally for a 
microscope. Eq.(34.23) does not apply here, and in any event gives the angular not the lateral magnification.34.114. IDENTIFY: For u and u′ as defined in Figure 34.64 in the textbook, uM
u
′
= . 
SET UP: 2f is negative. From Figure 34.64, the length of the telescope is 1 2f f+ . 
EXECUTE: (a) From the figure, 
1 2 2
and .y y yu u
f f f
′= = = − The angular magnification is 1
2
.u fM
u f
′
= = − 
(b) 1 12
2
95.0 cm 15.0 cm.
6.33
f fM f
f M
= − ⇒ = − = − = − 
(c) The length of the telescope is 95.0 cm 15.0 cm 80.0 cm,− = compared to the length of 110 cm for the telescope 
in Exercise 34.57. 
EVALUATE: An advantage of this construction is that the telescope is somewhat shorter. 
34.115. IDENTIFY: Use 1 1 1
s s f
+ =
′
 to calculate s′ (the distance of each point from the lens), for points A, B and C. 
SET UP: The object and lens are shown in Figure 34.115a. 
EXECUTE: (a) 1 1 1 1 1 1For point : 36.0 cm.
45.0 cm 20.0 cm
C s
s s f s
′+ = ⇒ + = ⇒ =
′ ′
 
36.0 (15.0 cm) 12.0 cm
45.0
sy y
s
′
′ = − = − = − , so the image of point C is 36.0 cm to the right of the lens, and 
12.0 cm below the axis. 
For point A: 45.0 cm 8.00 cm(cos45 ) 50.7 cms = + ° = . 
1 1 1 1 1 1 33.0 cm.
50.7 cm 20.0 cm
s
s s f s
′+ = ⇒ + = ⇒ =
′ ′
33.0 (15.0 cm 8.00 cm(sin 45 )) 6.10 cm,
45.0
sy y
s
′
′ = − = − − ° = − 
so the image of point A is 33.0 cm to the right of the lens, and 6.10 cm below the axis. 
34-36 Chapter 34 
For point B: 45.0 cm 8.00 cm(cos45 ) 39.3 cms = − ° = . 1 1 1 1 1 1 40.7 cm.
39.3 cm 20.0 cm
s
s s f s
′+ = ⇒ + = ⇒ =
′ ′
 
40.7 (15.0 cm 8.00 cm(sin 45 )) 21.4 cm,
39.3
sy y
s
′
′ = − = − + ° = − so the image of point B is 40.7 cm to the right of the 
lens, and 21.4 cm below the axis. The image is shown in Figure 34.115b. 
(b) The length of the pencil is the distance from point A to B: 
2 2 2 2( ) ( ) (33.0 cm 40.7 cm) (6.10 cm 21.4 cm) 17.1 cmA B A BL x x y y= − + − = − + − = 
EVALUATE: The image is below the optic axis and is larger than the object. 
 
Figure 34.115 
34.116. IDENTIFY and SET UP: Consider the ray diagram drawn in Figure 34.116. 
EXECUTE: (a) Using the diagram and law of sines, sin sin but sin sin
( )
h
R f g R
θ α θ α= = =
−
(law of 
reflection), ( ).g R f= − Bisecting the triangle: 2cos cos cos
( ) 2
R RR f
R f
θ θ θ= ⇒ − =
−
. 
0
1 12 2
2 cos cos
Rf f
θ θ
⎡ ⎤ ⎡ ⎤= − = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
. 0 2
Rf = is the value of f for θ near zero (incident ray near the axis). When θ 
increases, (2 1/ cos )θ− decreases and f decreases. 
(b) 0
0 0
10.02 0.98 so2 0.98
cos
f f f
f f θ
−
= − ⇒ = − = . 1cos 0.98
2 0.98
θ = =
−
 and 11.4 .θ = ° 
EVALUATE: For 45θ = °, 00.586f f= , and f approaches zero as θ approaches 60°. 
 
Figure 34.116 
Geometric Optics 34-37 
34.117. IDENTIFY: The distance between image and object can be calculated by taking the derivative of the separation 
distance and minimizing it. 
SET UP: For a real image 0s′ > and the distance between the object and the image is D s s′= + . For a real 
image must have s f> . 
EXECUTE: 
2
but sf sf sD s s s D s
s f s f s f
′ ′= + = ⇒ = + =
− − −
. 
2 2 2
2 2
2 2 0
( ) ( )
dD d s s s s sf
ds ds s f s f s f s f
⎛ ⎞ −
= = − = =⎜ ⎟− − − −⎝ ⎠
. 2 2 0s sf− = . ( 2 ) 0s s f− = . 2s f= is the solution for which 
s f> . For 2s f= , 2s f′ = . Therefore, the minimum separation is 2 2 4f f f+ = . 
(b) A graph of /D f versus /s f is sketched in Figure 34.117. Note that the minimum does occur for 4D f= . 
EVALUATE: If, for example, 3 / 2s f= , then 3s f′ = and 4.5D s s f′= + = , greater than the minimum value. 
 
Figure 34.117 
34.118. IDENTIFY and SET UP: For a plane mirror, s s′ = − . 
EXECUTE: (a) By the symmetry of image production, any image must be the same distance D as the object from 
the mirror intersection point. But if the images and the object are equal distances from the mirror intersection, they 
lie on a circle with radius equal to D. 
(b) The center of the circle lies at the mirror intersection as discussed above. 
(c) The diagram is sketched in Figure 34.118. 
EVALUATE: To see the image, light from the object must be able to reflect from each mirror and reach the 
person's eyes. 
 
Figure 34.118 
34.119. IDENTIFY: Apply a b b an n n n
s s R
−
+ =
′
 to refraction at the cornea to find where the object for the cornea must be in 
order for the image to be at the retina. Then use 1 1 1
s s f
+ =
′
 to calculate f so that the lens produces an image of a 
distant object at this point. 
SET UP: For refraction at the cornea, 1.33an = and 1.40bn = . The distance from the cornea to the retina in this 
model of the eye is 2.60 cm. From Problem 34.46, 0.71 cmR = . 
EXECUTE: (a) People with normal vision cannot focus on distant objects under water because the image is 
unable to be focused in a short enough distance to form on the retina. Equivalently, the radius of curvature of the 
normal eye is about five or six times too great for focusing at the retina to occur. 
(b) When introducing glasses, let�s first consider what happens at the eye: 
2
2 2 2
1.33 1.40 0.07 3.02 cm.
2.6 cm 0.71cm
a b b an n n n s
s s R s
−
+ = ⇒ + = ⇒ = −
′
 That is, the object for the cornea must be 3.02 cm 
behind the cornea. Now, assume the glasses are 2.00 cm in front of the eye, so 1 22.00 cm 5.02 cms s′ = + = . 
34-38 Chapter 34 
1 1 1
1 1 1
s s f
+ =
′ ′
 gives 
1
1 1 1
5.02 cm f
+ =
′∞
 and 1 5.02 cm.f ′= This is the focal length in water, but to get it in air, we use 
the formula from Problem 34.98: liq1 1
liq
1.52 1.333(5.02 cm) 1.35 cm
( 1) 1.333(1.52 1)
n n
f f
n n
⎡ ⎤− ⎡ ⎤−′= = =⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎣ ⎦⎣ ⎦
. 
EVALUATE: A converging lens is needed. 
35-1 
INTERFERENCE 
 35.1. IDENTIFY: Compare the path difference to the wavelength. 
SET UP: The separation between sources is 5.00 m, so for points between the sources the largest possible path 
difference is 5.00 m. 
EXECUTE: (a) For constructive interference the path difference is , 0, 1, 2, . . .m mλ = ± ± Thus only the path 
difference of zero is possible. This occurs midway between the two sources, 2.50 m from A. 
(b) For destructive interference the path difference is 12( ) , 0, 1, 2, . . .m mλ+ = ± ± 
A path difference of 2 3.00/λ± = m is possible but a path difference as large as 3 / 2 9.00λ = m is not possible. For 
a point a distance x from A and 5.00 fromx B− the path difference is 
(5.00 m ). (5.00 m ) 3.00 m gives 4.00 m.x x x x x− − − − = + = (5.00 m ) 3.00 m gives 1.00 mx x x− − = − = . 
EVALUATE: The point of constructive interference is midway between the points of destructive interference. 
 35.2. IDENTIFY: For destructive interference the path difference is 12( ) , 0, 1, 2,m mλ+ = ± ± … . The longest wavelength 
is for 0m = . For constructive interference the path difference is , 0, 1, 2,m mλ = ± ± … The longest wavelength is 
for 1m = . 
SET UP: The path difference is 120 m. 
EXECUTE: (a) For destructive interference 120 m 240 m.
2
λ λ= ⇒ = 
(b) The longest wavelength for constructive interference is 120 m.λ = 
EVALUATE: The path difference doesn't depend on the distance of point Q from B. 
 35.3. IDENTIFY: Use c f λ= to calculate the wavelength of the transmitted waves. Compare the difference in the 
distance from A to P and from B to P. For constructive interference this path difference is an integer multiple of the 
wavelength. 
SET UP: Consider Figure 35.3 
 
The distance of point P 
from each coherent source 
is Ar x= and 
9.00 m .Br x= − 
 
Figure 35.3 
EXECUTE: The path difference is 9.00 m 2 .B Ar r x− = − 
, 0, 1, 2, B Ar r m mλ− = = ± ± … 
8
6
2.998 10 m/s 2.50 m
120 10 Hz
c
f
λ ×= = =
×
 
Thus 9.00 m 2 (2.50 m)x m− = and 9.00 m (2.50 m) 4.50 m (1.25 m) .
2
mx m−= = − x must lie in the range 0 to 
9.00 m since P is said to be between the two antennas. 
0m = gives 4.50 mx = 
1m = + gives 4.50 m 1.25 m 3.25 mx = − = 
2m = + gives 4.50 m 2.50 m 2.00 mx = − = 
3m = + gives 4.50 m 3.75 m 0.75 mx = − = 
1m = − gives 4.50 m 1.25 m 5.75 mx = + = 
2m = − gives 4.50 m 2.50 m 7.00 mx = + = 
3m = − gives 4.50 m 3.75 m 8.25 mx= + = 
35
35-2 Chapter 35 
All other values of m give values of x out of the allowed range. Constructive interference will occur for 
0.75 m, 2.00 m, 3.25 m, 4.50 m, 5.75 m, 7.00 m, and 8.25 m.x = 
EVALUATE: Constructive interference occurs at the midpoint between the two sources since that point is the 
same distance from each source. The other points of constructive interference are symmetrically placed relative to 
this point. 
 35.4. IDENTIFY: For constructive interference the path difference d is related to λ by , 0,1,2,d m mλ= = … For 
destructive interference 12( ) , 0,1,2,d m mλ= + = … 
SET UP: 2040 nmd = 
EXECUTE: (a) The brightest wavelengths are when constructive interference occurs: 
3 4
5
2040 nm 2040 nm680 nm, 510 nm and
3 4
2040 nm 408 nm.
5
m m
dd m
m
λ λ λ λ
λ
= ⇒ = ⇒ = = = =
= =
 
(b) The path-length difference is the same, so the wavelengths are the same as part (a). 
(c) 12( ) md m λ= + so 1 1
2 2
2040 nm
m
d
m m
λ = =
+ +
. The visible wavelengths are 3 583 nmλ = and 4 453 nmλ = . 
EVALUATE: The wavelengths for constructive interference are between those for destructive interference. 
 35.5. IDENTIFY: If the path difference between the two waves is equal to a whole number of wavelengths, constructive 
interference occurs, but if it is an odd number of half-wavelengths, destructive interference occurs. 
SET UP: We calculate the distance traveled by both waves and subtract them to find the path difference. 
EXECUTE: Call P1 the distance from the right speaker to the observer and P2 the distance from the left speaker to 
the observer. 
(a) P1 = 8.0 m and 2 22 (6.0 m) (8.0 m) 10. 0 mP = + = . The path distance is 
2 1P P PΔ = − = 10.0 m � 8.0 m = 2.0 m 
(b) The path distance is one wavelength, so constructive interference occurs. 
(c) P1 = 17.0 m and 2 22 (6.0 m) (17.0 m) 18.0 mP = + = . The path difference is 18.0 m � 17.0 m = 1.0 m, which is 
one-half wavelength, so destructive interference occurs. 
EVALUATE: Constructive interference also occurs if the path difference 2λ , 3 λ , 4 λ , etc., and destructive 
interference occurs if it is λ /2, 3 λ /2, 5 λ /2, etc. 
 35.6. IDENTIFY: At an antinode the interference is constructive and the path difference is an integer number of 
wavelengths; path difference , 0, 1, 2,m mλ= = ± ± … at an antinode. 
SET UP: The maximum magnitude of the path difference is the separation d between the two sources. 
EXECUTE: (a) At 1 2 1, 4 ,S r r λ− = and this path difference stays the same all along the -axis,y so 
2 2 14. At , 4m S r r ,λ= + − = − and the path difference below this point, along the negative y-axis, stays the same, so 
4.m = − 
(b) The wave pattern is sketched in Figure 35.6. 
(c) The maximum and minimum m-values are determined by the largest integer less than or equal to .d
λ
 
(d) If 17 7 7,
2
d mλ= ⇒ − ≤ ≤ + so there will be a total of 15 antinodes between the sources. 
EVALUATE: We are considering points close to the two sources and the antinodal curves are not straight lines. 
 
Figure 35.6 
Interference 35-3 
 35.7. IDENTIFY: At an antinodal point the path difference is equal to an integer number of wavelengths. 
SET UP: For 3m = , the path difference is 3λ . 
EXECUTE: Measuring with a ruler from both 1 2andS S to the different points in the antinodal line labeled 3m = , 
we find that the difference in path length is three times the wavelength of the wave, as measured from one crest to 
the next on the diagram. 
EVALUATE: There is a whole curve of points where the path difference is 3λ . 
 35.8. IDENTIFY: The value of 20y is much smaller than R and the approximate expression m
my R
d
λ
= is accurate. 
SET UP: 320 10.6 10 my
−= × . 
EXECUTE: 
9
3
3
20
20 (20)(1.20 m)(502 10 m) 1.14 10 m 1.14 mm
10.6 10 m
Rd
y
λ − −
−
×
= = = × =
×
 
EVALUATE: 2020tan
y
R
θ = so 20 0.51θ = ° and the approximation 20 20sin tanθ θ≈ is very accurate. 
 35.9. IDENTIFY and SET UP: The dark lines correspond to destructive interference and hence are located by Eq.(35.5): 
1
1 2sin so sin , 0, 1, 2,
2
m
d m m
d
λ
θ λ θ
⎛ ⎞+⎜ ⎟⎛ ⎞ ⎝ ⎠= + = = ± ±⎜ ⎟
⎝ ⎠
… 
Solve for θ that locates the second and third dark lines. Use tany R θ= to find the distance of each of the dark 
lines from the center of the screen. 
EXECUTE: 1st dark line is for 0m = 
2nd dark line is for 1m = and 
9
3
1 3
3 3(500 10 m)sin 1.667 10
2 2(0.450 10 m)d
λθ
−
−
−
×
= = = ×
×
 and 31 1.667 10 radθ
−= × 
3rd dark line is for 2m = and 
9
3
2 3
5 5(500 10 m)sin 2.778 10
2 2(0.450 10 m)d
λθ
−
−
−
×
= = = ×
×
 and 32 2.778 10 radθ
−= × 
(Note that 1θ and 2θ are small so that the approximation sin tanθ θ θ≈ ≈ is valid.) The distance of each dark line 
from the center of the central bright band is given by tan ,my R θ= where 0.850 mR = is the distance to the 
screen. 
tan so m my Rθ θ θ≈ = 
3 3
1 1 (0.750 m)(1.667 10 rad) 1.25 10 my Rθ
− −= = × = × 
3 3
2 2 (0.750 m)(2.778 10 rad) 2.08 10 my Rθ
− −= = × = × 
3 3
2 1 2.08 10 m 1.25 10 m 0.83 mmy y y
− −Δ = − = × − × = 
EVALUATE: Since 1θ and 2θ are very small we could have used Eq.(35.6), generalized to destructive 
interference: 1 / .
2m
y R m dλ⎛ ⎞= +⎜ ⎟
⎝ ⎠
 
35.10. IDENTIFY: Since the dark fringes are eqully spaced, mR y" , the angles are small and the dark bands are located 
by 1
2
1
2( )
m
m
y R
d
λ
+
+
= . 
SET UP: The separation between adjacent dark bands is Ry
d
λ
Δ = . 
EXECUTE: 
7
4
3
(1.80 m) (4.50 10 m) 1.93 10 m 0.193 m.
4.20 10 m
R Ry d
d y
λ λ − −
−
×
Δ = ⇒ = = = × =
Δ ×
 
EVALUATE: When the separation between the slits decreases, the separation between dark fringes increases. 
35.11. IDENTIFY and SET UP: The positions of the bright fringes are given by Eq.(35.6): ( / ).my R m dλ= For each 
fringe the adjacent fringe is located at 1 ( 1) / .my R m dλ+ = + Solve for .λ 
EXECUTE: The separation between adjacent fringes is 1 / .m my y y R dλ+Δ = − = 
3 3
7(0.460 10 m)(2.82 10 m) 5.90 10 m 590 nm
2.20 m
d y
R
λ
− −
−Δ × ×= = = × = 
EVALUATE: Eq.(35.6) requires that the angular position on the screen be small. The angular position of bright 
fringes is given by sin / .m dθ λ= The slit separation is much larger than the wavelength 3( / 1.3 10 ),dλ −= × so θ 
is small so long as m is not extremely large. 
35-4 Chapter 35 
35.12. IDENTIFY: The width of a bright fringe can be defined to be the distance between its two adjacent destructive 
minima. Assuming the small angle formula for destructive interference 
1
2( )
m
m
y R
d
λ+
= . 
SET UP: 30.200 10 md −= × . 4.00 mR = . 
EXECUTE: The distance between any two successive minima 
is
9
1 3
(400 10 m)(4.00 m) 8.00 mm.
(0.200 10 m)m m
y y R
d
λ −
+ −
×
− = = =
×
 Thus, the answer to both part (a) and part (b) is that the 
width is 8.00 mm. 
EVALUATE: For small angles, when my R# , the interference minima are equally spaced. 
35.13. IDENTIFY and SET UP: The dark lines are located by 1sin .
2
d mθ λ⎛ ⎞= +⎜ ⎟
⎝ ⎠
 The distance of each line from the 
center of the screen is given by tan .y R θ= 
EXECUTE: First dark line is for 0m = and 1sin / 2.d θ λ= 
9
1 16
550 10 msin 0.1528 and 8.789 .
2 2(1.80 10 m)d
λθ θ
−
−
×
= = = = °
×
 Second dark line is for 1m = and 2sin 3 / 2.d θ λ= 
9
2 6
3 550 10 msin 3 0.4583
2 2(1.80 10 m)d
λθ
−
−
⎛ ⎞×
= = =⎜ ⎟×⎝ ⎠
 and 2 27.28 .θ = ° 
1 1tan (0.350 m) tan8.789 0.0541 my R θ= = ° = 
2 2tan (0.350 m) tan 27.28 0.1805 my R θ= = ° = 
The distance between the lines is 2 1 0.1805 m 0.0541 m 0.126 m 12.6 cm.y y yΔ = − = − = = 
EVALUATE: 1sin 0.1528θ = and 1 2tan 0.1546. sin 0.4583θ θ= = and 2tan 0.5157.θ = As the angle increases, 
sin tanθ θ≈ becomes a poorer approximation. 
35.14. IDENTIFY: Using Eq.(35.6) for small angles: m
my R
d
λ
= . 
SET UP: First-order means 1m = . 
EXECUTE: The distance between corresponding bright fringes is 
9
3
(5.00 m)(1) (660 470) (10 m) 3.17 mm.
(0.300 10 m)
Rmy
d
λ −−Δ = Δ = − × =×
 
EVALUATE: Theseparation between these fringes for different wavelengths increases when the slit separation 
decreases. 
35.15. IDENTIFY and SET UP: Use the information given about the bright fringe to find the distance d between the two 
slits. Then use Eq.(35.5) and tany R θ= to calculate λ for which there is a first-order dark fringe at this same 
place on the screen. 
EXECUTE: 
9
41 1
1 3
1
(3.00 m)(600 10 m), so 3.72 10 m.
4.84 10 m
R Ry d
d y
λ λ − −
−
×
= = = = ×
×
 (R is much greater than d, so Eq.35.6 
is valid.) The dark fringes are located by 1sin , 0, 1, 2,
2
d m mθ λ⎛ ⎞= + = ± ±⎜ ⎟
⎝ ⎠
… The first order dark fringe is located 
by 2sin / 2 ,dθ λ= where 2λ is the wavelength we are seeking. 
2tan sin
2
Ry R R
d
λθ θ= ≈ = 
We want 2λ such that 1.y y= This gives 1 22
R R
d d
λ λ
= and 2 12 1200 nm.λ λ= = 
EVALUATE: For 600 nmλ = the path difference from the two slits to this point on the screen is 600 nm. For this 
same path difference (point on the screen) the path difference is / 2λ when 1200 nm.λ = 
35.16. IDENTIFY: Bright fringes are located at m
my R
d
λ
= , when my R# . Dark fringes are at 12sin ( )d mθ λ= + and 
tany R θ= . 
SET UP: 
8
7
14
3.00 10 m/s 4.75 10 m
6.32 10 Hz
c
f
λ −×= = = ×
×
. For the third bright fringe (not counting the central bright 
spot), 3m = . For the third dark fringe, 2m = . 
Interference 35-5 
EXECUTE: (a) 
7
53(4.75 10 m)(0.850 m) 3.89 10 m 0.0389 mm
0.0311 mm
m Rd
y
λ − −×= = = × = 
(b) 
7
1
2 5
4.75 10 msin (2 ) (2.5) 0.0305
3.89 10 md
λθ
−
−
⎛ ⎞×
= + = =⎜ ⎟×⎝ ⎠
and 1.75θ = ° . tan (85.0 cm) tan1.75 2.60 cmy R θ= = =° . 
EVALUATE: The third dark fringe is closer to the center of the screen than the third bright fringe on one side of 
the central bright fringe. 
35.17. IDENTIFY: Bright fringes are located at angles θ given by sind mθ λ= . 
SET UP: The largest value sinθ can have is 1.00. 
EXECUTE: (a) sindm θ
λ
= . For sin 1θ = , 
3
7
0.0116 10 m 19.8
5.85 10 m
dm
λ
−
−
×
= = =
×
. Therefore, the largest m for fringes 
on the screen is 19m = . There are 2(19) 1 39+ = bright fringes, the central one and 19 above and 19 below it. 
(b) The most distant fringe has 19m = ± . 
7
3
5.85 10 msin 19 0.958
0.0116 10 m
m
d
λθ
−
−
⎛ ⎞×
= = ± = ±⎜ ⎟×⎝ ⎠
and 73.3θ = ± ° . 
EVALUATE: For small θ the spacing yΔ between adjacent fringes is constant but this is no longer the case for 
larger angles. 
35.18. IDENTIFY: At large distances from the antennas the equation sin , 0, 1, 2,d m mθ λ= = ± ± …gives the angles where 
maximum intensity is observed and 12sin ( ) , 0, 1, 2,d m mθ λ= + = ± ± … gives the angles where minimum intensity 
is observed. 
SET UP: 12.0 md = . c
f
λ = . 
EXECUTE: (a) 
8
6
3.00 10 m/s 2.78 m
107.9 10 Hz
c
f
λ ×= = =
×
. 2.78 msin (0.232)
12.0 m
m m m
d
λθ ⎛ ⎞= = =⎜ ⎟
⎝ ⎠
. 
13.4 , 27.6 , 44.1 , 68.1θ = ± ± ± ±° ° ° ° . 
(b) 1 12 2sin ( ) ( )(0.232)m md
λθ = + = + . 6.66 , 20.4 , 35.5 , 54.3θ = ± ± ± ±° ° ° ° . 
EVALUATE: The angles for zero intensity are approximately midway between those for maximum intensity. 
35.19. IDENTIFY: Eq.(35.10): 20 cos ( 2)I I φ= . Eq.(35.11): 2 1(2 / )( )r rφ π λ= − . 
SET UP: φ is the phase difference and 2 1( )r r− is the path difference. 
EXECUTE: (a) 20 0(cos 30.0 ) 0.750I I I= ° = 
(b) 60.0 ( /3) radπ° = . [ ]2 1( ) ( / 2 ) ( /3) / 2 / 6 80 nmr r φ π λ π π λ λ− = = = = . 
EVALUATE: 360 / 6φ = ° and 2 1( ) / 6r r λ− = . 
35.20. IDENTIFY: path difference
2
φ
π λ
Δ
= relates the path difference to the phase difference φΔ . 
SET UP: The sources and point P are shown in Figure 35.20. 
EXECUTE: 524 cm 486 cm2 119 radians
2 cm
φ π
⎛ ⎞−
Δ = =⎜ ⎟
⎝ ⎠
 
EVALUATE: The distances from B to P and A to P aren't important, only the difference in these distances. 
 
Figure 35.20 
35.21. IDENTIFY and SET UP: The phase difference φ is given by (2 / )sindφ π λ θ= (Eq.35.13.) 
EXECUTE: 3 9[2 (0.340 10 m)/(500 10 m)]sin 23.0 1670 radφ π − −= × × ° = 
EVALUATE: The mth bright fringe occurs when 2 ,mφ π= so there are a large number of bright fringes within 
23.0° from the centerline. Note that Eq.(35.13) gives φ in radians. 
35-6 Chapter 35 
35.22. IDENTIFY: The maximum intensity occurs at all the points of constructive interference. At these points, the path 
difference between waves from the two transmitters is an integral number of wavelengths. 
SET UP: For constructive interference, sin θ = mλ/d. 
EXECUTE: (a) First find the wavelength of the UHF waves: 
λ = c/f = (3.00 × 108 m/s)/(1575.42 MHz) = 0.1904 m 
For maximum intensity (πd sin θ )/λ = mπ, so 
sin θ = mλ/d = m[(0.1904 m)/(5.18 m)] = 0.03676m 
The maximum possible m would be for θ = 90°, or sin θ = 1, so 
mmax = d/λ = (5.18 m)/(0.1904 m) = 27.2 
which must be ±27 since m is an integer. The total number of maxima is 27 on either side of the central fringe, plus 
the central fringe, for a total of 27 + 27 + 1 = 55 bright fringes. 
(b) Using sin θ = mλ/d, where m = 0, ±1, ±2, and ±3, we have 
 sin θ = mλ/d = m[(0.1904 m)/(5.18 m)] = 0.03676m 
 m = 0: sin θ = 0, which gives θ = 0° 
m = ±1: sin θ = ±(0.03676)(1), which gives θ = ±2.11° 
m = ±2: sin θ = ±(0.03676)(2), which gives θ = ±4.22° 
m = ±3: sin θ = ±(0.03676)(3), which gives θ = ±6.33° 
(c) 20
sincos dI I π θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 = ( )2 2 (5.18 m)sin(4.65 )2.00 W/m cos
0.1904 m
π °⎡ ⎤
⎢ ⎥⎣ ⎦
 = 1.28 W/m2. 
EVALUATE: Notice that sinθ increases in integer steps, but θ only increases in integer steps for small θ. 
35.23. (a) IDENTIFY and SET UP: The minima are located at angles θ given by 1sin .
2
d mθ λ⎛ ⎞= +⎜ ⎟
⎝ ⎠
 The first minimum 
corresponds to 0.m = Solve for .θ Then the distance on the screen is tan .y R θ= 
EXECUTE: 
9
3
3
660 10 msin 1.27 10
2 2(0.260 10 m)d
λθ
−
−
−
×
= = = ×
×
 and 31.27 10 radθ −= × 
3(0.700 m) tan(1.27 10 rad) 0.889 mm.y −= × = 
(b) IDENTIFY and SET UP: Eq.(35.15) given the intensity I as a function of the position y on the screen: 
2
0 cos .
dyI I
R
π
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 Set 0 / 2I I= and solve for y. 
EXECUTE: 0
1
2
I I= says 2 1cos
2
dy
R
π
λ
⎛ ⎞ =⎜ ⎟
⎝ ⎠
 
1cos
2
dy
R
π
λ
⎛ ⎞ =⎜ ⎟
⎝ ⎠
 so rad
4
dy
R
π π
λ
= 
9
3
(660 10 m)(0.700 m) 0.444 mm
4 4(0.260 10 m)
Ry
d
λ −
−
×
= = =
×
 
EVALUATE: 0 / 2I I= at a point on the screen midway between where 0I I= and 0.I = 
35.24. IDENTIFY: Eq. (35.14): 20 cos sin .
dI I π θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 
SET UP: The intensity goes to zero when the cosine�s argument becomes an odd integer multiple of 
2
π 
EXECUTE: sin ( 1/ 2)d mπ θ π
λ
= + gives sin ( 1/ 2),d mθ λ= + which is Eq. (35.5). 
EVALUATE: Section 35.3 shows that the maximum-intensity directions from Eq.(35.14) agree with Eq.(35.4). 
35.25. IDENTIFY: The intensity decreases as we move away from the central maximum. 
SET UP: The intensity is given by 20 cos
dyI I
R
π
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 
EXECUTE: First find the wavelength: λ = c/f = (3.00 × 108 m/s)/(12.5 MHz) = 24.00 m 
At the farthest the receiver can be placed, I = I0/4, which gives 
20
0 cos4
I dyI
R
π
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 ⇒ 2 1cos
4
dy
R
π
λ
⎛ ⎞ =⎜ ⎟
⎝ ⎠
 ⇒ 1cos
2
dy
R
π
λ
⎛ ⎞ = ±⎜ ⎟
⎝ ⎠
 
Interference 35-7 
The solutions are πdy/λR = π/3 and 2π/3. Using π/3, we get 
y = λR/3d = (24.00 m)(500 m)/[3(56.0 m)] = 71.4 m 
It must remain within 71.4 m of point C. 
EVALUATE: Using πdy/λR = 2π/3 gives y = 142.8 m. But to reach this point, the receiver would have to go 
beyond 71.4 m from C, where the signal would be too weak, so this second point is not possible. 
35.26. IDENTIFY: The phase difference φ and the path difference 1 2r r− are related by 1 2
2 ( )r rπφ
λ
= − . The intensity is 
given by 20 cos 2
I I φ⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 
SET UP: 
8
8
3.00 10 m/s 2.50 m
1.20 10 Hz
c
f
λ ×= = =
×
. When the receiver measures zero intensity 0I , 0φ = . 
EXECUTE: (a) 1 2
2 2( ) (1.8 m) 4.52 rad.
2.50 m
r rπ πφ
λ
= − = = 
(b) 2 20 0 0
4.52 radcos cos 0.404 .
2 2
I I I Iφ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
EVALUATE: 1 2( )r r−is greater than / 2λ , so one minimum has been passed as the receiver is moved. 
35.27. IDENTIFY: Consider interference between rays reflected at the upper and lower surfaces of the film. Consider 
phase difference due to the path difference of 2t and any phase differences due to phase changes upon reflection. 
SET UP: Consider Figure 35.27. 
 
Both rays (1) and (2) 
undergo a 180° phase 
change on reflection, so 
these is no net phase 
difference introduced and 
the condition for 
destructive interference is 
12 .
2
t m λ⎛ ⎞= +⎜ ⎟
⎝ ⎠
 
 
Figure 35.27 
EXECUTE: 
1
2 ;
2
m
t
λ⎛ ⎞+⎜ ⎟
⎝ ⎠= thinnest film says 0m = so 
4
t λ= 
0
1.42
λλ = and 
9
70 650 10 m 1.14 10 m 114 nm
4(1.42) 4(1.42)
t λ
−
−×= = = × = 
EVALUATE: We compared the path difference to the wavelength in the film, since that is where the path 
difference occurs. 
35.28. IDENTIFY: Require destructive interference for light reflected at the front and rear surfaces of the film. 
SET UP: At the front surface of the film, light in air ( 1.00n = ) reflects from the film ( 2.62n = ) and there is a 
180° phase shift due to the reflection. At the back surface of the film, light in the film ( 2.62n = ) reflects from 
glass ( 1.62n = ) and there is no phase shift due to reflection. Therefore, there is a net 180° phase difference 
produced by the reflections. The path difference for these two rays is 2t, where t is the thickness of the film. The 
wavelength in the film is 505 nm
2.62
λ = . 
EXECUTE: (a) Since the reflection produces a net 180° phase difference, destructive interference of the reflected 
light occurs when 2t mλ= . 505 nm (96.4 nm)
2[2.62]
t m m⎛ ⎞= =⎜ ⎟
⎝ ⎠
. The minimum thickness is 96.4 nm. 
(b) The next three thicknesses are for 2m = , 3 and 4: 192 nm, 289 nm and 386 nm. 
EVALUATE: The minimum thickness is for / 2t nλ= . Compare this to Problem 35.27, where the minimum 
thickness for destructive interference is / 4t nλ= . 
35.29. IDENTIFY: The fringes are produced by interference between light reflected from the top and bottom surfaces of 
the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the top surface 
of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the air wedge do 
35-8 Chapter 35 
have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) is therefore 
1
22 ( )t m λ= + . 
SET UP: The geometry of the air wedge is sketched in Figure 35.29. At a distance x from the point of contact of 
the two plates, the thickness of the air wedge is t. 
EXECUTE: tan t
x
θ = so tant x θ= . 12( ) 2m
t m λ= + . 12( ) 2 tanm
x m λ
θ
= + and 31 2( ) 2 tanm
x m λ
θ+
= + . The 
distance along the plate between adjacent fringes is 1 2 tanm m
x x x λ
θ+
Δ = − = . 1.0015.0 fringes/cm
x
=
Δ
 and 
1.00 0.0667 cm
15.0 fringes/cm
xΔ = = . 
9
4
2
546 10 mtan 4.09 10
2 2(0.0667 10 m)x
λθ
−
−
−
×
= = = ×
Δ ×
. The angle of the wedge is 
44.09 10 rad 0.0234−× = ° . 
EVALUATE: The fringes are equally spaced; xΔ is independent of m. 
 
Figure 35.29 
35.30. IDENTIFY: The fringes are produced by interference between light reflected from the top and from the bottom 
surfaces of the air wedge. The refractive index of glass is greater than that of air, so the waves reflected from the 
top surface of the air wedge have no reflection phase shift and the waves reflected from the bottom surface of the 
air wedge do have a half-cycle reflection phase shift. The condition for constructive interference (bright fringes) 
therefore is 122 ( )t m λ= + . 
SET UP: The geometry of the air wedge is sketched in Figure 35.30. 
EXECUTE: 40.0800 mmtan 8.89 10
90.0 mm
θ −= = × . tan t
x
θ = so 4(8.89 10 )t x−= × . 12( ) 2m
t m λ= + . 
1
2 4( ) 2(8.89 10 )m
x m λ −= + ×
and 31 2 4( ) 2(8.89 10 )m
x m λ+ −= + ×
. The distance along the plate between adjacent fringes 
is 
9
4
1 4 4
656 10 m 3.69 10 m 0.369 mm
2(8.89 10 ) 2(8.89 10 )m m
x x x λ
−
−
+ − −
×
Δ = − = = = × =
× ×
. The number of fringes per cm is 
1.00 1.00 27.1 fringes/cm
0.0369 cmx
= =
Δ
. 
EVALUATE: As 0t → the interference is destructive and there is a dark fringe at the line of contact between the 
two plates. 
 
Figure 35.30 
35.31. IDENTIFY: The light reflected from the top of the TiO2 film interferes with the light reflected from the top of the 
glass surface. These waves are out of phase due to the path difference in the film and the phase differences caused 
by reflection. 
SET UP: There is a π phase change at the TiO2 surface but none at the glass surface, so for destructive 
interference the path difference must be mλ in the film. 
EXECUTE: (a) Calling T the thickness of the film gives 2T = mλ0/n, which yields T = mλ0/(2n). Substituting the 
numbers gives 
T = m (520.0 nm)/[2(2.62)] = 99.237m 
Interference 35-9 
T must be greater than 1036 nm, so m = 11, which gives T = 1091.6 nm, since we want to know the minimum 
thickness to add. 
ΔT = 1091.6 nm � 1036 nm = 55.6 nm 
(b) (i) Path difference = 2T = 2(1092 nm) = 2184 nm = 2180 nm. 
(ii) The wavelength in the film is λ = λ0/n = (520.0 nm)/2.62 = 198.5 nm. 
Path difference = (2180 nm)/[(198.5 nm)/wavelength] = 11.0 wavelengths 
EVALUATE: Because the path difference in the film is 11.0 wavelengths, the light reflected off the top of the film 
will be 180° out of phase with the light that traveled through the film and was reflected off the glass due to the 
phase change at reflection off the top of the film. 
35.32. IDENTIFY: Consider the phase difference produced by the path difference and by the reflections. For destructive 
interference the total phase difference is an integer number of half cycles. 
SET UP: The reflection at the top surface of the film produces a half-cycle phase shift. There is no phase shift at 
the reflection at the bottom surface. 
EXECUTE: (a) Since there is a half-cycle phase shift at just one of the interfaces, the minimum thickness for 
constructive interference is 0 550 nm 74.3 nm.
4 4 4(1.85)
t
n
λ λ
= = = = 
(b) The next smallest thickness for constructive interference is with another half wavelength thickness added: 
( )0 3 550 nm3 3 223 nm.
4 4 4(1.85)
t
n
λ λ
= = = = 
EVALUATE: Note that we must compare the path difference to the wavelength in the film. 
35.33. IDENTIFY: Consider the interference between rays reflected from the two surfaces of the soap film. Strongly 
reflected means constructive interference. Consider phase difference due to the path difference of 2t and any phase 
difference due to phase changes upon reflection. 
(a) SET UP: Consider Figure 35.33. 
 
There is a 180° phase 
change when the light is 
reflected from the outside 
surface of the bubble and 
no phase change when the 
light is reflected from the 
inside surface. 
 
Figure 35.33 
EXECUTE: The reflections produce a net 180° phase difference and for there to be constructive interference the 
path difference 2t must correspond to a half-integer number of wavelengths to compensate for the / 2λ shift due to 
the reflections. Hence the condition for constructive interference is 0
12 ( / ), 0,1,2,
2
t m n mλ⎛ ⎞= + =⎜ ⎟
⎝ ⎠
… Here 0λ is 
the wavelength in air and 0( / )nλ is the wavelength in the bubble, where the path difference occurs. 
0
2 2(290 nm)(1.33) 771.4 nm
1 1 1
2 2 2
tn
m m m
λ = = =
+ + +
 
for 0, 1543 nm;m λ= = for 1, 514 nm;m λ= = for 2, 308 nm;m λ= = … Only 514 nm is in the visible region; 
the color for this wavelength is green. 
(b) 0
2 2(340 nm)(1.33) 904.4 nm
1 1 1
2 2 2
tn
m m m
λ = = =
+ + +
 
for 0, 1809 nm;m λ= = for 1, 603 nm;m λ= = for 2, 362 nm;m λ= = … Only 603 nm is in the visible region; 
the color for this wavelength is orange. 
EVALUATE: The dominant color of the reflected light depends on the thickness of the film. If the bubble has 
varying thickness at different points, these points will appear to be differentcolors when the light reflected from 
the bubble is viewed. 
35.34. IDENTIFY: The number of waves along the path is the path length divided by the wavelength. The path 
difference and the reflections determine the phase difference. 
SET UP: The path length is 62 17.52 10 mt −= × . The wavelength in the film is 0
n
λλ = . 
35-10 Chapter 35 
EXECUTE: (a) 648 nm 480 nm
1.35
λ = = . The number of waves is 
6
9
2 17.52 10 m 36.5
480 10 m
t
λ
−
−
×
= =
×
. 
(b) The path difference introduces a / 2λ , or 180° , phase difference. The ray reflected at the top surface of the 
film undergoes a 180° phase shift upon reflection. The reflection at the lower surface introduces no phase shift. 
Both rays undergo a 180° phase shift, one due to reflection and one due to reflection. The two effects cancel and 
the two rays are in phase as they leave the film. 
EVALUATE: Note that we must use the wavelength in the film to determine the number of waves in the film. 
35.35. IDENTIFY: Require destructive interference between light reflected from the two points on the disc. 
SET UP: Both reflections occur for waves in the plastic substrate reflecting from the reflective coating, so they 
both have the same phase shift upon reflection and the condition for destructive interference (cancellation) is 
1
22 ( )t m λ= + , where t is the depth of the pit. 
0
n
λλ = . The minimum pit depth is for 0m = . 
EXECUTE: 2
2
t λ= . 0 790 nm 110 nm 0.11 m
4 4 4(1.8)
t
n
λ λ μ= = = = = . 
EVALUATE: The path difference occurs in the plastic substrate and we must compare the wavelength in the 
substrate to the path difference. 
35.36. IDENTIFY: Consider light reflected at the front and rear surfaces of the film. 
SET UP: At the front surface of the film, light in air ( 1.00n = ) reflects from the film ( 2.62n = ) and there is a 
180° phase shift due to the reflection. At the back surface of the film, light in the film ( 2.62n = ) reflects from 
glass ( 1.62n = ) and there is no phase shift due to reflection. Therefore, there is a net 180° phase difference 
produced by the reflections. The path difference for these two rays is 2t, where t is the thickness of the film. The 
wavelength in the film is 505 nm
2.62
λ = . 
EXECUTE: (a) Since the reflection produces a net 180° phase difference, destructive interference of the reflected 
light occurs when 2t mλ= . 505 nm (96.4 nm)
2[2.62]
t m m⎛ ⎞= =⎜ ⎟
⎝ ⎠
. The minimum thickness is 96.4 nm. 
(b) The next three thicknesses are for 2m = , 3 and 4: 192 nm, 289 nm and 386 nm. 
EVALUATE: The minimum thickness is for / 2t nλ= . Compare this to Problem 34.27, where the minimum 
thickness for destructive interference is / 4t nλ= . 
35.37. IDENTIFY and SET UP: Apply Eq.(35.19) and calculate y for 1800.m = 
EXECUTE: Eq.(35.19): 9 4( / 2) 1800(633 10 m) / 2 5.70 10 m 0.570 mmy m λ − −= = × = × = 
EVALUATE: A small displacement of the mirror corresponds to many wavelengths and a large number of fringes 
cross the line. 
35.38. IDENTIFY: Apply Eq.(35.19). 
SET UP: 818m = . Since the fringes move in opposite directions, the two people move the mirror in opposite 
directions. 
EXECUTE: (a) For Jan, the total shift was 
7
41
1
818(6.06 10 m) 2.48 10 m.
2 2
my λ
−
−×= = = × For Linda, the total shift 
was 
7
42
2
818(5.02 10 m) 2.05 10 m.
2 2
my λ
−
−×= = = × 
(b) The net displacement of the mirror is the difference of the above values: 
1 2 0.248 mm 0.205 mm 0.043 mm.y y yΔ = − = − = 
EVALUATE: The person using the larger wavelength moves the mirror the greater distance. 
35.39. IDENTIFY: Consider the interference between light reflected from the top and bottom surfaces of the air film 
between the lens and the glass plate. 
SET UP: For maximum intensity, with a net half-cycle phase shift due to reflections, 12
2
t m λ⎛ ⎞= +⎜ ⎟
⎝ ⎠
. 
2 2t R R r= − − . 
EXECUTE: 2 2 2 2(2 1) (2 1)
4 4
m mR R r R r Rλ λ+ += − − ⇒ − = − 
2 2
2 2 2 (2 1) (2 1) (2 1) (2 1)
4 2 2 4
(2 1) , for .
2
m m R m R mR r R r
m Rr R
λ λ λ λ
λ λ
+ + + +⎡ ⎤ ⎡ ⎤⇒ − = + − ⇒ = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
+
⇒ ≈ "
 
Interference 35-11 
The second bright ring is when 1:m = 
7
4(2(1) 1) (5.80 10 m) (0.952 m) 9.10 10 m 0.910 mm.
2
r
−
−+ ×≈ = × = 
So the diameter of the second bright ring is 1.82 mm. 
EVALUATE: The diameter of the thm ring is proportional to 2 1m + , so the rings get closer together as m 
increases. This agrees with Figure 35.17b in the textbook. 
35.40. IDENTIFY: As found in Problem 35.39, the radius of the thm bright ring is (2 1) ,
2
m Rr λ+≈ for .R λ" 
SET UP: Introducing a liquid between the lens and the plate just changes the wavelength from to 
n
λλ , where n 
is the refractive index of the liquid. 
EXECUTE: (2 1) 0.850 mm( ) 0.737 mm.
2 1.33
m R rr n
n n
λ+
≈ = = = 
EVALUATE: The refractive index of the water is less than that of the glass plate, so the phase changes on 
reflection are the same as when air is in the space. 
35.41. IDENTIFY: The liquid alters the wavelength of the light and that affects the locations of the interference minima. 
SET UP: The interference minima are located by 12sin ( )d mθ λ= + . For a liquid with refractive index n, 
air
liq n
λλ = . 
EXECUTE: 
1
2( )sin constant
m
d
θ
λ
+
= = , so liqair
air liq
sinsin θθ
λ λ
= . liqair
air air
sinsin
/ n
θθ
λ λ
= and air
liq
sin sin35.20 1.730
sin sin19.46
n θ
θ
= = =
°
°
. 
EVALUATE: In the liquid the wavelength is shorter and 12sin ( )m d
λθ = + gives a smaller θ than in air, for the 
same m. 
35.42. IDENTIFY: As the brass is heated, thermal expansion will cause the two slits to move farther apart. 
SET UP: For destructive interference, d sin θ = λ/2. The change in separation due to thermal expansion is dw = 
αw0 dT, where w is the distance between the slits. 
EXECUTE: The first dark fringe is at d sin θ = λ/2 ⇒ sin θ = λ/2d. 
Call d ≡ w for these calculations to avoid confusion with the differential. sin θ = λ/2w 
Taking differentials gives d(sin θ) = d(λ/2w) and cosθ dθ = − λ/2 dw/w2. 
For thermal expansion, dw = αw0 dT, which gives 02
0 0
cos 
2 2
w dT dTd
w w
λ α λαθ θ = − = − . Solving for dθ gives 
0 02 cos
dTd
w
λαθ
θ
= − . Get λ: w0 sin θ0 = λ/2 → λ = 2w0 sinθ0. Substituting this quantity into the equation for dθ gives 
0 0
0
0 0
2 sin tan 
2 cos
w dTd dT
w
θ αθ θ α
θ
= − = − . 
5 1tan(32.5 )(2.0 10 K )(115 K) 0.001465 rad 0.084dθ − −= − × = − = −° ° 
The minus sign tells us that the dark fringes move closer together. 
EVALUATE: We can also see that the dark fringes move closer together because sinθ is proportional to 1/d, so as 
d increases due to expansion, θ decreases. 
35.43. IDENTIFY: Both frequencies will interfere constructively when the path difference from both of them is an 
integral number of wavelengths. 
SET UP: Constructive interference occurs when sinθ = mλ/d. 
EXECUTE: First find the two wavelengths. 
λ1 = v/f1 = (344 m/s)/(900 Hz) = 0.3822 m 
λ2 = v/f2 = (344 m/s)/(1200 Hz) = 0.2867 m 
To interfere constructively at the same angle, the angles must be the same, and hence the sines of the angles must 
be equal. Each sine is of the form sin θ = mλ/d, so we can equate the sines to get 
 m1λ1/d = m2λ2/d 
 m1(0.3822 m) = m2(0.2867 m) 
 m2 = 4/3 m1 
35-12 Chapter 35 
Since both m1 and m2 must be integers, the allowed pairs of values of m1 and m2 are 
 m1 = m2 = 0 
 m1 = 3, m2 = 4 
 m1 = 6, m2 = 8 
 m1 = 9, m2 = 12 
 etc. 
For m1 = m2 = 0, we have θ = 0. 
For m1 = 3, m2 = 4, we have sin θ1 = (3)(0.3822 m)/(2.50 m), giving θ1 = 27.3° 
For m1 = 6, m2 = 8, we have sin θ1 = (6)(0.3822 m)/(2.50 m), giving θ1 = 66.5° 
For m1 = 9, m2 = 12, we have sin θ1 = (9)(0.3822 m)/(2.50 m) = 1.38 > 1, so no angle is possible. 
EVALUATE: At certain other angles, one frequency will interfere constructively, but the other will not. 
35.44. IDENTIFY: For destructive interference, 2 1
1
2
d r r m λ⎛ ⎞= − = +⎜ ⎟
⎝ ⎠
. 
SET UP: 2 22 1 (200 m)r r x x− = + −EXECUTE: 
2
2 2 2 1 1(200 m) 2
2 2
x x m x mλ λ⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + + + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
. 
220,000 m 1 1 .
1 2 2
2
x m
m
λ
λ
⎛ ⎞= − +⎜ ⎟⎛ ⎞ ⎝ ⎠+⎜ ⎟
⎝ ⎠
 The wavelength is calculated by 
8
6
3.00 10 m s 51.7 m.
5.80 10 Hz
c
f
λ ×= = =
×
 
0 : 761 m; 1: 219 m; 2 : 90.1 m; 3; 20.0 m.m x m x m x m x= = = = = = = = 
EVALUATE: For 3m = , 3.5 181 md λ= = . The maximum possible path difference is the separation of 200 m 
between the sources. 
35.45. IDENTIFY: The two scratches are parallel slits, so the light that passes through them produces an interference 
pattern. However the light is traveling through a medium (plastic) that is different from air. 
SET UP: The central bright fringe is bordered by a dark fringe on each side of it. At these dark fringes, d sin θ = 
½ λ/n, where n is the refractive index of the plastic. 
EXECUTE: First use geometry to find the angles at which the two dark fringes occur. At the first dark fringe 
tanθ = [(5.82 mm)/2]/(3250 mm), giving θ = ±0.0513° 
For destructive interference, we have d sin θ = ½ λ/n and 
n = λ/(2dsin θ) = (632.8 nm)/[2(0.000225 m)(sin 0.0513°)] = 1.57 
EVALUATE: The wavelength of the light in the plastic is reduced compared to what it would be in air. 
35.46. IDENTIFY: Interference occurs due to the path difference of light in the thin film. 
SET UP: Originally the path difference was an odd number of half-wavelengths for cancellation to occur. If the 
path difference decreases by ½ wavelength, it will be a multiple of the wavelength, so constructive interference 
will occur. 
EXECUTE: Calling ΔT the thickness that must be removed, we have 
path difference = 2ΔT = ½ λ/n and ΔT = λ/4n = (525 nm)/[4(1.40)] = 93.75 nm, 
At 4.20 nm/yr, we have (4.20 nm/yr)t = 93.75 nm and t = 22.3 yr. 
EVALUATE: If you were giving a warranty on this film, you certainly could not give it a �lifetime guarantee�! 
35.47. IDENTIFY and SET UP: If the total phase difference is an integer number of cycles the interference is constructive 
and if it is a half-integer number of cycles it is destructive. 
EXECUTE: (a) If the two sources are out of phase by one half-cycle, we must add an extra half a wavelength to 
the path difference equations Eq.(35.1) and Eq.(35.2). This exactly changes one for the other, for 
1 1
2 2and ,m m m m→ + + → since m in any integer. 
(b) If one source leads the other by a phase angleφ , the fraction of a cycle difference is .
2
φ
π
 Thus the path length 
difference for the two sources must be adjusted for both destructive and constructive interference, by this amount. 
So for constructive inference: 1 2 ( 2 ) ,r r m φ π λ− = + and for destructive interference, 1 2 ( 1 2 2 )r r m φ π λ− = + + , 
where in each case 0, 1, 2,m = ± ± … 
EVALUATE: If 0φ = these results reduce to Eqs.(35.1) and (35.2). 
35.48. IDENTIFY: Follow the steps specified in the problem. 
SET UP: Use cos( / 2) cos( )cos( / 2) sin( )sin( / 2)t t tω φ ω φ ω φ+ = − . Then 
22cos( / 2)cos( / 2) 2cos( )cos ( / 2) 2sin( )sin( / 2)cos( / 2)t t tφ ω φ ω φ ω φ φ+ = − . Then use 2 1 cos( )cos ( / 2)
2
φφ += and 
Interference 35-13 
2sin( / 2)cos( / 2) sinφ φ φ= . This gives cos( ) (cos( )cos( ) sin( )sin( )) cos( ) cos( )t t t t tω ω φ ω φ ω ω φ+ − = + + , using 
again the trig identity for the cosine of the sum of two angles. 
EXECUTE: (a) The electric field is the sum of the two fields and can be written as 
2 1( ) ( ) ( ) cos( ) cos( )PE t E t E t E t E tω ω φ= + = + + . ( ) 2 cos( / 2)cos( / 2)PE t E tφ ω φ= + . 
(b) ( ) cos( / 2),pE t A tω φ= + so comparing with part (a), we see that the amplitude of the wave (which is always 
positive) must be 2 | cos( / 2) | .A E φ= 
(c) To have an interference maximum, 2
2
mφ π= . So, for example, using 1,m = the relative phases are 
2 1: 0; : 4 ; : 22p
E E E φφ π π= = , and all waves are in phase. 
(d) To have an interference minimum, 1 .
2 2
mφ π ⎛ ⎞= +⎜ ⎟
⎝ ⎠
 So, for example using 0,m = relative phases are 
2 1: 0; : ; : /2 /2,pE E Eφ π φ π= = and the resulting wave is out of phase by a quarter of a cycle from both of the 
original waves. 
(e) The instantaneous magnitude of the Poynting vector is 
2 2 2 2
0 0| | ( ) (4 cos ( 2)cos ( 2)).pcE t c E tε ε φ ω φ= = +S
$%
 
For a time average, 2 2 2av 0
1cos ( 2) , so 2 cos ( 2).
2
t S cEω φ ε φ+ = = 
EVALUATE: The result of part (e) shows that the intensity at a point depends on the phase difference φ at that 
point for the waves from each source. 
35.49. IDENTIFY: Follow the steps specified in the problem. 
SET UP: The definition of hyperbola is the locus of points such that the difference between 2 1to and toP S P S is 
a constant. 
EXECUTE: (a) r mλΔ = . 2 21 ( )r x y d= + − and 
2 2
2 ( )r x y d= + + . 
2 2 2 2( ) ( )r x y d x y d mλΔ = + + − + − = . 
(b) For a given andm λ , rΔ is a constant and we get a hyperbola. Or, in the case of all m for a given λ , a family 
of hyperbolas. 
(c) 2 2 2 2 12( ) ( ) ( ) .x y d x y d m λ+ + − + − = + 
EVALUATE: The hyperbolas approach straight lines at large distances from the source. 
35.50. IDENTIFY: Follow the derivation of Eq.(35.7), but with different amplitudes for the two waves. 
SET UP: cos( ) cosπ φ φ− = − 
EXECUTE: (a) 2 2 2 2 2 21 2 1 22 cos( ) 4 4 cospE E E E E E E Eπ φ φ= + − − = + + =
2 25 4 cosE E φ+ 
2 2 2 2
0 0 0 0
1 5 4 9cos . 0 .
2 2 2 2p
I cE c E E I cEε ε φ φ ε⎡ ⎤⎛ ⎞ ⎛ ⎞= = + = ⇒ =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
 Therefore, 0
5 4 cos .
9 9
I I φ⎡ ⎤= +⎢ ⎥⎣ ⎦
 
(b) The graph is shown in Figure 35.50. min 0
1 which occurs when ( odd).
9
I I n nφ π= = 
EVALUATE: The maxima and minima occur at the same points on the screen as when the two sources have the 
same amplitude, but when the amplitudes are different the intensity is no longer zero at the minima. 
 
Figure 35.50 
35-14 Chapter 35 
35.51. IDENTIFY and SET UP: Consider interference between rays reflected from the upper and lower surfaces of the 
film to relate the thickness of the film to the wavelengths for which there is destructive interference. The thermal 
expansion of the film changes the thickness of the film when the temperature changes. 
EXECUTE: For this film on this glass, there is a net / 2λ phase change due to reflection and the condition for 
destructive interference is 2 ( / ),t m nλ= where 1.750.n = 
Smallest nonzero thickness is given by / 2 .t nλ= 
At 020.0 C, (582.4 nm) /[(2)(1.750)] 166.4 nm.t° = = 
At 0170 C, (588.5 nm) /[(2)(1.750)] 168.1 nm.t° = = 
0 (1 )t t Tα= + Δ so 
5 1
0 0( ) /( ) (1.7 nm) /[(166.4 nm)(150 C)] 6.8 10 (C )t t t Tα
− −= − Δ = ° = × ° 
EVALUATE: When the film is heated its thickness increases, and it takes a larger wavelength in the film to equal 
2t.The value we calculated for α is the same order of magnitude as those given in Table 17.1. 
35.52. IDENTIFY and SET UP: At the 3m = bright fringe for the red light there must be destructive interference at this 
same θ for the other wavelength. 
EXECUTE: For constructive interference: 1sin sin 3(700 nm) 2100 nm.d m dθ λ θ= ⇒ = = For destructive 
interference: 2 2 1 1
2 2
1 sin 2100 nmsin .
2
dd m
m m
θθ λ λ⎛ ⎞= + ⇒ = =⎜ ⎟ + +⎝ ⎠
 So the possible wavelengths are 
2 2600 nm, for 3, and 467 nm, for 4.m mλ λ= = = = 
EVALUATE: Both andd θ drop out of the calculation since their combination is just the path difference, which is 
the same for both types of light. 
35.53. IDENTIFY: Apply 0 cos sin
dI I π θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 
SET UP: 0 / 2I I= when sin
dπ θ
λ
 is rad
4
π , 3 rad
4
π ,�. 
EXECUTE: First we need to find the angles at which the intensity drops by one-half from the value of the thm 
bright fringe. 2 00 cos sin sin ( 1 2) .2 2
md I d dI I mπ π π θ πθ θ
λ λ λ
⎛ ⎞= = ⇒ ≈ = +⎜ ⎟
⎝ ⎠
 
30 : ; 1:
4 4 2m m m
m m
d d d
λ λ λθ θ θ θ θ− += = = = = = ⇒ Δ = . 
EVALUATE: There is no dependence on the m-value of the fringe, so all fringes at small angles have the same 
half-width. 
34.54. IDENTIFY: Consider the phase difference produced by the path difference and by the reflections. 
SETUP: There is just one half-cycle phase change upon reflection, so for constructive interference 
1 1
1 1 2 22 22 ( ) ( )t m mλ λ= + = + , where these wavelengths are in the glass. The two different wavelengths differ by just 
one 2 1-value, 1.m m m= − 
EXECUTE: 1 2 1 21 1 1 2 1 2 1 1
2 1
1 1 ( )
2 2 2 2( )
m m m mλ λ λ λλ λ λ λ
λ λ
+ +⎛ ⎞ ⎛ ⎞+ = − ⇒ − = ⇒ =⎜ ⎟ ⎜ ⎟ −⎝ ⎠ ⎝ ⎠
. 
01
1
477.0 nm 540.6 nm 1 17(477.0 nm)8. 2 8 1334 nm.
2(540.6 nm 477.0 nm) 2 4(1.52)
m t t
n
λ+ ⎛ ⎞= = = + ⇒ = =⎜ ⎟− ⎝ ⎠
 
EVALUATE: Now that we have t we can calculate all the other wavelengths for which there is constructive 
interference. 
35.55. IDENTIFY: Consider the phase difference due to the path difference and due to the reflection of one ray from the 
glass surface. 
(a) SET UP: Consider Figure 35.55 
 
path 
difference = 
2 22 / 4h x x+ − = 
2 24h x x+ − 
 
Figure 35.55 
Interference 35-15 
Since there is a 180° phase change for the reflected ray, the condition for constructive interference is path 
difference 1
2
m λ⎛ ⎞= +⎜ ⎟
⎝ ⎠
 and the condition for destructive interference is path difference .mλ= 
(b) EXECUTE: Constructive interference: 2 21 4
2
m h x xλ⎛ ⎞+ = + −⎜ ⎟
⎝ ⎠
 and 
2 24 .1
2
h x x
m
λ + −=
+
 Longest λ is for 
0m = and then ( ) ( )2 2 2 22 4 2 4(0.24 m) (0.14 m) 0.14 m 0.72 mh x xλ = + − = + − = 
EVALUATE: For 0.72 mλ = the path difference is / 2.λ 
35.56. IDENTIFY: Require constructive interference for the reflection from the top and bottom surfaces of each 
cytoplasm layer and each guanine layer. 
SET UP: At the water (or cytoplasm) to guanine interface, there is a half-cycle phase shift for the reflected light, 
but there is not one at the guanine to cytoplasm interface. Therefore there will always be one half-cycle phase 
difference between two neighboring reflected beams, just due to the reflections. 
EXECUTE: For the guanine layers: 
g g
g 1 1 1
g 2 2 2
21 2(74 nm) (1.80) 266 nm2 ( ) 533 nm ( 0).
2 ( ) ( ) ( )
t n
t m m
n m m m
λ λ λ= + ⇒ = = = ⇒ = =
+ + +
 
For the cytoplasm layers: 
c c
c 1 1 1
c 2 2 2
1 2 2(100 nm) (1.333) 267 nm2 533 nm ( 0).
2 ( ) ( ) ( )
t nt m m
n m m m
λ λ λ⎛ ⎞= + ⇒ = = = ⇒ = =⎜ ⎟ + + +⎝ ⎠
 
(b) By having many layers the reflection is strengthened, because at each interface some more of the transmitted 
light gets reflected back, increasing the total percentage reflected. 
(c) At different angles, the path length in the layers changes (always to a larger value than the normal incidence 
case). If the path length changes, then so do the wavelengths that will interfere constructively upon reflection. 
EVALUATE: The thickness of the guanine and cytoplasm layers are inversely proportional to their refractive 
indices 100 1.80
74 1.333
⎛ ⎞=⎜ ⎟
⎝ ⎠
, so both kinds of layers produce constructive interference for the same wavelength in air. 
35.57. IDENTIFY: The slits will produce an interference pattern, but in the liquid, the wavelength of the light will be less 
than it was in air. 
SET UP: The first bright fringe occurs when d sin θ = λ/n. 
EXECUTE: In air: dsin18.0° = λ. In the liquid: dsin12.6° = λ/n. Dividing the equations gives 
n = (sin 18.0°)/(sin 12.6°) = 1.42 
EVALUATE: It was not necessary to know the spacing of the slits, since it was the same in both air and the liquid. 
35.58. IDENTIFY: Consider light reflected at the top and bottom surfaces of the film. Wavelengths that are predominant 
in the transmitted light are those for which there is destructive interference in the reflected light. 
SET UP: For the waves reflected at the top surface of the oil film there is a half-cycle reflection phase shift. For 
the waves reflected at the bottom surface of the oil film there is no reflection phase shift. The condition for 
constructive interference is 122 ( )t m λ= + . The condition for destructive interference is 2t mλ= . The range of 
visible wavelengths is approximately 400 nm to 700 nm. In the oil film, 0
n
λλ = . 
EXECUTE: (a) 01 12 22 ( ) ( )t m m n
λλ= + = + . 0 1 1 1
2 2 2
2 2(380 nm)(1.45) 1102 nmtn
m m m
λ = = =
+ + +
. 
0m = : 0 2200 nmλ = . 1m = : 0 735 nmλ = . 2m = : 0 441 nmλ = . 3m = : 0 315 nmλ = . The visible 
wavelength for which there is constructive interference in the reflected light is 441 nm. 
(b) 02t m m
n
λλ= = . 0
2 1102 nmtn
m m
λ = = . 1m = : 0 1102 nmλ = . 2m = : 0 551 nmλ = . 3m = : 0 367 nmλ = . 
The visible wavelength for which there is destructive interference in the reflected light is 551 nm. This is the 
visible wavelength predominant in the transmitted light. 
EVALUATE: At a particular wavelength the sum of the intensities of the reflected and transmitted light equals the 
intensity of the incident light. 
35.59. (a) IDENTIFY: The wavelength in the glass is decreased by a factor of 1/ ,n so for light through the upper slit a 
shorter path is needed to produce the same phase at the screen. Therefore, the interference pattern is shifted 
downward on the screen. 
(b) SET UP: Consider the total phase difference produced by the path length difference and also by the different 
wavelength in the glass. 
35-16 Chapter 35 
EXECUTE: At a point on the screen located by the angle θ the difference in path length is sin .d θ This 
introduces a phase difference of 
0
2 ( sin ),dπφ θ
λ
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
 where 0λ is the wavelength of the light in air or vacuum. 
In the thickness L of glass the number of wavelengths is 
0
.L nL
λ λ
= A corresponding length L of the path of the ray 
through the lower slit, in air, contains 0/L λ wavelengths. The phase difference this introduces is 
0 0
2 nL Lφ π
λ λ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
 and 02 ( 1)( / ).n Lφ π λ= − The total phase difference is the sum of these two, 
0 0
0
2 ( sin ) 2 ( 1)( / ) (2 / )( sin ( 1)).d n L d L nπ θ π λ π λ θ
λ
⎛ ⎞
+ − = + −⎜ ⎟
⎝ ⎠
 Eq.(35.10) then gives 
2
0
0
cos ( sin ( 1)) .I I d L nπ θ
λ
⎡ ⎤⎛ ⎞
= + −⎢ ⎥⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦
 
(c) Maxima means cos / 2 1φ = ± and / 2 , 0, 1, 2,m mφ π= = ± ± … 0( / )( sin ( 1))d L n mπ λ θ π+ − = 
0sin ( 1)d L n mθ λ+ − = 
0 ( 1)sin m L n
d
λθ − −= 
EVALUATE: When 0L → or 1n → the effect of the plate goes away and the maxima are located by Eq.(35.4). 
35.60. IDENTIFY: Dark fringes occur because the path difference is one-half of a wavelength. 
SET UP: At the first dark fringe, dsinθ = λ/2. The intensity at any angle θ is given by 20
sincos dI I π θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 
(a) At the first dark fringe, we have 
d sin θ = λ/2 
d/λ = 2/(2 sin 15.0°) = 1.93 
(b) 2 00
sincos
10
d II I π θ
λ
⎛ ⎞= =⎜ ⎟
⎝ ⎠
 ⇒ sin 1cos
10
dπ θ
λ
⎛ ⎞ =⎜ ⎟
⎝ ⎠
 
sin 1arccos
10
dπ θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 = 71.57° = 1.249 rad 
Using the result from part (a), that d/λ = 1.93, we have π(1.93)sin θ = 1.249. sin θ = 0.2060 and 
θ = ±11.9° 
EVALUATE: Since the first dark fringes occur at ±15.0°, it is reasonable that at ≈12° the intensity is reduced to 
only 1/10 of its maximum central value. 
35.61. IDENTIFY: There are two effects to be considered: first, the expansion of the rod, and second, the change in the 
rod�s refractive index. 
SET UP: 0
n
λλ = and 5 10 (2.50 10 (C ) )n n T
− −Δ = × Δ° . 6 10 (5.00 10 (C ) )L L T
− −Δ = × Δ° . 
EXECUTE: The extra length of rod replaces a little of the air so that the change in the number of wavelengths due 
to this is given by: glass glass 0air1
0 0 0
2 2( 1)2n L n L Tn LN
α
λ λ λ
Δ − ΔΔ
Δ = − = and 
6
1 7
2(1.48 1)(0.030 m)(5.00 10 C )(5.00 C ) 1.22.
5.89 10 m
N
−
−
− × ° °
Δ = =
×
 
The change in the number of wavelengths due to the change in refractive index of the rod is: 
5
glass 0
2 7
0
2 2(2.50 10 C )(5.00 C min)(1.00 min)(0.0300 m) 12.73.
5.89 10 m
n L
N
λ
−
−
Δ × ° °
Δ = = =
×
 
So, the total change in the number of wavelengths as the rod expands is 12.73 1.22 14.0NΔ = + = fringes/minute. 
EVALUATE: Both effects increase the number of wavelengths along the length of the rod. Both LΔ and 
glassnΔ are very small and the two effects can be considered separately. 
35.62. IDENTIFY:Apply Snell's law to the refraction at the two surfaces of the prism. 1S and 2S serve as coherent 
sources so the fringe spacing is Ry
d
λ
Δ = , where d is the distance between 1S and 2S . 
Interference 35-17 
SET UP: For small angles, sinθ θ≈ , withθ expressed in radians. 
EXECUTE: (a) Since we can approximate the angles of incidence on the prism as being small, Snell�s Law tells 
us that an incident angle of θ on the flat side of the prism enters the prism at an angle of ,nθ where n is the 
index of refraction of the prism. Similarly on leaving the prism, the in-going angle is / n Aθ − from the normal, 
and the outgoing angle, relative to the prism, is ( ).n n Aλ − So the beam leaving the prism is at an angle of 
( )n n A Aθ θ′ = − + from the optical axis. So ( 1) .n Aθ θ′− = − At the plane of the source 0S , we can calculate the 
height of one image above the source: tan( ) ( ) ( 1) 2 ( 1).
2
d a a n Aa d aA nθ θ θ θ′ ′= − ≈ − = − ⇒ = − 
(b) To find the spacing of fringes on a screen, we use 
7
3
3
(2.00 m 0.200 m) (5.00 10 m) 1.57 10 m.
2 ( 1) 2(0.200 m) (3.50 10 rad) (1.50 1.00)
R Ry
d aA n
λ λ − −
−
+ ×
Δ = = = = ×
− × −
 
EVALUATE: The fringe spacing is proportional to the wavelength of the light. The biprism serves as an 
alternative to two closely spaced narrow slits.
36-1 
DIFFRACTION 
 36.1. IDENTIFY: Use tany x θ= to calculate the angular position θ of the first minimum. The minima are located by 
Eq.(36.2): sin ,
m
a
λθ = 1, 2,m = ± ± … First minimum means 1m = and 1sin / aθ λ= and 1sin .aλ θ= Use this 
equation to calculate .λ 
SET UP: The central maximum is sketched in Figure 36.1. 
 
EXECUTE: 1 1tany x θ= 
1
1tan
y
x
θ = = 
3
31.35 10 m 0.675 10
2.00 m
−
−× = × 
3
1 0.675 10 radθ
−= × 
 
Figure 36.1 
3 3
1sin (0.750 10 m)sin(0.675 10 rad) 506 nmaλ θ
− −= = × × = 
EVALUATE: 1θ is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been 
used. 
 36.2. IDENTIFY: The angle is small, so m
m
y x
a
λ= . 
SET UP: 1 10.2 mmy = 
EXECUTE: 
7
5
1 3
1
(0.600 m)(5.46 10 m)
3.21 10 m.
10.2 10 m
x x
y a
a y
λ λ − −
−
×= ⇒ = = ×
×
 
EVALUATE: The diffraction pattern is observed at a distance of 60.0 cm from the slit. 
 36.3. IDENTIFY: The dark fringes are located at angles θ that satisfy sin , 1, 2, ....m m
a
λθ = = ± ± 
SET UP: The largest value of sinθ is 1.00. 
EXECUTE: (a) Solve for m that corresponds to sin 1θ = : 
3
9
0.0666 10 m
113.8
585 10 m
a
m
λ
−
−
×= = =
×
. The largest value m 
can have is 113. 1m = ± , 2± , …, 113± gives 226 dark fringes. 
(b) For 113m = ± , 
9
3
585 10 m
sin 113 0.9926
0.0666 10 m
θ
−
−
⎛ ⎞×= ± = ±⎜ ⎟×⎝ ⎠
 and 83.0θ = ± °. 
EVALUATE: When the slit width a is decreased, there are fewer dark fringes. When a λ< there are no dark 
fringes and the central maximum completely fills the screen. 
 36.4. IDENTIFY and SET UP: / aλ is very small, so the approximate expression m
m
y R
a
λ= is accurate. The distance 
between the two dark fringes on either side of the central maximum is 12y . 
EXECUTE: 
9
3
1 3
(633 10 m)(3.50 m)
2.95 10 m 2.95 mm
0.750 10 m
R
y
a
λ − −
−
×= = = × =
×
. 12 5.90 mmy = . 
EVALUATE: When a is decreased, the width 12y of the central maximum increases. 
 36.5. IDENTIFY: The minima are located by sin
m
a
λθ = 
SET UP: 12.0 cma = . 40.0 cmx = . 
36
36-2 Chapter 36 
EXECUTE: The angle to the first minimum is θ = arcsin
a
λ⎛ ⎞
⎜ ⎟
⎝ ⎠
 = arcsin
9.00 cm
48.6 .
12.00 cm
⎛ ⎞
= °⎜ ⎟
⎝ ⎠
 
So the distance from the central maximum to the first minimum is just 1 tany x θ= = 
(40.0 cm) tan(48.6 ) 45.4 cm.° = ± 
EVALUATE: 2 / aλ is greater than 1, so only the 1m = minimum is seen. 
 36.6. IDENTIFY: The angle that locates the first diffraction minimum on one side of the central maximum is given by 
sin
a
λθ = . The time between crests is the period T. 1f
T
= and v
f
λ = . 
SET UP: The time between crests is the period, so 1.0 hT = . 
EXECUTE: (a) 1
1 1
1.0 h
1.0 h
f
T
−= = = . 
1
800 km/h
800 km
1.0 h
v
f
λ −= = = . 
(b) Africa-Antarctica: 
800 km
sin
4500 km
θ = and 10.2θ = ° . 
Australia-Antarctica: 
800 km
sin
3700 km
θ = and 12.5θ = ° . 
EVALUATE: Diffraction effects are observed when the wavelength is about the same order of magnitude as the 
dimensions of the opening through which the wave passes. 
 36.7. IDENTIFY: We can model the hole in the concrete barrier as a single slit that will produce a single-slit diffraction 
pattern of the water waves on the shore. 
SET UP: For single-slit diffraction, the angles at which destructive interference occurs are given by sinθm = mλ/a, 
where m = 1, 2, 3, …. 
EXECUTE: (a) The frequency of the water waves is f = 75.0 1 1min 1.25 s 1.25− −= = Hz, so their wavelength is λ = v/f = 
(15.0 cm/s)/(1.25 Hz) = 12.0 cm. 
At the first point for which destructive interference occurs, we have 
tan θ = (0.613 m)/(3.20 m) ⇒ θ = 10.84°. a sin θ = λ and 
a = λ/sin θ = (12.0 cm)/(sin 10.84°) = 63.8 cm. 
(b) First find the angles at which destructive interference occurs. 
sin θ2 = 2λ/a = 2(12.0 cm)/(63.8 cm) → θ2 = ±22.1° 
sin θ3 = 3λ/a = 3(12.0 cm)/(63.8 cm) → θ3 = ±34.3° 
sin θ4 = 4λ/a = 4(12.0 cm)/(63.8 cm) → θ4 = ±48.8° 
sin θ5 = 5λ/a = 5(12.0 cm)/(63.8 cm) → θ5 = ±70.1° 
EVALUATE: These are large angles, so we cannot use the approximation that θm ≈ mλ/a. 
 36.8. IDENTIFY: The minima are located by sin
m
a
λθ = . For part (b) apply Eq.(36.7). 
SET UP: For the first minimum, 1m = . The intensity at 0θ = is 0I . 
EXECUTE: (a) sin sin90.0 1
m m
a a a
λ λ λθ = = ° = = = . Thus 4580 nm 5.80 10 mm.a λ −= = = × 
(b) According to Eq.(36.7), 
[ ] [ ]2 2
0
sin (sin ) sin (sin / 4)
0.128.
(sin ) (sin / 4)
aI
I a
π θ λ π π
π θ λ π π
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= = =⎨ ⎬ ⎨ ⎬
⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
 
EVALUATE: If / 2a λ= , for example, then at 45θ = °, [ ]
2
0
sin ( / 2)(sin / 4)
0.81
( / 2)(sin / 4)
I
I
π π
π π
⎧ ⎫⎪ ⎪= =⎨ ⎬
⎪ ⎪⎩ ⎭
. As /a λ decreases, 
the screen becomes more uniformly illuminated. 
 36.9. IDENTIFY and SET UP: v f λ= gives .λ The person hears no sound at angles corresponding to diffraction 
minima. The diffraction minima are located by sin / , 1, 2,m a mθ λ= = ± ± … Solve for .θ 
EXECUTE: / (344 m/s) /(1250 Hz) 0.2752 m;v fλ = = = 1.00 m. 1,a m= = ± 16.0 ;θ = ± ° 2, m = ± 
33.4 ; 3, 55.6 ;mθ θ= ± ° = ± = ± ° no solution for larger m 
EVALUATE: / 0.28aλ = so for the large wavelength sound waves diffraction by the doorway is a large effect. 
Diffraction would not be observable for visible light because its wavelength is much smaller and / 1.aλ V 
Diffraction 36-3 
36.10. IDENTIFY: Compare yE to the expression max sin( )yE E kx tω= − and determine k, and from that calculate λ . 
/f c λ= . The dark bands are located by sin m
a
λθ = . 
SET UP: 83.00 10 m/sc = × . The first dark band corresponds to 1m = . 
EXECUTE: (a) max sin( )E E kx tω= − . 
7
7 1
2 2 2
5.24 10 m
1.20 10 m
k
k
π π πλ
λ
−
−= ⇒ = = = ××
. 
8
14
7
3.0 10 m s
5.73 10 Hz
5.24 10 m
c
f c fλ
λ −
×= ⇒ = = = ×
×
. 
(b) sina θ λ= . 
7
65.24 10 m 1.09 10 m
sin sin 28.6
a
λ
θ
−
−×= = = ×
°
. 
(c) sin ( 1, 2, 3, . . .)a m mθ λ= = . 
7
2 6
5.24 10 msin 2 2
1.09 10 ma
λθ
−
−
×= ± = ±
×
 and 2 74θ = ± . 
EVALUATE: For 3m = , m
a
λ
 is greater than 1 so only the first and second dark bands appear. 
36.11. IDENTIFY and SET UP: sin / aθ λ= locates the first minimum. tany x θ= . 
EXECUTE: tan (36.5 cm) (40.0 cm) and 42.38y xθ θ= = = ° . 
9sin (620 10 m) (sin 42.38 ) 0.920 ma λ θ μ−= = × ° = 
EVALUATE: 0.74 radθ = and sin 0.67θ = , so the approximation sinθ θ≈ would not be accurate. 
36.12. IDENTIFY: The angle is small, so m
m
y x
a
λ= applies. 
SET UP: The width of the central maximum is 12y , so 1 3.00 mmy = . 
EXECUTE: (a) 
7
4
1 3
1
(2.50 m)(5.00 10 m)
4.17 10 m.
3.00 10 m
x x
y a
a y
λ λ − −
−
×= ⇒ = = = ×
×
 
(b) 
5
2
3
1
(2.50 m)(5.00 10 m)
4.17 10 m4.2 cm.
3.00 10 m
x
a
y
λ − −
−
×= = = × =
×
 
(c) 
10
7
3
1
(2.50 m)(5.00 10 m)
4.17 10 m.
3.00 10 m
x
a
y
λ − −
−
×= = = ×
×
 
EVALUATE: The ratio /a λ stays constant, so a is smaller when λ is smaller. 
36.13. IDENTIFY: Calculate the angular positions of the minima and use tany x θ= to calculate the distance on the 
screen between them. 
(a) SET UP: The central bright fringe is shown in Figure 36.13a. 
 
EXECUTE: The first 
minimum is located by 
1sin a
λθ = =
9
3
3
633 10 m
1.809 10
0.350 10 m
−
−
−
× = ×
×
 
3
1 1.809 10 radθ
−= × 
 
Figure 36.13a 
3 3
1 1tan (3.00 m) tan(1.809 10 rad) 5.427 10 my x θ
− −= = × = × 
3 2
12 2(5.427 10 m) 1.09 10 m 10.9 mmw y
− −= = × = × = 
(b) SET UP: The first bright fringe on one side of the central maximum is shown in Figure 36.13b. 
 
EXECUTE: 2 1w y y= − 
3
1 5.427 10 m (part (a))y
−= × 
3
2
2
sin 3.618 10
a
λθ −= = × 
3
2 3.618 10 radθ
−= × 
2
2 2tan 1.085 10 my x θ
−= = × 
 
Figure 36.13b 
2 3
2 1 1.085 10 m 5.427 10 m 5.4 mmw y y
− −= − = × − × = 
EVALUATE: The central bright fringe is twice as wide as the other bright fringes. 
36-4 Chapter 36 
36.14. IDENTIFY: 
2
0
sin( / 2)
/ 2
I I
β
β
⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 
2
sina
πβ θ
λ
= . 
SET UP: The angle θ is small, so sin tan /y xθ θ≈ ≈ . 
EXECUTE: 
4
1
7
2 2 2 (4.50 10 m)
sin (1520 m ) .
(6.20 10 m)(3.00 m)
a a y
y y
x
π π πβ θ
λ λ
−
−
−
×= ≈ = =
×
 
(a) 
1 3
3 (1520 m )(1.00 10 m)1.00 10 m : 0.760.
2 2
y
β − −− ×= × = = 
2 2
0 0 0
sin( 2) sin(0.760)
0.822
2 0.760
I I I I
β
β
⎛ ⎞ ⎛ ⎞⇒ = = =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 
(b) 
1 3
3 (1520 m )(3.00 10 m)3.00 10 m : 2.28.
2 2
y
β − −− ×= × = = 
2 2
0 0 0
sin( 2) sin(2.28)
0.111 .
2 2.28
I I I I
β
β
⎛ ⎞ ⎛ ⎞⇒ = = =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 
(c) 
1 3
3 (1520 m )(5.00 10 m)5.00 10 m : 3.80.
2 2
y
β − −− ×= × = = 
2 2
0 0 0
sin( 2) sin(3.80)
0.0259 .
2 3.80
I I I I
β
β
⎛ ⎞ ⎛ ⎞⇒ = = =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 
EVALUATE: The first minimum occurs at 
7
1 4
(6.20 10 m)(3.00 m)
4.1 mm
4.50 10 m
x
y
a
λ −
−
×= = =
×
. The distances in parts (a) 
and (b) are within the central maximum. 5.00 mmy = is within the first secondary maximum. 
36.15. (a) IDENTIFY: Use Eq.(36.2) with 1m = to locate the angular position of the first minimum and then use 
tany x θ= to find its distance from the center of the screen. 
SET UP: The diffraction pattern is sketched in Figure 36.15. 
 
1sin a
λθ = = 
9
3
3
540 10 m
2.25 10
0.240 10 m
−
−
−
× = ×
×
 
3
1 2.25 10 radθ
−= × 
 
Figure 36.15 
3 3
1 1tan (3.00 m) tan(2.25 10 rad) 6.75 10 m 6.75 mmy x θ
− −= = × = × = 
(b) IDENTIFY and SET UP: Use Eqs.(36.5) and (36.6) to calculate the intensity at this point. 
EXECUTE: Midway between the center of the central maximum and the first minimum implies 
31 (6.75 mm) 3.375 10 m.
2
y −= = × 
3
3 33.375 10 mtan 1.125 10 ; 1.125 10 rad
3.00 m
y
x
θ θ
−
− −×= = = × = × 
The phase angle β at this point on the screen is 
3 3
9
2 2
sin (0.240 10 m)sin(1.125 10 rad) .
540 10 m
a
π πβ θ π
λ
− −
−
⎛ ⎞= = × × =⎜ ⎟ ×⎝ ⎠
 
Then 
2 2
6 2
0
sin / 2 sin / 2
(6.00 10 W/m )
/ 2 / 2
I I
β π
β π
−⎛ ⎞ ⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 
6 2 6 2
2
4
(6.00 10 W/m ) 2.43 10 W/m .I
π
− −⎛ ⎞= × = ×⎜ ⎟
⎝ ⎠
 
EVALUATE: The intensity at this point midway between the center of the central maximum and the first 
minimum is less than half the maximum intensity. Compare this result to the corresponding one for the two-slit 
pattern, Exercise 35.23. 
Diffraction 36-5 
36.16. IDENTIFY: In the single-slit diffraction pattern, the intensity is a maximum at the center and zero at the dark 
spots. At other points, it depends on the angle at which one is observing the light. 
SET UP: Dark fringes occur when sin θm = mλ/a, where m = 1, 2, 3, …, and the intensity is given by 
2
0
sin / 2
/ 2
I
β
β
⎛ ⎞
⎜ ⎟
⎝ ⎠
, where 
sin
/ 2
aπ θβ
λ
= . 
EXECUTE: (a) At the maximum possible angle, θ = 90°, so 
mmax = (asin90°)/λ = (0.0250 mm)/(632.8 nm) = 39.5 
Since m must be an integer and sin θ must be ≤ 1, mmax = 39. The total number of dark fringes is 39 on each side of 
the central maximum for a total of 78. 
(b) The farthest dark fringe is for m = 39, giving 
sinθ39 = (39)(632.8 nm)/(0.0250 mm) ⇒ θ39 = ±80.8° 
(c) The next closer dark fringe occurs at sinθ38 = (38)(632.8 nm)/(0.0250 mm) ⇒θ38 = 74.1°. 
The angle midway these two extreme fringes is (80.8° + 74.1°)/2 = 77.45°, and the intensity at this angle is I = 
2
0
sin / 2
/ 2
I
β
β
⎛ ⎞
⎜ ⎟
⎝ ⎠
, where 
sin (0.0250 mm)sin(77.45 )
/ 2
632.8 nm
aπ θ πβ
λ
°= = = 121.15 rad, which 
gives ( )
2
2 sin(121.15 rad)8.50 W/m
121.15 rad
I
⎡ ⎤= ⎢ ⎥⎣ ⎦
 = 5.55 × 10-4 W/m2 
EVALUATE: At the angle in part (c), the intensity is so low that the light would be barely perceptible. 
36.17. IDENTIFY and SET UP: Use Eq.(36.6) to calculate λ and use Eq.(36.5) to calculate I. 3.25 ,θ = ° 
356.0 rad, 0.105 10 m.aβ −= = × 
(a) EXECUTE: 
2
sina
πβ θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 so 
32 sin 2 (0.105 10 m)sin3.25
668 nm
56.0 rad
aπ θ πλ
β
−× °= = = 
(b) 
2
2 2 5
0 0 0 02 2
sin / 2 4 4
(sin( / 2)) [sin(28.0 rad)] 9.36 10
/ 2 (56.0 rad)
I I I I I
β β
β β
−⎛ ⎞ ⎛ ⎞= = = = ×⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
EVALUATE: At the first minimum 2β π= rad and at the point considered in the problem 17.8β π= rad, so the 
point is well outside the central maximum. Since β is close to mπ with 18,m = this point is near one of the 
minima. The intensity here is much less than 0.I 
36.18. IDENTIFY: Use 
2
sin
aπβ θ
λ
= to calculate β . 
SET UP: The total intensity is given by drawing an arc of a circle that has length 0E and finding the length of the 
chord which connects the starting and ending points of the curve. 
EXECUTE: (a) 
2 2
sin .
2
a a
a
π π λβ θ π
λ λ
= = = From Figure 36.18a, 0 0
2
.
2
p
p
E
E E Eπ
π
= ⇒ = 
The intensity is 
2
0
0 02
2 4
0.405 .
I
I I I
π π
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
 This agrees with Eq.(36.5). 
(b)
2 2
sin 2 .
a a
a
π π λβ θ π
λ λ
= = = From Figure 36.18b, it is clear that the total amplitude is zero, as is the intensity. 
This also agrees with Eq.(36.5). 
(c)
2 2 3
sin 3 .
2
a a
a
π π λβ θ π
λ λ
= = = From Figure 36.18c, 0 0
2
3 .
2 3
p
p
E
E E Eπ
π
= ⇒ = The intensity is 
2
0 02
2 4
.
3 9
I I I
π π
⎛ ⎞= =⎜ ⎟
⎝ ⎠
 This agrees with Eq.(36.5). 
36-6 Chapter 36 
EVALUATE: In part (a) the point is midway between the center of the central maximum and the first minimum. 
In part (b) the point is at the first maximum and in (c) the point is approximately at the location of the first 
secondary maximum. The phasor diagrams help illustrate the rapid decrease in intensity at successive maxima. 
 
Figure 36.18 
36.19. IDENTIFY: The space between the skyscrapers behaves like a single slit and diffracts the radio waves. 
SET UP: Cancellation of the waves occurs when a sin θ = mλ, m = 1, 2, 3, …, and the intensity of the waves is 
given by 
2
0
sin / 2
/ 2
I
β
β
⎛ ⎞
⎜ ⎟
⎝ ⎠
, where 
sin
/ 2
aπ θβ
λ
= . 
EXECUTE: (a) First find the wavelength of the waves: 
λ = c/f = (3.00 × 108 m/s)/(88.9 MHz) = 3.375 m 
For no signal, a sin θ = mλ. 
m = 1: sin θ1 = (1)(3.375 m)/(15.0 m) ⇒ θ1 = ±13.0° 
m = 2: sin θ2 = (2)(3.375 m)/(15.0 m) ⇒ θ2 = ±26.7° 
m = 3: sin θ3 = (3)(3.375 m)/(15.0 m) ⇒ θ3 = ±42.4° 
m = 4: sin θ4 = (4)(3.375 m)/(15.0 m) ⇒ θ4 = ±64.1° 
(b) 
2
0
sin / 2
/ 2
I
β
β
⎛ ⎞
⎜ ⎟
⎝ ⎠
, where 
sin (15.0 m)sin(5.00 )
/ 2
3.375 m
aπ θ πβ
λ
°= = = 1.217 rad 
( )
2
2 sin(1.217 rad)3.50 W/m
1.217 rad
I
⎡ ⎤= ⎢ ⎥⎣ ⎦
 = 2.08 W/m2 
EVALUATE: The wavelength of the radio waves is very long compared to that of visible light, but it is still 
considerably shorter than the distance between the buildings. 
36.20. IDENTIFY: The net intensity is the product of the factor due to single-slit diffraction and the factor due to double 
slit interference. 
SET UP: The double-slit factor is 2DS 0 cos 2
I I
φ⎛ ⎞= ⎜ ⎟
⎝ ⎠
 and the single-slit factor is 
2
SS
sin / 2
/ 2
Iβ
β
⎛ ⎞= ⎜ ⎟
⎝ ⎠
. 
EXECUTE: (a) d sinθ = mλ ⇒ sinθ = mλ/d. 
sinθ1 = λ/d, sinθ2 = 2λ/d, sinθ3 = 3λ/d, sinθ4 = 4λ/d 
(b) At the interference bright fringes, cos2φ/2 = 1 and sin ( /3)sin/ 2 a dπ θ π θβ
λ λ
= = . 
At θ1, sin θ1 = λ/d, so 
( /3)( / )
/ 2 /3
d dπ λβ π
λ
= = . The intensity is therefore 
2
2
1 0
sin / 2
cos
2 / 2
I I
φ β
β
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
 = 
2
0
sin /3
(1)
/3
I
π
π
⎛ ⎞
⎜ ⎟
⎝ ⎠
 = 0.684 I0 
At θ2, sin θ2 = 2λ/d, so 
( /3)(2 / )
/ 2 2 /3
d dπ λβ π
λ
= = . Using the same procedure as for θ1, we have I2 = 
2
0
sin 2 /3
(1)
2 /3
I
π
π
⎛ ⎞
⎜ ⎟
⎝ ⎠
= 0.171 I0 
At θ3, we get / 2β π= , which gives I3 = 0 since sin π = 0. 
At θ4, sin θ4 = 4λ/d, so / 2 4 /3β π= , which gives 
2
4 0
sin 4 /3
4 /3
I I
π
π
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 = 0.0427 I0 
(c) Since d = 3a, every third interference maximum is missing. 
(d) In Figure 36.12c in the textbook, every fourth interference maximum at the sides is missing because d = 4a. 
Diffraction 36-7 
EVALUATE: The result in this problem is different from that in Figure 36.12c because in this case d = 3a, so 
every third interference maximum at the sides is missing. Also the “envelope” of the intensity function decreases 
more rapidly here than in Figure 36.12c because the first diffraction minimum is reached sooner, and the decrease 
in intensity from one interference maximum to the next is faster for a = d/3 than for a = d/4. 
36.21. (a) IDENTIFY and SET UP: The interference fringes (maxima) are located by sin ,d mθ λ= with 
0, 1, 2, .m = ± ± … The intensity I in the diffraction pattern is given by 
2
0
sin / 2
,
/ 2
I I
β
β
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
 with 
2
sin .a
πβ θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 
We want 3m = ± in the first equation to give θ that makes 0I = in the second equation. 
EXECUTE: sind mθ λ= gives 2 3 2 (3 / ).a a d
d
π λβ π
λ
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
0I = says sin / 2 0
/ 2
β
β
= so 2β π= and then 2 2 (3 / )a dπ π= and ( / ) 3.d a = 
(b) IDENTIFY and SET UP: Fringes 0, 1, 2m = ± ± are within the central diffraction maximum and the 3m = ± 
fringes coincide with the first diffraction minimum. Find the value of m for the fringes that coincide with the 
second diffraction minimum. 
EXECUTE: Second minimum implies 4 .β π= 
2 2
sin 2 ( / ) 2 ( /3)
m
a a m a d m
d
π π λβ θ π π
λ λ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
 
Then 4β π= says 4 2 ( /3)mπ π= and 6.m = Therefore the 4m = ± and 5m = + fringes are contained within the 
first diffraction maximum on one side of the central maximum; two fringes. 
EVALUATE: The central maximum is twice as wide as the other maxima so it contains more fringes. 
36.22. IDENTIFY and SET UP: Use Figure 36.14b in the textbook. There is totally destructive interference between slits 
whose phasors are in opposite directions. 
EXECUTE: By examining the diagram, we see that every fourth slit cancels each other. 
EVALUATE: The total electric field is zero so the phasor diagram corresponds to a point of zero intensity. The 
first two maxima are at 0φ = and φ π= , so this point is not midway between two maxima. 
36.23. (a) IDENTIFY and SET UP: If the slits are very narrow then the central maximum of the diffraction pattern for 
each slit completely fills the screen and the intensity distribution is given solely by the two-slit interference. The 
maxima are given by 
sind mθ λ= so sin / .m dθ λ= Solve for .θ 
EXECUTE: 1st order maximum: 1,m = so 
9
3
3
580 10 m
sin 1.094 10 ;
0.530 10 md
λθ
−
−
−
×= = = ×
×
 0.0627θ = ° 
2nd order maximum: 2,m = so 32sin 2.188 10 ;
d
λθ −= = × 0.125θ = ° 
(b) IDENTIFY and SET UP: The intensity is given by Eq.(36.12): 
2
2
0
sin / 2
cos ( / 2) .
/ 2
I I
βφ
β
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 Calculate φ and 
β at each θ from part (a). 
EXECUTE: 
2 2
sin 2 ,
d d m
m
d
π π λφ θ π
λ λ
⎛ ⎞ ⎛ ⎞⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
 so 2 2cos ( / 2) cos ( ) 1mφ π= = 
(Since the angular positions in part (a) correspond to interference maxima.) 
2 2 0.320 mm
sin 2 ( / ) 2 (3.794 rad)
0.530 mm
a a m
m a d m m
d
π π λβ θ π π
λ λ
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
 
1st order maximum: 1,m = so 
2
0 0
sin(3.794/ 2) rad
(1) 0.249
(3.794 / 2) rad
I I I
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
 
2nd order maximum: 2,m = so 
2
0 0
sin3.794 rad
(1) 0.0256
3.794 rad
I I I
⎛ ⎞= =⎜ ⎟
⎝ ⎠
 
EVALUATE: The first diffraction minimum is at an angle θ given by sin / aθ λ= so 0.104 .θ = ° The first order 
fringe is within the central maximum and the second order fringe is inside the first diffraction maximum on one 
side of the central maximum. The intensity here at this second fringe is much less than 0.I 
36.24. IDENTIFY: A double-slit bright fringe is missing when it occurs at the same angle as a double-slit dark fringe. 
SET UP: Single-slit diffraction dark fringes occur when a sin θ = mλ, and double-slit interference bright fringes 
occur when d sin θ = m′ λ. 
36-8 Chapter 36 
EXECUTE: (a) The angles are the same for cancellation, so dividing the equations gives 
d/a = m′ /m ⇒ m′ /m = 7 ⇒ m′ = 7m 
When m = 1, m′ = 7; when m = 2, m′ = 14, and so forth, so every 7th bright fringe is missing from the double-slit 
interference pattern. 
EVALUATE: (b) The result is independent of the wavelength, so every 7th fringe will be cancelled for all 
wavelengths. But the bright interference fringes occur when d sin θ = mλ, so the location of the cancelled fringes 
does depend on the wavelength. 
36.25. IDENTIFY and SET UP: The phasor diagrams are similar to those in Fig.36.14. An interference minimum occurs 
when the phasors add to zero. 
EXECUTE: (a) The phasor diagram is given in Figure 36.25a 
 
Figure 36.25a 
There is destructive interference between the light through slits 1 and 3 and between 2 and 4. 
(b) The phasor diagram is given in Figure 36.25b. 
 
Figure 36.25b 
There is destructive interference between the light through slits 1 and 2 and between 3 and 4. 
(c) The phasor diagram is given in Figure 36.25c. 
 
Figure 36.25c 
There is destructive interference between light through slits 1 and 3 and between 2 and 4. 
EVALUATE: Maxima occur when 0, 2 , 4 , etc.φ π π= Our diagrams show that there are three minima between 
the maxima at 0φ = and 2 .φ π= This agrees with the general result that for N slits there are 1N − minima 
between each pair of principal maxima. 
36.26. IDENTIFY: A double-slit bright fringe is missing when it occurs at the same angle as a double-slit dark fringe. 
SET UP: Single-slit diffraction dark fringes occur when a sin θ = mλ, and double-slit interference bright fringes 
occur when d sin θ = m′ λ. 
EXECUTE: (a) The angle at which the first bright fringe occurs is given by 
tan θ1 = (1.53 mm)/(2500 mm) ⇒ θ1 = 0.03507°. d sin θ1 = λ and 
d = λ/(sinθ1) = (632.8 nm)/sin(0.03507°) = 0.00103 m = 1.03 mm 
(b) The 7th double-slit interference bright fringe is just cancelled by the 1st diffraction dark fringe, so sinθdiff = λ/a 
and sinθinterf = 7λ/d 
The angles are equal, so λ/a = 7λ/d → a = d/7 = (1.03 mm)/7 = 0.148 mm. 
EVALUATE: We can generalize that if d = na, where n is a positive integer, then every nth double-slit bright fringe 
will be missing in the pattern. 
36.27. IDENTIFY: The diffraction minima are located by dsin
m
a
λθ = and the two-slit interference maxima are located 
by isin .
m
d
λθ = The third bright band is missing because the first order single slit minimum occurs at the same 
angle as the third order double slit maximum. 
SET UP: The pattern is sketched in Figure 36.27. 
3 cm
tan
90 cm
θ = , so 1.91θ = ° . 
EXECUTE: Single-slit dark spot: sina θ λ= and 4500 nm 1.50 10 nm 15.0 m (width)
sin sin1.91
a
λ μ
θ
= = = × =
°
 
Double-slit bright fringe: sin 3d θ λ= and 43 3(500 nm) 4.50 10 nm 45.0 m (separation)
sin sin1.91
d
λ μ
θ
= = = × =
°
. 
Diffraction 36-9 
EVALUATE: Note that / 3.0d a = . 
 
Figure 36.27 
36.28. IDENTIFY: The maxima are located by sind mθ λ= . 
SET UP: The order corresponds to the values of m. 
EXECUTE:First-order: 1sind θ λ= . Fourth-order: 4sin 4d θ λ= . 
4
4 1 4
1
sin 4
, sin 4sin 4sin8.94 and 38.4
sin
d
d
θ λ θ θ θ
θ λ
= = = ° = ° . 
EVALUATE: We did not have to solve for d. 
36.29. IDENTIFY and SET UP: The bright bands are at angles θ given by sin .d mθ λ= Solve for d and then solve for θ 
for the specified order. 
EXECUTE: (a) 78.4θ = ° for 3m = and 681 nm,λ = so 4/ sin 2.086 10 cmd mλ θ −= = × 
The number of slits per cm is 1/ 4790 slits/cmd = 
(b) 1st order: 1,m = so 9 6sin / (681 10 m) /(2.086 10 m)dθ λ − −= = × × and 19.1θ = ° 
2nd order: 2,m = so sin 2 / dθ λ= and 40.8θ = ° 
(c) For 4, sin 4 /m dθ λ= = is greater than 1.00, so there is no 4th-order bright band. 
EVALUATE: The angular position of the bright bands for a particular wavelength increases as the order increases. 
36.30. IDENTIFY: The bright spots are located by sind mθ λ= . 
SET UP: Third-order means 3m = and second-order means 2m = . 
EXECUTE: constant
sin
m
d
λ
θ
= = , so r r v v
r vsin sin
m mλ λ
θ θ
= . 
v v
v r
r r
2 400 nm
sin sin (sin 65.0 ) 0.345
3 700 nm
m
m
λθ θ
λ
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
° and v 20.2θ = ° . 
EVALUATE: The third-order line for a particular λ occurs at a larger angle than the second-order line. In a given 
order, the line for violet light (400 nm) occurs at a smaller angle than the line for red light (700 nm). 
36.31. IDENTIFY and SET UP: Calculate d for the grating. Use Eq.(36.13) to calculate θ for the longest wavelength in 
the visible spectrum and verify that θ is small. Then use Eq.(36.3) to relate the linear separation of lines on the 
screen to the difference in wavelength. 
EXECUTE: (a) 5
1
 cm 1.111 10 m
900
d −
⎛ ⎞= = ×⎜ ⎟
⎝ ⎠
 
For 2700 nm, / 6.3 10 .dλ λ −= = × The first-order lines are located at sin / ;dθ λ= sinθ is small enough for 
sinθ θ≈ to be an excellent approximation. 
(b) / ,y x dλ= where 2.50 m.x = 
The distance on the screen between 1st order bright bands for two different wavelengths is ( ) / ,y x dλΔ = Δ so 
5 3( ) / (1.111 10 m)(3.00 10 m) /(2.50 m) 13.3 nmd y xλ − −Δ = Δ = × × = 
EVALUATE: The smaller d is (greater number of lines per cm) the smaller the λΔ that can be measured. 
36.32. IDENTIFY: The maxima are located by sind mθ λ= . 
SET UP: 6
5 1
1
350 slits mm 2.86 10 m
3.50 10 m
d −−⇒ = = ××
 
36-10 Chapter 36 
EXECUTE: 
7
400 6
4.00 10 m
1: arcsin arcsin 8.05
2.86 10 m
m
d
λθ
−
−
⎛ ⎞×⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
. 
7
700 6
7.00 10 m
arcsin arcsin 14.18
2.86 10 md
λθ
−
−
⎛ ⎞×⎛ ⎞= = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
. 1 14.18 8.05 6.13θΔ = ° − ° = °. 
7
400 6
3 3(4.00 10 m)
3: arcsin arcsin 24.8
2.86 10 m
m
d
λθ
−
−
⎛ ⎞×⎛ ⎞= = = = °⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
. 
7
700 6
3 3(7.00 10 m)
arcsin arcsin 47.3
2.86 10 md
λθ
−
−
⎛ ⎞×⎛ ⎞= = = °⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
. 1 47.3 24.8 22.5θΔ = ° − ° = °. 
EVALUATE: θΔ is larger in third order. 
36.33. IDENTIFY: The maxima are located by sind mθ λ= . 
SET UP: 61.60 10 md −= × 
EXECUTE: 
7
6
[6.328 10 m]
arcsin arcsin arcsin([0.396] )
1.60 10 m
m m
m
d
λθ
−
−
⎛ ⎞×⎛ ⎞= = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
. For 11, 23.3m θ= = ° . For 
2m = , 2 52.3θ = ° . There are no other maxima. 
EVALUATE: The reflective surface produces the same interference pattern as a grating with slit separation d. 
36.34. IDENTIFY: The maxima are located by sind mθ λ= . 
SET UP: 6
5 1
1
5000 slits cm 2.00 10 m.
5.00 10 m
d −−⇒ = = ××
 
EXECUTE: (a) 
6
7sin (2.00 10 m)sin13.5 4.67 10 m.
1
d
m
θλ
−
−×= = = × 
(b) 
7
6
2(4.67 10 m)
2 : arcsin arcsin 27.8 .
2.00 10 m
m
m
d
λθ
−
−
⎛ ⎞×⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
 
EVALUATE: Since the angles are fairly small, the second-order deviation is approximately twice the first-order 
deviation. 
36.35. IDENTIFY: The maxima are located by sind mθ λ= . 
SET UP: 6
5 1
1
350 slits mm 2.86 10 m
3.50 10 m
d −−⇒ = = ××
 
EXECUTE: 
7
6
(5.20 10 m)
arcsin arcsin arcsin((0.182) )
2.86 10 m
m m
m
d
λθ
−
−
⎛ ⎞×⎛ ⎞= = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
. 
1: 10.5 ; 2 : 21.3 ; 3 : 33.1 .m m mθ θ θ= = ° = = ° = = ° 
EVALUATE: The angles are not precisely proportional to m, and deviate more from being proportional as the 
angles increase. 
36.36. IDENTIFY: The resolution is described by R Nm
λ
λ
= =
Δ
. Maxima are located by sind mθ λ= . 
SET UP: For 500 slits/mm, 1 1(500 slits mm) (500,000 slits m)d − −= = . 
EXECUTE: (a) 
7
7 7
6.5645 10 m
1820 slits.
2(6.5645 10 m 6.5627 10 m)
N
m
λ
λ
−
− −
×= = =
Δ × − ×
 
(b) 1sin
m
d
λθ − ⎛ ⎞= ⇒⎜ ⎟
⎝ ⎠
1 7 1
1 sin ((2)(6.5645 10 m)(500,000 m )) 41.0297θ
− − −= × = ° and 
1 7 1
2 sin ((2)(6.5627 10 m)(500,000 m )) 41.0160θ
− − −= × = ° . 0.0137θΔ = ° 
EVALUATE: cos /d d Nθ θ λ= , so for 1820 slits the angular interval θΔ between each of these maxima and the 
first adjacent minimum is 
7
6
6.56 10 m
0.0137 .
cos (1820)(2.0 10 m)cos41Nd
λθ
θ
−
°
−
×Δ = = =
× °
 This is the same as the angular 
separation of the maxima for the two wavelengths and 1820 slits is just sufficient to resolve these two wavelengths 
in second order. 
36.37. IDENTIFY: The resolving power depends on the line density and the width of the grating. 
SET UP: The resolving power is given by R = Nm = = λ/Δλ. 
EXECUTE: (a) R = Nm = (5000 lines/cm)(3.50 cm)(1) = 17,500 
(b) The resolving power needed to resolve the sodium doublet is 
R = λ/Δλ = (589 nm)/(589.59 nm – 589.00 nm) = 998 
Diffraction 36-11 
so this grating can easily resolve the doublet. 
(c) (i) R = λ/Δλ. Since R = 17,500 when m = 1, R = 2 × 17,500 = 35,000 for m = 2. Therefore 
Δλ = λ/R = (587.8 nm)/35,000 = 0.0168 nm 
λmin = λ +Δλ = 587.8002 nm + 0.0168 nm = 587.8170 nm 
(ii) max 587.8002 nm 0.0168 nm 587.7834 nm λ λ λ= − Δ = − = 
EVALUATE: (iii) Therefore the range of resolvable wavelengths is 587.7834 nm < λ < 587.8170 nm. 
36.38. IDENTIFY and SET UP: Nm
λ
λ
=
Δ
 
EXECUTE: 
587.8002 nm 587.8002
3302 slits
(587.9782 nm 587.8002 nm) 0.178
N
m
λ
λ
= = = =
Δ −
. 
3302 slits
2752 .
1.20 cm 1.20 cm cm
N = = 
EVALUATE: A smaller number of slits would be needed to resolve these two lines in higher order. 
36.39. IDENTIFY and SET UP: The maxima occur at angles θ given by Eq.(36.16), 2 sin ,d mθ λ= where d is the 
spacing between adjacent atomic planes. Solve for d. 
EXECUTE: second order says 2.m = 
9
102(0.0850 10 m) 2.32 10 m 0.232 nm
2sin 2sin 21.5
m
d
λ
θ
−
−×= = = × =
°
 
EVALUATE: Our result is similar to d calculated in Example 36.5. 
36.40. IDENTIFY: The maxima are given by 2 sind mθ λ= , 1m = , 2, … 
SET UP: 103.50 10 md −= × . 
EXECUTE: (a) 1m = and 10 102 sin 2(3.50 10 m)sin15.0 1.81 10 m 0.181 nmd
m
θλ − −= = × = × =° . This is an x ray. 
(b) 
10
10
1.81 10 m
sin (0.2586)
2 2[3.50 10 m]
m m m
d
λθ
−
−
⎛ ⎞×⎛ ⎞= = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
. 2m = : 31.1θ = ° . 3m = : 50.9θ = ° . The equation 
doesn't have any solutions for 3m > . 
EVALUATE: In this problem / 0.52.dλ = 
36.41. IDENTIFY: Rayleigh's criterion says sin 1.22
D
λθ = 
SET UP: The best resolution is 0.3 arcseconds, which is about 5(8.33 10 )−× ° . 
EXECUTE: (a) 
7
5
1.22 1.22(5.5 10 m)
0.46 m
sin sin(8.33 10 )
D
λ
θ
−
−
×= = =
× °
 
EVALUATE: (b) The Keck telescopes are able to gather more light than the Hale telescope, and 
hence they can detect fainter objects. However, their larger size does not allow them to have greater 
resolution⎯atmospheric conditions limit the resolution. 
36.42. IDENTIFY: Apply sin 1.22
D
λθ = . 
SET UP: (1/ 60)θ = ° 
EXECUTE: 
7
31.22 1.22(5.5 10 m) 2.31 10 m 2.3 mm
sin sin(1/ 60)
D
λ
θ
−
−×= = = × = 
EVALUATE: The larger the diameter the smaller the angle that can be resolved. 
36.43. IDENTIFY: Apply sin 1.22
D
λθ = . 
SET UP: 
W
h
θ = , where 28 kmW = and 1200 kmh = . θ is small, so sinθ θ≈ . 
EXECUTE: 
6
4
1.22 1.2 10 m
1.22 1.22(0.036 m) 1.88 m
sin 2.8 10 m
h
D
W
λ λ
θ
×= = = =
×
 
EVALUATE: D must be significantly larger than the wavelength, so a much larger diameter is needed for 
microwaves than for visible wavelengths. 
36.44.IDENTIFY: Apply sin 1.22
D
λθ = . 
SET UP: θ is small, so 8sin 1.00 10 radθ θ −≈ = × . 
EXECUTE: 
6 8sin (8.00 10 m)(1.00 10 )
0.0656 m 6.56 cm
1.22 1.22 1.22
D Dθ θλ
−× ×= ≈ = = = 
EVALUATE: λ corresponds to microwaves. 
36-12 Chapter 36 
36.45. IDENTIFY and SET UP: The angular size of the first dark ring is given by 1sin 1.22 / Dθ λ= (Eq.36.17). Calculate 
1,θ and then the diameter of the ring on the screen is 12(4.5 m) tan .θ 
EXECUTE: 
9
1 6
620 10 m
sin 1.22 0.1022;
7.4 10 m
θ
−
−
⎛ ⎞×= =⎜ ⎟×⎝ ⎠
 1 0.1024 radθ = 
The radius of the Airy disk (central bright spot) is 1(4.5 m) tan 0.462 m.r θ= = The diameter is 
2 0.92 m 92 cm.r = = 
EVALUATE: / 0.084.Dλ = For this small D the central diffraction maximum is broad. 
36.46. IDENTIFY: Rayleigh’s criterion limits the angular resolution. 
SET UP: Rayleigh’s criterion is sin θ ≈ θ = 1.22 λ/D. 
EXECUTE: (a) Using Rayleigh’s criterion 
sinθ ≈ θ = 1.22 λ/D = (1.22)(550 nm)/(135/4 mm) = 1.99 × 10–5 rad 
On the bear this angle subtends a distance x. θ = x/R and 
x = Rθ = (11.5 m)(1.99 × 10–5 rad) = 2.29 × 10–4 m = 0.23 mm 
(b) At f/22, D is 4/22 times as large as at f/4. Since θ is proportional to 1/D, and x is proportional to θ, x is 
1/(4/22) = 22/4 times as large as it was at f/4. x = (0.229 mm)(22/4) = 1.3 mm 
EVALUATE: A wide-angle lens, such as one having a focal length of 28 mm, would have a much smaller opening 
at f/22 and hence would have an even less resolving ability. 
36.47. IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means angular separation θ of the objects equals 
1.22 / .Dλ The angular separation θ of the objects is their linear separation divided by their distance from the 
telescope. 
EXECUTE: 
3
11
250 10 m
,
5.93 10 m
θ ×=
×
 where 115.93 10 m× is the distance from earth to Jupiter. Thus 74.216 10 .θ −= × 
Then 1.22
D
λθ = and 
9
7
1.22 1.22(500 10 m)
1.45 m
4.216 10
D
λ
θ
−
−
×= = =
×
 
EVALUATE: This is a very large telescope mirror. The greater the angular resolution the greater the diameter the 
lens or mirror must be. 
36.48. IDENTIFY: Rayleigh’s criterion says res 1.22 D
λθ = . 
SET UP: 7.20 cmD = . res
y
s
θ = , where s is the distance of the object from the lens and 4.00 mmy = . 
EXECUTE: 1.22
y
s D
λ= . 
3 2
9
(4.00 10 m)(7.20 10 m)
429 m
1.22 1.22(550 10 m)
yD
s
λ
− −
−
× ×= = =
×
. 
EVALUATE: The focal length of the lens doesn’t enter into the calculation. In practice, it is difficult to achieve 
resolution that is at the diffraction limit. 
36.49. IDENTIFY and SET UP: Let y be the separation between the two points being resolved and let s be their distance 
from the telescope. Then the limit of resolution corresponds to 1.22
y
D s
λ = . 
EXECUTE: (a) Let the two points being resolved be the opposite edges of the crater, so y is the diameter of the 
crater. For the moon, 83.8 10 m.s = × 1.22y s Dλ= . 
Hubble: D = 2.4 m and 400λ = nm gives the maximum resolution, so y = 77 m 
Arecibo: D = 305 m and 60.75 m; 1.1 10yλ = = × m 
(b) 
1.22
yD
s
λ
= . Let 0.30y ≈ (the size of a license plate). 9(0.30 m)(2.4 m) [(1.22)(400 10 m)] 1500 kms −= × = . 
EVALUATE: /D λ is much larger for the optical telescope and it has a much larger resolution even though the 
diameter of the radio telescope is much larger. 
36.50. IDENTIFY: Apply sin 1.22
D
λθ = . 
SET UP: θ is small, so sinθ θ≈ . Smallest resolving angle is for short-wavelength light (400 nm). 
EXECUTE: 
9
8400 10 m1.22 (1.22) 9.61 10 rad
5.08 mD
λθ
−
−×≈ = = × . 10,000 mi
R
θ = , where R is the distance to the star. 
11
8
10,000 mi 16,000 km
1.7 10 km
9.6 10 rad
R
θ −
= = = ×
×
. 
EVALUATE: This is less than a light year, so there are no stars this close. 
Diffraction 36-13 
36.51. IDENTIFY: Let y be the separation between the two points being resolved and let s be their distance from the 
telescope. The limit of resolution corresponds to 1.22 D y sλ = . 
SET UP: 164.28 ly 4.05 10 ms = = × . Assume visible light, with 400 mλ = . 
EXECUTE: 9 16 91.22 1.22(400 10 m)(4.05 10 m (10.0 m) 2.0 10 my s Dλ −= = × × = × 
EVALUATE: The diameter of Jupiter is 81.38 10 m,× so the resolution is insufficient, by about one order of 
magnitude. 
36.52. IDENTIFY: If the apparatus of Exercise 36.4 is placed in water, then all that changes is the wavelength 
n
λλ λ′→ = . 
SET UP: For y x<< , the distance between the two dark fringes on either side of the central maximum is 
2D y′ ′= . Let 2D y= be the separation of 35.91 10 m−× found in Exercise 36.4. 
EXECUTE: 
3
3
1
2 2 5.91 10 m
2 4.44 10 m 4.44 mm.
1.33
x x D
y
a an n
λ λ − −′ ×′ = = = = = × = 
EVALUATE: The water shortens the wavelength and this decreases the width of the central maximum. 
36.53. (a) IDENTIFY and SET UP: The intensity in the diffraction pattern is given by Eq.(36.5):
2
0
sin / 2
,
/ 2
I I
β
β
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 where 
2
sin .a
πβ θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 Solve for θ that gives 0
1
.
2
I I= The angles θ+ and θ− are shown in Figure 36.53. 
EXECUTE: 0
1
2
I I= so sin / 2 1
/ 2 2
β
β
= 
Let / 2;x β= the equation for x is sin 1 0.7071.
2
x
x
= = 
Use trial and error to find the value of x that is a solution to this equation. 
(sin ) /
1.0 rad 0.841
1.5 rad 0.665
1.2 rad 0.777
1.4 rad 0.7039
1.39 rad 0.7077; thus 1.39 rad and 2 2.78 rad
x x x
x xβ= = =
 
 
2θ θ θ θ+ − +Δ = − = 
sin
2 a
λβθ
π+
= = 
2.78 rad
0.4425
2 rada a
λ λ
π
⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
 
Figure 36.53 
(i) For 
1
2, sin 0.4425 0.2212; 12.78 ; 2 25.6
2
a θ θ θ θ
λ + + +
⎛ ⎞= = = = ° Δ = = °⎜ ⎟
⎝ ⎠
 
(ii) For 
1
5, sin 0.4425 0.0885; 5.077 ; 2 10.2
5
a θ θ θ θ
λ + + +
⎛ ⎞= = = = ° Δ = = °⎜ ⎟
⎝ ⎠
 
(iii) For 
1
10, sin 0.4425 0.04425; 2.536 ; 2 5.1
10
a θ θ θ θ
λ + + +
⎛ ⎞= = = = ° Δ = = °⎜ ⎟
⎝ ⎠
 
(b) IDENTIFY and SET UP: 0sin a
λθ = locates the first minimum. Solve for 0.θ 
EXECUTE: (i) For 0 0 0
1
2, sin ; 30.0 ; 2 60.0
2
a θ θ θ
λ
= = = ° = ° 
(ii) For 0 0 0
1
5, sin ; 11.54 ; 2 23.1
5
a θ θ θ
λ
= = = ° = ° 
36-14 Chapter 36 
(iii) For 0 0 0
1
10, sin ; 5.74 ; 2 11.5
10
a θ θ θ
λ
⎛ ⎞= = = ° = °⎜ ⎟
⎝ ⎠
 
EVALUATE: Either definition of the width shows that the central maximum gets narrower as the slit gets wider. 
36.54. IDENTIFY: The two holes behave like double slits and cause the sound waves to interfere after they pass through 
the holes. The motion of the speakers causes a Doppler shift in the wavelength of the sound. 
SET UP: The wavelength of the sound that strikes the wall is λ = λ0 – vsTs, and destructive interference first 
occurs where sin θ = λ/2. 
EXECUTE: (a) First find the wavelength of the sound that strikes the openings in the wall. 
λ = λ0 – vsTs = v/ fs – vs/ fs = (v – vs)/ fs = (344 m/s – 80.0 m/s)/(1250 Hz) = 0.211 m 
Destructive interference first occurs where d sin θ = λ/2, which gives 
d = λ/(2 sinθ) = (0.211 m)/(2 sin 12.7°) = 0.480 m 
(b) λ = v/f = (344 m/s)/(1250 Hz) = 0.275 m 
sinθ = λ/2d = (0.275 m)/[2(0.480 m)] → θ = ±16.7° 
EVALUATE: The moving source produces sound of shorter wavelength than the stationary source, so the angles at 
which destructive interference occurs are smaller for the moving source than for the stationary source. 
36.55. IDENTIFY and SET UP: sin / aθ λ= locates the first dark band. In the liquid the wavelength changes and this 
changes the angular position of the first diffraction minimum. 
EXECUTE: liquidairair liquidsin ; sina a
λλθ θ= = 
liquid
liquid air
air
sin
0.4836
sin
θ
λ λ
θ
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
 
air / nλ λ= (Eq.33.5), so air liquid/ 1/ 0.4836 2.07n λ λ= = = 
EVALUATE: Light travels faster in air and n must be 1.00.> The smaller λ in the liquid reduces θ that located 
the first dark band. 
36.56. IDENTIFY: 
1
d
N
= , so the bright fringes are located by 1 sin
N
θ λ= 
SET UP: Red: R
1 sin 700 nm
N
λ = . Violet: V1 sin 400 nmN λ = . 
EXECUTE: R
V
sin 7
sin 4
θ
θ
= . R V R V15 15θ θ θ θ− = ° → = + °. V
V
sin( 15 ) 7
.
sin 4
θ
θ
+° = Using a trig identify from Appendix B, 
V V
V
sin cos15 cos sin15
7 4
sin
θ θ
θ
° + ° = . Vcos15 cot sin15 7 4θ° + ° = . 
V Vtan 0.330 18.3θ θ= ⇒ = °and R V 15 18.3 15 33.3θ θ= + ° = ° + ° = °. Then R1 sin 700 nmN θ = gives 
5R
9
sin sin 33.3
7.84 10 lines m 7840 lines cm
700 nm 700 10 m
N
θ
−= = = × =×
. The spectrum begins at 18 .3 and ends at 33.3 . 
EVALUATE: As N is increased, the angular range of the visible spectrum increases. 
36.57. (a) IDENTIFY and SET UP: The angular position of the first minimum is given by sina mθ λ= (Eq.36.2), with 
1.m = The distance of the minimum from the center of the pattern is given by tan .y x θ= 
9
3 3
3
540 10 m
sin 1.50 10 ; 1.50 10 rad
0.360 10 ma
λθ θ
−
− −
−
×= = = × = ×
×
 
3 3
1 tan (1.20 m) tan(1.50 10 rad) 1.80 10 m 1.80 mm.y x θ
− −= = × = × = 
(Note that θ is small enough for sin tan ,θ θ θ≈ ≈ and Eq.(36.3) applies.) 
(b) IDENTIFY and SET UP: Find the phase angle β where 0 / 2.I I= Then use Eq.(36.6) to solve for θ and 
tany x θ= to find the distance. 
EXECUTE: From part (a) of Problem 36.53, 0
1
2
I I= when 2.78 rad.β = 
2
sina
πβ θ
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 (Eq.(36.6)), so sin .
2 a
βλθ
π
= 
9
4
3
(2.78 rad)(540 10 m)(1.20 m)
tan sin 7.96 10 m 0.796 mm
2 2 (0.360 10 m)
x
y x x
a
βλθ θ
π π
−
−
−
×= ≈ ≈ = = × =
×
 
EVALUATE: The point where 0 / 2I I= is not midway between the center of the central maximum and the first 
minimum; see Exercise 36.15. 
Diffraction 36-15 
36.58. IDENTIFY: 
2
0
sin
I I
γ
γ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
. The maximum intensity occurs when the derivative of the intensity function with 
respect to γ is zero. 
SET UP: 
sin
cos
d
d
γ γ
γ
= . 
2
1 1d
dγ γ γ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
. 
EXECUTE: 
2
0 2
sin sin cos sin
2 0
dI d
I
d d
γ γ γ γ
γ γ γ γ γ γ
⎛ ⎞ ⎛ ⎞⎛ ⎞= = − =⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
. 
2
cos sin
cos sin tan
γ γ γ γ γ γ γ
γ γ
− ⇒ = ⇒ = . 
(b) The graph in Figure 36.58 is a plot of f( γ ) = γ − tan γ . When ( )f γ equals zero, there is an intensity 
maximum. Getting estimates from the graph, and then using trial and error to narrow in on the value, we find that 
the three smallest γ -values are γ = 4.49 rad 7.73 rad, and 10.9 rad. 
EVALUATE: 0γ = is the central maximum. The three values of γ we found are the locations of the first three 
secondary maxima. The first four minima are at 3.14 radγ = , 6.28 rad, 9.42 rad, and 12.6 rad. The maxima are 
between adjacent minima, but not precisely midway between them. 
 
Figure 36.58 
36.59. IDENTIFY and SET UP: Relate the phase difference between adjacent slits to the sum of the phasors for all slits. The 
phase difference between adjacent slits is 
2 2
sin
d dπ π θφ θ
λ λ
= ≈ when θ is small and sin .θ θ≈ Thus .
2 d
λφθ
π
= 
EXECUTE: A principal maximum occurs when max 2 ,mφ φ π= = where m is an integer, since then all the phasors 
add. The first minima on either side of the mth principal maximum occur when min 2 (2 / )m Nφ φ π π
±= = ± and the 
phasor diagram for N slits forms a closed loop and the resultant phasor is zero. The angular position of a principal 
maximum is max .2 d
λθ φ
π
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 The angular position of the adjacent minimum is min min.2 d
λθ φ
π
± ±⎛ ⎞= ⎜ ⎟
⎝ ⎠
 
min max
2 2
2 2d N d N Nd
λ π λ π λθ φ θ θ
π π
+ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= + = + = +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
 
min max
2
2 d N Nd
λ π λθ φ θ
π
− ⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
 
The angular width of the principal maximum is min min
2
,
Nd
λθ θ+ −− = as was to be shown. 
EVALUATE: The angular width of the principal maximum decreases like 1/ N as N increases. 
36.60. IDENTIFY: The change in wavelength of the Hα line is due to a Doppler shift in the wavelength due to the motion 
of the galaxy. 
SET UP: From Equation 16.30, the Doppler effect formula for light is R S
c v
f f
c v
−=
+
. 
EXECUTE: First find the wavelength of the light using the grating information. 
λ = d sin θ1 = [1/(575,800 lines/m)] sin 23.41° = 6.900 × 10–7 m = 690.0 nm 
Using Equation 16.30, we have R S
c v
f f
c v
−=
+
. In this case, fR is the frequency of the 690.0-nm light that the 
cosmologist measures, and fS is the frequency of the 656.3-nm light of the Hα line obtained in the laboratory. 
36-16 Chapter 36 
Solving for v gives 
( )
( )
2
R S
2
R S
1 /
1 /
f f
v
f f
−
=
+
c. Since fλ = c, f = c/λ, which gives fR/fS = λS/λR. Substituting this into the 
equation for v, we get 
2
S
R
2
S
R
1
1
v
λ
λ
λ
λ
⎛ ⎞
− ⎜ ⎟
⎝ ⎠=
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
c = ( )
2
8
2
656.3 nm
1
690.0 nm 3.00 10 m/s
656.3 nm
1
690.0 nm
⎛ ⎞− ⎜ ⎟
⎝ ⎠ ×
⎛ ⎞+ ⎜ ⎟
⎝ ⎠
 = 1.501 × 107 m/s, 
which is 5.00% the speed of light. 
EVALUATE: Since v is positive, the galaxy is moving away from us. We can also see this because the wavelength 
has increased due to the motion. 
36.61. IDENTIFY and SET UP: Draw the specified phasor diagrams. There is totally destructive interference between 
two slits when their phasors are in opposite directions. 
EXECUTE: (a) For eight slits, the phasor diagrams must have eight vectors. The diagrams for each specified 
value of φ are sketched in Figure 36.61a. In each case the phasors all sum to zero. 
(b) The additional phasor diagrams for 3 / 2φ π= and 3 / 4π are sketched in Figure 36.61b. 
For
3 5 7
, , and ,
4 4 4
π π πφ φ φ= = = totally destructive interference occurs between slits four apart. For 3 ,
2
πφ = 
totally destructive interference occurs with every second slit. 
EVALUATE: At a minimum the phasors for all slits sum to zero. 
 
Figure 36.61 
36.62. IDENTIFY: Maxima are given by 2 sind mθ λ= . 
SET UP: d is the separation between crystal planes. 
EXECUTE: (a) 
0.125 nm
arcsin arcsin arcsin(0.2216 )
2 2(0.282 nm)
m
m m
d
λθ ⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
. 
For 1: 12.8 , 2 : 26.3 , 3: 41.7 , and 4 : 62.4 .m m m mθ θ θ θ= = ° = = ° = = ° = = ° No larger m values yield answers. 
(b) If the separation 
2
, then arcsin arcsin(0.3134 ).
22
a m
d m
a
λθ
⎛ ⎞
= = =⎜ ⎟⎜ ⎟
⎝ ⎠
 
So for 1: 18.3 , 2: 38.8 , and 3: 70.1 .m m mθ θ θ= = ° = = ° = = ° No larger m values yield answers. 
EVALUATE: In part (b), where d is smaller, the maxima for each m are at larger θ 
36.63. IDENTIFY and SET UP: In each case consider the relevant phasor diagram. 
EXECUTE: (a) For the maxima to occur for N slits, the sum of all the phase differences between the slits must 
add to zero (the phasor diagram closes on itself). This requires that, adding up all the relative phase shifts, 
2 ,N mφ π= for some integer m . Therefore 2 ,m
N
πφ = for m not an integer multiple of ,N which would give a 
maximum. 
(b) The sum of N phase shifts 
2 m
N
πφ = brings you full circle back to the maximum, so only the 1N − previous 
phases yield minima between each pair of principal maxima. 
EVALUATE: The 1N − minima between each pair of principal maxima cause the maxima to become sharper as N 
increases. 
36.64. IDENTIFY: Set d a= in the expressions for φ and β and use the results in Eq.(36.12). 
SET UP: Figure 36.64 shows a pair of slits whose width and separation are equal 
Diffraction 36-17 
EXECUTE: Figure 36.64 shows that the two slits are equivalent to a single slit of width 2a . 
2 2
sin , so sin .
d aπ πφ θ β θ φ
λ λ
= = = So then the intensity is 
2 2 2 2
2
0 0 0 02 2 2 2
sin ( /2) (2sin( /2)cos( /2)) sin sin ( /2)
cos ( /2)
( /2) ( /2)
I I I I I
β β β β ββ
β β β β
′⎛ ⎞
= = = =⎜ ⎟ ′⎝ ⎠
, where
2 (2 )
sin ,
aπβ θ
λ
′ = 
which is Eq. (35.5) with double the slit width. 
EVALUATE: In Chapter 35 we considered the limit where a d<< . a d> is not possible. 
 
Figure 36.64 
36.65. IDENTIFY and SET UP: The condition for an intensity maximum is sin , 0, 1, 2,d m mθ λ= = ± ± … Third order 
means 3.m = The longest observable wavelength is the one that gives 90θ = ° and hence 1.θ = 
EXECUTE: 6500 lines/cm so 56.50 10× lines/m and 6
5
1
m 1.538 10 m
6.50 10
d −= = ×
×
 
6
7sin (1.538 10 m)(1) 5.13 10 m 513 nm
3
d
m
θλ
−
−×= = = × = 
EVALUATE: The longestwavelength that can be obtained decreases as the order increases. 
36.66. IDENTIFY and SET UP: As the rays first reach the slits there is already a phase difference between adjacent slits of 
2 sin
.
dπ θ
λ
′
 This, added to the usual phase difference introduced after passing through the slits, yields the 
condition for an intensity maximum. For a maximum the total phase difference must equal 2 mπ . 
EXECUTE: 
2 sin 2 sin
2 (sin sin )
d d
m d m
π θ π θ π θ θ λ
λ λ
′ ′+ = ⇒ + = 
(b) 600 6
5 1
1
slits mm 1.67 10 m.
6.00 10 m
d −−⇒ = = ××
 
7
6
7
6
For 0 ,
0 : arcsin(0) 0.
6.50 10 m
1: arcsin arcsin 22.9 .
1.67 10 m
6.50 10 m
1: arcsin arcsin 22.9 .
1.67 10 m
For 20.0 ,
0 : arcsin( sin 20.0 ) 20.0 .
m
m
d
m
d
m
θ
θ
λθ
λθ
θ
θ
−
−
−
−
′ =
= = =
⎛ ⎞×⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
⎛ ⎞×⎛ ⎞= − = − = − = −⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠
′ =
= = − = −
7
6
7
6
6.50 10 m
1: arcsin sin 20.0 2.71 .
1.67 10 m
6.50 10 m
1: arcsin sin 20.0 47.0 .
1.67 10 m
m
m
θ
θ
−
−
−
−
⎛ ⎞×= = − =⎜ ⎟×⎝ ⎠
⎛ ⎞×= − = − − = −⎜ ⎟×⎝ ⎠
 
EVALUATE: When 0θ′ > , the maxima are shifted downward on the screen, toward more negative angles. 
36.67. IDENTIFY: The maxima are given by sind mθ λ= . We need sin 1m
d
λθ = ≤ in order for all the visible 
wavelengths are to be seen. 
SET UP: For 650 6
5 1
1
slits mm 1.53 10 m.
6.50 10 m
d −−⇒ = = ××
 
36-18 Chapter 36 
EXECUTE: 7 1 1 11
2 3
4.00 10 m : 1: 0.26; 2 : 0.52; 3: 0.78.m m m
d d d
λ λ λλ −= × = = = = = = 
7 2 2 2
2
2 3
7.00 10 m : 1: 0.46; 2 : 0.92; 3 : 1.37.m m m
d d d
λ λ λλ −= × = = = = = = So, the third order does not contain the violet 
end of the spectrum, and therefore only the first and second order diffraction patterns contain all colors of the spectrum. 
EVALUATE: θ for each maximum is larger for longer wavelengths. 
36.68. IDENTIFY: Apply sin 1.22
D
λθ = . 
SET UP: θ is small, so sin x
R
θ Δ≈ , where xΔ is the size of the detail and 87.2 10 lyR = × . 121 ly 9.41 10 km= × . /c fλ = 
EXECUTE: sin 
5 8
3 9
1.22 (1.22) (1.22)(3.00 10 km s)(7.2 10 ly)
1.22 2.06 ly
(77.000 10 km)(1.665 10 Hz)
x R cR
x
D R D Df
λ λθ Δ × ×= ≈ ⇒ Δ = = = =
× ×
. 
12 13(9.41 10 km ly)(2.06 ly) 1.94 10 km.× = × 
EVALUATE: 18 cmλ = . / Dλ is very small, so x
R
Δ
is very small. Still, R is very large and xΔ is many orders of 
magnitude larger than the diameter of the sun. 
36.69. IDENTIFY and SET UP: Add the phases between adjacent sources. 
EXECUTE: (a) stsin . Place1 maximum at or 90 .d mθ λ θ= ∞ = d .λ= If ,d λ< this puts the first maximum 
“beyond .∞ ” Thus, if d λ< there is only a single principal maximum. 
(b) At a principal maximum when 0δ = , the phase difference due to the path difference between adjacent slits 
is path
sin
2 . This just scales 2
d θπ π
λ
⎛ ⎞Φ = ⎜ ⎟
⎝ ⎠
radians by the fraction the wavelength is of the path difference between 
adjacent sources. If we add a relative phase δ between sources, we still must maintain a total phase difference of 
zero to keep our principal maximum. 
1
path
2 sin
0 or sin
2
d
d
π θ δλδ δ θ
λ π
− ⎛ ⎞Φ ± = ⇒ = ± = ⎜ ⎟
⎝ ⎠
 
(c) 
0.280 m
0.0200 m
14
d = = (count the number of spaces between 15 points). Let 45 . Also recall , sof cθ λ= = 
9
max 8
2 (0.0200 m)(8.800 10 Hz)sin 45
2.61 radians.
(3.00 10 m s)
πδ ×= ± = ±
×
 
EVALUATE: δ must vary over a wider range in order to sweep the beam through a greater angle. 
36.70. IDENTIFY: The wavelength of the light is smaller under water than it is in air, which will affect the resolving 
power of the lens, by Rayleigh’s criterion. 
SET UP: The wavelength under water is λ = λ0/n, and for small angles Rayleigh’s criterion is θ = 1.22λ/D. 
EXECUTE: (a) In air the wavelength is λ0 = c/f = (3.00 × 108 m/s)/(6.00 × 1014 Hz) = 5.00 × 10–7 m. In water the 
wavelength is λ = λ0/n = (5.00 × 10–7 m)/1.33 = 3.76 × 10–7 m. With the lens open all the way, we have D = f/2.8 = 
(35.0 mm)/2.80 = (0.0350 m)/2.80. In the water, we have 
sin θ ≈ θ = 1.22 λ/D = (1.22)(3.76 × 10–7 m)/[(0.0350 m)/2.80] = 3.67 × 10–5 rad 
Calling w the width of the resolvable detail, we have 
θ = w/R → w = Rθ = (2750 mm)(3.67 × 10–5 rad) = 0.101 mm 
(b) θ = 1.22 λ/D = (1.22)(5.00 × 10–7 m)/[(0.0350 m)/2.80] = 4.88 × 10–5 rad 
w = Rθ = (2750 mm)(4.88 × 10–5 rad) = 0.134 mm 
EVALUATE: Due to the reduced wavelength underwater, the resolution of the lens is better under water than in air. 
36.71. IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means the angular separation θ of the objects is given 
by 1.22 / . / ,D y sθ λ θ= = where 75.0 my = is the distance between the two objects and s is their distance from 
the astronaut (her altitude). 
EXECUTE: 1.22
y
s D
λ= 
3
5
9
(75.0 m)(4.00 10 m)
4.92 10 m 492 km
1.22 1.22(500 10 m)
yD
s
λ
−
−
×= = = × =
×
 
EVALUATE: In practice, this diffraction limit of resolution is not achieved. Defects of vision and distortion by the 
earth’s atmosphere limit the resolution more than diffraction does. 
Diffraction 36-19 
36.72. IDENTIFY: Apply sin 1.22
D
λθ = . 
SET UP: θ is small, so sin x
R
θ Δ≈ , where xΔ is the size of the details and R is the distance to the earth. 
151 ly 9.41 10 m= × . 
EXECUTE: (a) 
6 5
17
5
(6.00 10 m)(2.50 10 m)
1.23 10 m 13.1 ly
1.22 (1.22)(1.0 10 m)
D x
R
λ −
Δ × ×= = = × =
×
 
(b) 
5 15
81.22 (1.22)(1.0 10 m)(4.22 ly)(9.41 10 m ly) 4.84 10 km
1.0 m
R
x
D
λ −× ×Δ = = = × . This is about 10,000 times the 
diameter of the earth! Not enough resolution to see an earth-like planet! xΔ is about 3 times the distance from the 
earth to the sun. 
(c)
5 15
6
6
(1.22)(1.0 10 m)(59 ly)(9.41 10 m ly)
1.13 10 m 1130 km.
6.00 10 m
x
−× ×Δ = = × =
×
 
3
5
planet
1130 km
8.19 10 ;
1.38 10 km
x
x
D
−Δ = = × Δ
×
 is small compared to the size of the planet. 
EVALUATE: The very large diameter of Planet Imager allows it to resolve planet-sized detail at great distances. 
36.73. IDENTIFY and SET UP: Follow the steps specified in the problem. 
EXECUTE: (a) From the segment ,dy′ the fraction of the amplitude of 0E that gets through is 
0 0 sin( ).
dy dy
E dE E kx t
a a
ω
′ ′⎛ ⎞ ⎛ ⎞⇒ = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
(b) The path difference between each little piece is 
0sin ( sin ) sin( ( sin ) ).
E dy
y kx k D y dE k D y t
a
θ θ θ ω
′′ ′ ′⇒ = − ⇒ = − − This can be rewritten as 
0 (sin( )cos( sin ) sin( sin )cos( )).
E dy
dE kD t ky ky kD t
a
ω θ θ ω
′ ′ ′= − + − 
(c) So the total amplitude is given by the integral over the slit of the above. 
2 2
0
2 2
(sin( ) cos( sin ) sin( sin )
a a
a a
E
E dE dy kD t ky ky
a
ω θ θ
− −
′ ′ ′⇒ = = − +∫ ∫ cos( )).kD tω− 
But the second term integrates to zero, so we have: 
[ ]
[ ]
2
2
0
02
2
0 0
0
sin( sin )
sin( ) (cos( sin )) sin ( )
sin 2
sin( (sin ) 2) sin( (sin ) )
sin( ) sin( ) .
(sin ) 2 (sin )
sin . . .
At 0, 1 sin(
. . .
a
a
a
a
E ky
E kD t dy ky E kD t
a ka
ka a
E E kD t E kD t
ka a
E E
θω θ ω
θ
θ π θ λω ω
θ π θ λ
θ
−
−
⎡ ⎤′⎛ ⎞′ ′= − = − ⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
⎛ ⎞ ⎛ ⎞
⇒ = − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= = ⇒ =
∫
).kD tω−
 
(d) Since 
2 2
2
0 0
sin( (sin )/2) sin( 2)
,
(sin )/2 2
ka
I E I I I
ka
θ β
θ β
⎛ ⎞ ⎛ ⎞
∝ ⇒ = =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 where we have used 2 20 0 sin ( ).I E kx tω= − 
EVALUATE: The same result for ( )I θ is obtained as was obtained using phasors. 
36.74. IDENTIFY and SET UP: Follow the steps specified in the problem. 
EXECUTE: (a) Each source can be thought of as a traveling wave evaluated at x R= with a maximum amplitude 
of 0.E However, each successive source will pick up an extra phase from its respective pathlength to point 
sin
2
d
P .
θφ π
λ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
 which is just 2 ,π the maximum phase, scaled by whatever fraction the path difference, 
sin ,d θ is of the wavelength, λ . By adding up the contributions from each source (including the accumulating 
phase difference) this gives the expression provided. 
(b) ( ) cos( ) sin( ).i kR t ne kR t n i kR t nω φ ω φ ω φ− + = − + + − + The real part is just cos ( ).kR t nω φ− +So, 
1 1
( )
0 0
0 0
Re e cos( ).
N N
i kR t n
n n
E E kR t nω φ ω φ
− −
− +
= =
⎡ ⎤ = − +⎢ ⎥
⎣ ⎦
∑ ∑ (Note: Re means “the real part of . . . .”). But this is just 
0 0 0 0cos( ) cos( ) cos( 2 ) cos( ( 1) )E kR t E kR t E kR t E kR t Nω ω φ ω φ ω φ− + − + + − + + + − + − 
36-20 Chapter 36 
(c) 
1 1 1
( ) ( )
0 0 0
0 0 0
e e e e e e .
N N N
i kR t n i t ikR in i kR t in
n n n
E E Eω φ ω φ ω φ
− − −
− + − + −
= = =
= =∑ ∑ ∑ 
1
0 0
e (e ) .
N
in i n
n n
φ φ
∞ −
= =
=∑ ∑ But recall 
1
0
1
1
NN
n
n
x
x
x
−
=
−=
−∑ . 
/ 2 / 2 / 2 / 2 / 21
( 1) / 2
/ 2 / 2 / 2 / 2 / 2
0
e 1 1 ( ) ( )
Let e so (e ) (nice trick!). But .
e 1 1 ( ) ( )
iN iN iN iN iN iN iNN
i i n i N
i i i i i i i
n
e e e e e e
x e
e e e e e e
φ φ φ φ φ φ φ
φ φ φ
φ φ φ φ φ φ φ
− −−
−
− −
=
− − − −= = = =
− − − −∑ 
Putting everything together: 
[ ]
/ 2 / 21
( ) ( ( 1) / 2)
0 0 / 2 / 2
0
0
( )
e
( )
cos /2 sin /2 cos /2 sin /2
cos( ( 1) /2) sin( ( 1) / 2)
cos /2 sin /2 cos /2 sin /2
iN iNN
i kR t n i kR t N
i i
n
e e
E e E
e e
N i N N i N
E kR t N i kR t N
i i
φ φ
ω φ ω φ
φ φ
φ φ φ φω φ ω φ
φ φ φ φ
−−
− + − + −
−
=
−=
−
⎡ ⎤+ − += − + − + − + − ⎢ ⎥+ − +⎣ ⎦
∑
 
Taking only the real part gives 0
sin( /2)
cos( ( 1) /2) .
sin /2
N
E kR t N E
φω φ
φ
⇒ − + − = 
(d) 
2
2
0 2av
sin ( / 2)
.
sin ( / 2)
N
I E I
φ
φ
= = (The 2cos term goes to 12 in the time average and is included in the definition of 0.)I 
2
0
0 .2
E
I ∝ 
EVALUATE: (e) 
2 2
20
0 02 2
sin (2 / 2) (2sin / 2cos / 2)
2. 4 cos .
sin / 2 sin / 2 2
I
N I I I
φ φ φ φ
φ φ
= = = = Looking at Eq.(35.9), 
2
2 0 0
0 0 02 but for us .2 4
E I
I E I
′′ ∝ ∝ = 
36.75. IDENTIFY and SET UP: From Problem 36.74, 
2
0 2
sin ( / 2)
sin / 2
N
I I
φ
φ
= . Use this result to obtain each result specified 
in the problem. 
EXECUTE: (a) 
0
0
lim
0
I
φ →
→ . 
0 0
sin ( / 2) 2 cos( / 2)
ˆUse l'Hopital's rule: lim lim .
sin / 2 1 2 cos( / 2)
N N N
N
φ φ
φ φ
φ φ→ →
⎛ ⎞
= =⎜ ⎟
⎝ ⎠
 So 2 0
0
lim .I N I
φ →
= 
(b) The location of the first minimum is when the numerator first goes to zero at min min
2
or .
2
N
N
πφ π φ= = The 
width of the central maximum goes like min2 ,φ so it is proportional to 
1
.
N
 
(c) Whenever 
2
N
n
φ π= where n is an integer, the numerator goes to zero, giving a minimum in intensity. That is, 
I is a minimum wherever 
2
.
n
N
πφ = This is true assuming that the denominator doesn’t go to zero as well, which 
occurs when ,
2
m
φ π= where m is an integer. When both go to zero, using the result from part(a), there is a 
maximum. That is, if 
n
N
is an integer, there will be a maximum. 
(d) From part (c), if 
n
N
 is an integer we get a maximum. Thus, there will be 1N − minima. (Places where n
N
 is 
not an integer for fixed N and integer n .) For example, 0n = will be a maximum, but 1,2. . ., 1n N= − will be 
minima with another maximum at .n N= 
(e) Between maxima 
2
φ
 is a half-integer multiple of 
3
i.e. , , etc.)
2 2
π ππ ⎛ ⎞⎜ ⎟
⎝ ⎠
 and if N is odd then 
2
02
sin ( / 2)
1, so .
sin / 2
N
I I
φ
φ
→ → 
EVALUATE: These results show that the principal maxima become sharper as the number of slits is increased. 
37-1 
RELATIVITY 
 37.1. IDENTIFY and SET UP: Consider the distance A to O′ and B to O′ as observed by an observer on the ground 
(Figure 37.1). 
 
Figure 37.1 
EXECUTE: Simultaneous to observer on train means light pulses from and A B′ ′ arrive at O′ at the same time. 
To observer at O light from A′ has a longer distance to travel than light from B′ so O will conclude that the pulse 
from ( )A A′ started before the pulse at ( ).B B′ To observer at O bolt A appeared to strike first. 
EVALUATE: Section 37.2 shows that if they are simultaneous to the observer on the ground then an observer on 
the train measures that the bolt at B′ struck first. 
 37.2. (a) 
2
1
γ 2.29.
1 (0.9)
= =
−
 6 6γ (2.29) (2.20 10 s) 5.05 10 s.t τ − −= = × = × 
(b) 8 6 3(0.900) (3.00 10 m s) (5.05 10 s) 1.36 10 m 1.36 km.d vt −= = × × = × = 
 37.3. IDENTIFY and SET UP: The problem asks for u such that 0
1
/ .
2
t tΔ Δ = 
EXECUTE: 0
2 21 /
t
t
u c
ΔΔ =
−
 gives ( )
2
2 8 8
0
1
1 / (3.00 10 m/s) 1 2.60 10 m/s
2
u c t t
⎛ ⎞= − Δ Δ = × − = ×⎜ ⎟
⎝ ⎠
; 0.867
u
c
= 
Jet planes fly at less than ten times the speed of sound, less than about 3000 m/s. Jet planes fly at much lower 
speeds than we calculated for u. 
 37.4. IDENTIFY: Time dilation occurs because the rocket is moving relative to Mars. 
SET UP: The time dilation equation is 0t tγΔ = Δ , where t0 is the proper time. 
EXECUTE: (a) The two time measurements are made at the same place on Mars by an observer at rest there, so 
the observer on Mars measures the proper time. 
(b) 0 2
1
(75.0 s) 435 s
1 (0.985)
t tγ μ μΔ = Δ = =
−
 
EVALUATE: The pulse lasts for a shorter time relative to the rocket than it does relative to the Mars observer. 
 37.5. (a) IDENTIFY and SET UP: 8 70 2.60 10 s; 4.20 10 s.t t
− −Δ = × Δ = × In the lab frame the pion is created and decays 
at different points, so this time is not the proper time. 
EXECUTE: 
22
0 0
22 2
 says 1
1 /
t u t
t
c tu c
Δ Δ⎛ ⎞Δ = − = ⎜ ⎟Δ⎝ ⎠−
 
22 8
0
7
2.60 10 s
1 1 0.998; 0.998
4.20 10 s
u t
u c
c t
−
−
⎛ ⎞Δ ×⎛ ⎞= − = − = =⎜ ⎟⎜ ⎟Δ ×⎝ ⎠ ⎝ ⎠
 
EVALUATE: ,u c< as it must be, but u/c is close to unity and the time dilation effects are large. 
(b) IDENTIFY and SET UP: The speed in the laboratory frame is 0.998 ;u c= the time measured in this frame is 
,tΔ so the distance as measured in this frame is d u t= Δ 
EXECUTE: 8 7(0.998)(2.998 10 m/s)(4.20 10 s) 126 md −= × × = 
EVALUATE: The distance measured in the pion’s frame will be different because the time measured in the pion’s 
frame is different (shorter). 
37
37-2 Chapter 37 
 37.6. γ 1.667= 
(a) 
8
0
1.20 10 m
0.300 s.
γ γ(0.800 )
t
t
c
Δ ×Δ = = = 
(b) 7(0.300 s) (0.800 ) 7.20 10 m.c = × 
(c) 0 0.300 s γ 0.180 s.tΔ = = (This is what the racer measures your clock to read at that instant.) At your origin 
you read the original 
8
8
1.20 10 m
0.5 s.
(0.800) (3 10 m s)
× =
×
 Clearly the observers (you and the racer) will not agree on the 
order of events! 
 37.7. IDENTIFY and SET UP: A clock moving with respect to an observer appears to run more slowly than a clock at rest 
in the observer’s frame. The clock in the spacecraft measurers the proper time 0.tΔ 365 days 8760 hours.tΔ = = 
EXECUTE: The clock on the moving spacecraft runs slow and shows the smaller elapsed time. 
2 2 6 8 2
0 1 / (8760 h) 1 (4.80 10 /3.00 10 ) 8758.88 ht t u cΔ = Δ − = − × × = . The difference in elapsed times is 
8760 h 8758.88 h 1.12 h− = . 
 37.8. IDENTIFY and SET UP: The proper time is measured in the frame where the two events occur at the same point. 
EXECUTE: (a) The time of 12.0 ms measured by the first officer on the craft is the proper time. 
(b) 0
2 21 /
t
t
u c
ΔΔ =
−
gives 2 3 201 ( / ) 1 (12.0 10 / 0.190) 0.998u c t t c c
−= − Δ Δ = − × = . 
EVALUATE: The observer at rest with respect to the searchlight measures a much shorter duration for the event. 
 37.9. IDENTIFY and SET UP: 2 20 1 / .l l u c= − The length measured when the spacecraft is moving is 074.0 m; l l= is 
the length measured in a frame at rest relative to the spacecraft. 
EXECUTE: 0 2 2 2
74.0 m
92.5 m.
1 / 1 (0.600 / )
l
l
u c c c
= = =
− −
 
EVALUATE: 0 .l l> The moving spacecraft appears to an observer on the planet to be shortened along the 
direction of motion. 
37.10. IDENTIFY and SET UP: When the meterstick is at rest with respect to you, you measure its length to be 1.000 m, 
and that is its proper length, 0l . 0.3048 ml = . 
EXECUTE: 2 20 1 /l l u c= − gives 
2 2 8
01 ( / ) 1 (0.3048/1.00) 0.9524 2.86 10 m/su c l l c c= − = − = = × . 
37.11. IDENTIFY and SET UP: The 2.2 μs lifetime is Δt0 and the observer on earth measures Δt. The atmosphere is 
moving relative to the muon so in its framethe height of the atmosphere is l and l0 is 10 km. 
EXECUTE: (a) The greatest speed the muon can have is c, so the greatest distance it can travel in 62.2 10 s−× is 
8 6(3.00 10 m/s)(2.2 10 s) 660 m 0.66 kmd vt −= = × × = = . 
(b) 
6
50
2 2 2
2.2 10 s
4.9 10 s
1 / 1 (0.999)
t
t
u c
−
−Δ ×Δ = = = ×
− −
 
8 5(0.999)(3.00 10 m/s)(4.9 10 s) 15 kmd vt −= = × × = 
In the frame of the earth the muon can travel 15 km in the atmosphere during its lifetime. 
(c) 2 2 20 1 / (10 km) 1 (0.999) 0.45 kml l u c= − = − = 
In the frame of the muon the height of the atmosphere is less than the distance it moves during its lifetime. 
37.12. IDENTIFY and SET UP: The scientist at rest on the earth’s surface measures the proper length of the separation 
between the point where the particle is created and the surface of the earth, so 0 45.0 kml = . The transit time 
measured in the particle’s frame is the proper time, 0tΔ . 
EXECUTE: (a) 
3
40
8
45.0 10 m
1.51 10 s
(0.99540)(3.00 10 m/s)
l
t
v
−×= = = ×
×
 
(b) 2 2 20 1 / (45.0 km) 1 (0.99540) 4.31 kml l u c= − = − = 
(c) time dilation formula: 2 2 4 2 50 1 / (1.51 10 s) 1 (0.99540) 1.44 10 st t u c
− −Δ = Δ − = × − = × 
from lΔ : 
3
5
8
4.31 10 m
1.44 10 s
(0.99540)(3.00 10 m/s)
l
t
v
−×= = = ×
×
 
The two results agree. 
37.13. (a) 0 3600 ml = . 
2 7 2
0 02 8 2
(4.00 10 m s)
1 (3600 m) 1 (3600 m)(0.991) 3568 m.
(3.00 10 m s)
u
l l l
c
×= − = − = =
×
 
Relativity 37-3 
(b) 500 7
3600 m
9.00 10 s.
4.00 10 m s
l
t
u
−Δ = = = ×
×
 
(c) 5
7
3568 m
8.92 10 s.
4.00 10 m s
l
t
u
−Δ = = = ×
×
 
37.14. Multiplying the last equation of (37.21) by u and adding to the first to eliminate t gives 
2
2
1
1 ,
u
x ut x x
c
γ
γ
⎛ ⎞′ ′+ = − =⎜ ⎟
⎝ ⎠
 
and multiplying the first by 
2
u
c
 and adding to the last to eliminate x gives 
2
2 2
1
γ 1 ,
γ
u u
t x t t
c c
⎛ ⎞′ ′+ = − =⎜ ⎟
⎝ ⎠
 
2so γ( ) and γ( ),x x ut t t ux c′ ′ ′ ′= + = + which is indeed the same as Eq. (37.21) with the primed coordinates 
replacing the unprimed, and a change of sign of u. 
37.15. (a) 
2
0.400 0.600
0.806
1 1 (0.400) (0.600)
v u c c
v c
uv c
′ + += = =
′+ +
 
(b) 
2
0.900 0.600
0.974
1 1 (0.900)(0.600)
v u c c
v c
uv c
′ + += = =
′+ +
 
(c) 
2
0.990 0.600
0.997 .
1 1 (0.990)(0.600)
v u c c
v c
uv c
′ + += = =
′+ +
 
37.16. γ 1.667(γ 5 3 if (4 5) ).u c= = = 
(a) In Mavis’s frame the event “light on” has space-time coordinates 0x′ = and 5.00t′ = s, so from the result of 
Exercise 37.14 or Example 37.7, γ( )x x ut′ ′= + and 9
2
γ γ 2.00 10 m, γ 8.33 s
ux
t t x ut t t
c
′⎛ ⎞′ ′ ′= + ⇒ = = × = =⎜ ⎟
⎝ ⎠
. 
(b) The 5.00-s interval in Mavis’s frame is the proper time 0tΔ in Eq.(37.6), so 0γ 8.33 s,t tΔ = Δ = as in part (a). 
(c) 9(8.33 s) (0.800 ) 2.00 10c = × m, which is the distance x found in part (a). 
37.17. IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light. 
SET UP: The relativistic velocity addition formula is 
21
x
x
x
v u
v
uv
c
−′ =
−
. 
EXECUTE: (a) For the pursuit ship to catch the cruiser, the distance between them must be decreasing, so the 
velocity of the cruiser relative to the pursuit ship must be directed toward the pursuit ship. 
(b) Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship. We want the velocity v′ of the 
cruiser knowing the velocity of the primed frame u and the velocity of the cruiser v in the unprimed frame (Tatooine). 
2
0.600 0.800
0.385
1 (0.600) (0.800)1
x
x
x
v u c c
v c
uv
c
− −′ = = = −
−−
 
The result implies that the cruiser is moving toward the pursuit ship at 0.385 .c 
EVALUATE: The nonrelativistic formula would have given –0.200c, which is considerably different from the 
correct result. 
37.18. Let yu be the y-component of the velocity of S′ relative to S. Following the steps used in the derivation of 
Eq.(37.23) we get 
21 /
y y
y
y y
v u
v
u v c
′ +
=
′+
. 
37.19. IDENTIFY and SET UP: Reference frames S and S′ are shown in Figure 37.19. 
 
Frame S is at rest in the 
laboratory. Frame S′ is 
attached to particle 1. 
 
Figure 37.19 
u is the speed of S′ relative to S; this is the speed of particle 1 as measured in the laboratory. Thus 0.650 .u c= + 
The speed of particle 2 in S′ is 0.950c. Also, since the two particles move in opposite directions, 2 moves in the 
x′− direction and 0.950 .xv c′ = − We want to calculate ,xv the speed of particle 2 in frame S; use Eq.(37.23). 
37-4 Chapter 37 
EXECUTE: 
2 2
0.950 0.650 0.300
0.784 .
1 / 1 (0.950 )( 0.650 ) / 1 0.6175
x
x
x
v u c c c
v c
uv c c c c
′ + − + −= = = = −
′+ + − −
 The speed of the second particle, 
as measured in the laboratory, is 0.784c. 
EVALUATE: The incorrect Galilean expression for the relative velocity gives that the speed of the second particle 
in the lab frame is 0.300c. The correct relativistic calculation gives a result more than twice this. 
37.20. IDENTIFY and SET UP: Let S be the laboratory frame and let S′ be the frame of one of the particles, as shown in 
Figure 37.20. Let the positive x direction for both frames be from particle 1 to particle 2. In the lab frame particle 1 
is moving in the +x direction and particle 2 is moving in the x− direction. Then 0.9520u c= and 0.9520v c= − . 
v′ is the velocity of particle 2 relative to particle 1. 
EXECUTE: 
2 2
0.9520 0.9520
0.9988
1 / 1 (0.9520 )( 0.9520 ) /
v u c c
v c
uv c c c c
− − −′ = = = −
− − −
. The speed of particle 2 relative to particle 1 
is 0.9988c . 0v′ < shows particle 2 is moving toward particle 1. 
 
Figure 37.20 
37.21. IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light. 
SET UP: The relativistic velocity addition formula is 
21
x
x
x
v u
v
uv
c
−′ =
−
. 
EXECUTE: In the relativistic velocity addition formula for this case, xv ′ is the relative speed of particle 1 with 
respect to particle 2, v is the speed of particle 2 measured in the laboratory, and u is the speed of particle 1 
measured in the laboratory, u = – v. 
2 2 2
( ) 2
1 ( ) 1x
v v v
v
v v c v c
− −′ = =
− − +
. 2
2
2 0x x
v
v v v
c
′ ′− + = and 2 2 3(0.890 ) 2 (0.890 ) 0c v c v c− + = . 
This is a quadratic equation with solution v = 0.611c (v must be less than c). 
EVALUATE: The nonrelativistic result would be 0.445c, which is considerably different from this result. 
37.22. IDENTIFY and SET UP: Let the starfighter’s frame be S and let the enemy spaceship’s frame be S′ . Let the 
positive x direction for both frames be from the enemy spaceship toward the starfighter. Then 0.400u c= + . 
0.700v c′ = + . v is the velocity of the missile relative to you. 
EXECUTE: (a) 
2
0.700 0.400
0.859
1 / 1 (0.400)(0.700)
v u c c
v c
uv c
′ + += = =
′+ +
 
(b) Use the distance it moves as measured in your frame and the speed it has in your frame to calculate the time it 
takes in your frame. 
9
8
8.00 10 m
31.0 s
(0.859)(3.00 10 m/s)
t
×= =
×
. 
37.23. IDENTIFY and SET UP: The reference frames are shown in Figure 37.23. 
 
S = Arrakis frame 
S′ = spaceship frame 
The object is the rocket. 
 
Figure 37.23 
u is the velocity of the spaceship relative to Arrakis. 
0.360 ; 0.920x xv c v c′= + = + 
(In each frame the rocket is moving in the positive coordinate direction.) 
Relativity 37-5 
Use the Lorentz velocity transformation equation, Eq.(37.22): 
2
.
1 /
x
x
x
v u
v
uv c
−′ =
−
 
EXECUTE: 
2 2 2
 so and 1
1 /
x x x x x
x x x x x
x
v u v v v v
v v u v u u v v
uv c c c
′ ′− ⎛ ⎞ ⎛ ⎞′ ′ ′= − = − − = −⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠
 
2 2
0.360 0.920 0.560
0.837
1 / 1 (0.360 )(0.920 ) / 0.6688
x x
x x
v v c c c
u c
v v c c c c
′− −= = = − = −
′− −
 
The speed of the spacecraft relative to Arrakis is 80.837 2.51 10 m/s.c = × The minus sign in our result for u means 
that the spacecraft is moving in the -direction,x− so it is moving away from Arrakis. 
EVALUATE: The incorrect Galilean expressionalso says that the spacecraft is moving away from Arrakis, but 
with speed 0.920c – 0.360c = 0.560c. 
37.24. IDENTIFY: We need to use the relativistic Doppler shift formula. 
SET UP: The relativistic Doppler shift formula, Eq.(37.25), is 0
c u
f f
c u
+=
−
. 
EXECUTE: 2 20
c u
f f
c u
+=
−
. 2 20( ) ( )c u f c u f− = + . 
2 2 2 2
0 0cf uf cf uf− = + . 
2 2 2 2
0 0cf cf uf uf− = + and 
2 2 2
0 0
2 2 2
0 0
( ) ( / ) 1
( / ) 1
c f f f f
u c
f f f f
− −= =
+ +
. 
(a) For f/f0 = 0.95, u = – 0.051c moving away from the source. 
(b) For f/f0 = 5.0, u = 0.923c moving towards the source. 
EVALUATE: Note that the speed required to achieve a 10 times greater Doppler shift is not 10 times the original 
speed. 
37.25. IDENTIFY and SET UP: Source and observer are approaching, so use Eq.(37.25): 0.
c u
f f
c u
+=
−
 Solve for u, the 
speed of the light source relative to the observer. 
(a) EXECUTE: 2 20
c u
f f
c u
+⎛ ⎞= ⎜ ⎟−⎝ ⎠
 
2 2 2
2 2 0 0
0 2 2 2
0 0
( ) ( / ) 1
( ) ( ) and 
( / ) 1
c f f f f
c u f c u f u c
f f f f
⎛ ⎞− −− = + = = ⎜ ⎟+ +⎝ ⎠
 
0 675 nm, 575 nmλ λ= = 
2
8 7
2
(675 nm/575 nm) 1
0.159 (0.159)(2.998 10 m/s) 4.77 10 m/s;
(675 nm/575 nm) 1
u c c
⎛ ⎞−= = = × = ×⎜ ⎟+⎝ ⎠
 definitely speeding 
(b) 7 7 84.77 10 m/s (4.77 10 m/s)(1 km/1000 m)(3600 s/1 h) 1.72 10 km/h.× = × = × Your fine would be 8$1.72 10× 
(172 million dollars). 
EVALUATE: The source and observer are approaching, so 0 0and .f f λ λ> < Our result gives ,u c< as it must. 
37.26. Using ( )0.600 3 5u c c= − = − in Eq.(37.25) gives 
( )
( ) 0 0 0
1 3 5 2 5
2.
1 3 5 8 5
f f f f
−
= = =
+
 
37.27. IDENTIFY and SET UP: If F is parallel to then v F changes the magnitude of v and not its direction. 
2 21 /
dp d mv
F
dt dt v c
⎛ ⎞
= = ⎜ ⎟
−⎝ ⎠
 
Use the chain rule to evaluate the derivative: ( ( )) .
d df dv
f v t
dt dv dt
= 
EXECUTE: (a) 
2 2 1/ 2 2 2 3/ 2 2
1 2
(1 / ) (1 / ) 2
m dv mv v dv
F
v c dt v c c dt
⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= + − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟− −⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
 
2 2
2 2 3 / 2 2 2 2 2 3 / 2
1
(1 / ) (1 / )
dv m v v dv m
F
dt v c c c dt v c
⎛ ⎞
= − + =⎜ ⎟− −⎝ ⎠
 
But 2 2 3 / 2, so ( / )(1 / ) .
dv
a a F m v c
dt
= = − 
EVALUATE: Our result agrees with Eq.(37.30). 
37-6 Chapter 37 
(b) IDENTIFY and SET UP: If F is perpendicular to then v F changes the direction of v and not its magnitude. 
2 2
.
1 /
d m
dt v c
⎛ ⎞
⎜ ⎟
−⎝ ⎠
v
F = 
/d dta v= but the magnitude of v in the denominator of Eq.(37.29) is constant. 
EXECUTE: 
2 21 /
ma
F
v c
=
−
 and 2 2 1/ 2( / )(1 / ) .a F m v c= − 
EVALUATE: This result agrees with Eq.(37.33). 
37.28. IDENTIFY and SET UP: 
2 2
1
1 /v c
γ =
−
. If γ is 1.0% greater than 1 then 1.010γ = , if γ is 10% greater than 1 
then 1.10γ = and if γ is 100% greater than 1 then 2.00γ = . 
EXECUTE: 21 1/v c γ= − 
(a) 21 1/(1.010) 0.140v c c= − = 
(b) 21 1/(1.10) 0.417v c c= − = 
(c) 21 1/(2.00) 0.866v c c= − = 
37.29. (a) 
2 2
2
1
mv
p mv
v c
= =
−
. 
2
2 2 2 2
2
1 3 3
1 2 1 1 0.866 .
4 4 2
v
v c v c v c c
c
⇒ = − ⇒ = − ⇒ = ⇒ = = 
(b) 3 3 1/ 3 2 / 3 2 / 32
2
1
γ 2 γ 2 γ (2) so 2 1 2 0.608
1
v
F ma ma
v c
c
−= = ⇒ = ⇒ = = ⇒ = − =
−
 
37.30. The force is found from Eq.(37.32) or Eq.(37.33). 
(a) Indistinguishable from 0.145 N.F ma= = 
(b) 3γ 1.75 N.ma = 
(c) 3γ 51.7 N.ma = 
(d) γ 0.145 N,ma = 0.333 N,1.03 N. 
37.31. (a) 
2
2 2
2 21
mc
K mc mc
v c
= − =
−
 
2
22 2
1 1 3
2 1 0.866 .
4 41
v
v c c
cv c
⇒ = ⇒ = − ⇒ = =
−
 
(b) 
2
2
22 2
1 1 35
5 6 1 0.986 .
36 361
v
K mc v c c
cv c
= ⇒ = ⇒ = − ⇒ = =
−
 
37.32. 2 27 8 2 10 9= 2 = 2(1.67×10 kg)(3.00×10 m s) = 3.01×10 J = 1.88×10 eV.E mc − − 
37.33. IDENTIFY and SET UP: Use Eqs.(37.38) and (37.39). 
EXECUTE: (a) 2 2 2 10, so 4.00 means 3.00 4.50 10 JE mc K E mc K mc −= + = = = × 
(b) 2 2 2 2 2 2 2 2( ) ( ) ; 4.00 , so 15.0( ) ( )E mc pc E mc mc pc= + = = 
1815 1.94 10 kg m/sp mc −= = × ⋅ 
(c) 2 2 2/ 1 /E mc v c= − 
2 2 24.00 gives 1 / 1/16 and 15/16 0.968E mc v c v c c= − = = = 
EVALUATE: The speed is close to c since the kinetic energy is greater than the rest energy. Nonrelativistic 
expressions relating E, K, p and v will be very inaccurate. 
37.34. (a) 2 3 2f(γ 1) (4.07 10 ) .W K mc mc
−= Δ = − = × 
(b) ( 2 2f iγ γ ) 4.79 .mc mc− = 
(c) The result of part (b) is far larger than that of part (a). 
37.35. IDENTIFY: Use 2E mc= to relate the mass increase to the energy increase. 
(a) SET UP: Your total energy E increases because your gravitational potential energy mgy increases. 
Relativity 37-7 
EXECUTE: E mg yΔ = Δ 
2 2 2( ) so / ( ) /E m c m E c mg y cΔ = Δ Δ = Δ = Δ 
2 2 8 2 13/ ( ) / (9.80 m/s )(30 m)/(2.998 10 m/s) 3.3 10 %m m g y c −Δ = Δ = × = × 
This increase is much, much too small to be noticed. 
(b) SET UP: The energy increases because potential energy is stored in the compressed spring. 
EXECUTE: 2 4 21 12 2 (2.00 10 N/m)(0.060 m) 36.0 JE U kxΔ = Δ = = × = 
2 16( ) / 4.0 10 kgm E c −Δ = Δ = × 
Energy increases so mass increases. The mass increase is much, much too small to be noticed. 
EVALUATE: In both cases the energy increase corresponds to a mass increase. But since 2c is a very large 
number the mass increase is very small. 
37.36. (a) 20 0E m c= . 
2 2
02 2E mc m c= = . Therefore, 
0
0 02 2
2 2
1 /
m
m m m
v c
= ⇒ =
−
. 
2 2
8
2 2
1 3
1 3 4 0.866 2.60 10 m s
4 4
v v
v c c
c c
= − ⇒ = ⇒ = = = × 
(b) 2 2 200 2 210 1
m
m c mc c
v c
= =
−
. 
2 2
8
2 2
1 99 99
1 . 0.995 2.98 10 m s
100 100 100
v v
v c c
c c
− = ⇒ = = = = × . 
37.37. IDENTIFY and SET UP: The energy equivalent of mass is 2E mc= . 3 3 37.86 g/cm 7.86 10 kg/mρ = = × . For a 
cube, 3V L= . 
EXECUTE: (a) 
20
3
2 8 2
1.0 10 J
1.11 10 kg
(3.00 10 m/s)
E
m
c
×= = = ×
×
 
(b) 
m
V
ρ = so 
3
3
3 3
1.11 10 kg
0.141 m
7.86 10 kg/m
m
V
ρ
×= = =
×
. 1/ 3 0.521 m 52.1 cmL V= = = 
EVALUATE: Particle/antiparticle annihilation has been observed in the laboratory, but only with small quantities 
of antimatter. 
37.38. 27 8 2 10(5.52 10 kg)(3.00 10 m s) 4.97 10 J 3105 MeV.− −× × = × = 
37.39. IDENTIFY and SET UP: The total energy is given in terms of the momentum by Eq.(37.39). In terms of the total 
energy E, the kinetic energy K is 2K E mc= − (from Eq.37.38). The rest energy is 2.mc 
EXECUTE: (a) 2 2 2( ) ( )E mc pc= + = 
27 8 2 2 18 8 2[(6.64 10 )(2.998 10 ) ] [(2.10 10 )(2.998 10 )] J− −× × + × × 
108.67 10 JE −= × 
(b) 2 27 8 2 10(6.64 10 kg)(2.998 10 m/s) 5.97 10 Jmc − −= × × = × 
2 10 10 108.67 10 J 5.97 10 J 2.70 10 JK E mc − − −= − = × − × = × 
(c) 
10
2 10
2.70 10 J
0.452
5.97 10 J
K
mc
−
−
×= =
×
 
EVALUATE: The incorrect nonrelativistic expressions for K and p give 2 10/ 2 3.3 10 J;K p m −= = × the correct 
relativistic value is less than this. 
37.40. 
1 22
2 4 2 2 1 2 2( ) 1
p
E m c p c mc
mc
⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
2 2
2 2 2 2
2 2
1 1
1
2 2 2
p p
E mc mc mc mv
m c m
⎛ ⎞
≈ + = + = +⎜ ⎟
⎝ ⎠
, the sum of the rest mass energy and the classical kinetic energy. 
37.41. (a) 7
2 2
1
8 10 m s γ 1.0376
1
v
v c
= × ⇒ = =
−
. For pm m= , 
2 12
nonrel
1
5.34 10 J
2
K mv −= = × . 
2 12 rel
rel
nonrel
(γ 1) 5.65 10 J. 1.06.
K
K mc
K
−= − = × = 
(b) 82.85 10 m s; γ 3.203.v = × = 
2 11 2 10
rel rel rel nonrel
1
6.78 10 J; (γ 1) 3.31 10 J; 4.88.
2
K mv K mc K K− −= = × = − = × = 
37-8 Chapter 37 
37.42. IDENTIFY: Since the speeds involved are close to that of light, we must use the relativistic formula for kinetic energy. 
SET UP: The relativistic kinetic energy is 2 2
2 2
1
( 1) 1
1 /
K mc mc
v c
γ
⎛ ⎞
= − = −⎜ ⎟
−⎝ ⎠
. 
(a) 
( )
2 2 27 8 2
2 2 2
1 1
( 1) 1 (1.67 10 kg)(3.00 10 m s) 1
1 / 1 0.100 /
K mc mc
v c c c
γ −
⎛ ⎞⎛ ⎞ ⎜ ⎟= − = − = × × −⎜ ⎟ ⎜ ⎟−⎝ ⎠ −⎝ ⎠
 
10 131(1.50 10 J) 1 7.56 10 J 4.73 MeV
1 0.0100
K − −
⎛ ⎞= × − = × =⎜ ⎟−⎝ ⎠
 
(b) 10 11
2
1(1.50 10 J) 1 2.32 10 J 145 MeV
1 (0.500)
K − −
⎛ ⎞
⎜ ⎟= × − = × =
⎜ ⎟−⎝ ⎠
 
(c) 10 10
2
1(1.50 10 J) 1 1.94 10J 1210 MeV
1 (0.900)
K − −
⎛ ⎞
⎜ ⎟= × − = × =
⎜ ⎟−⎝ ⎠
 
(d) 11 13 112.32 10 J 7.56 10 J 2.24 10 J 140 MeVE − − −Δ = × − × = × = 
(e) 10 11 101.94 10 J 2.32 10 J 1.71 10 J 1070 MeVE − − −Δ = × − × = × = 
(f) Without relativity, 2
1
2
K mv= . The work done in accelerating a proton from 0.100c to 0.500c in the 
nonrelativistic limit is 2 2 11
1 1
(0.500 ) (0.100 ) 1.81 10 J 113 MeV
2 2
E m c m c −Δ = − = × = . 
The work done in accelerating a proton from 0.500c to 0.900c in the nonrelativistic limit is 
2 2 111 1(0.900 ) (0.500 ) 4.21 10 J 263 MeV
2 2
E m c m c −Δ = − = × = . 
EVALUATE: We see in the first case the nonrelativistic result is within 20% of the relativistic result. In the second 
case, the nonrelativistic result is very different from the relativistic result since the velocities are closer to c. 
37.43. IDENTIFY and SET UP: Use Eq.(23.12) and conservation of energy to relate the potential difference to the kinetic 
energy gained by the electron. Use Eq.(37.36) to calculate the kinetic energy from the speed. 
EXECUTE: (a) K q V e V= Δ = Δ 
2 2 13
2 2
1
1 4.025 3.295 10 J 2.06 MeV
1 /
K mc mc
v c
−⎛ ⎞= − = = × =⎜ ⎟
−⎝ ⎠
 
6/ 2.06 10 VV K eΔ = = × 
(b) From part (a), 133.30 10 J 2.06 MeVK −= × = 
EVALUATE: The speed is close to c and the kinetic energy is four times the rest mass. 
37.44. (a) According to Eq.(37.38) and conservation of mass-energy 
2 2 2 9.752 2 1 1 1.292.
2 2(16.7)
m
Mc mc Mc
M
γ γ+ = ⇒ = + = + = 
Note that since 
2 2
1 ,
1 v c
γ =
−
we have that 
2 2
1 1
1 1 0.6331.
(1.292)
v
c γ
= − = − = 
(b) According to Eq.(37.36), the kinetic energy of each proton is 
2 27 8 2
13
1.00 MeV
( 1) (1.292 1)(1.67 10 kg)(3.00 10 m s) 274 MeV.
1.60 10 J
K Mcγ − −
⎛ ⎞
= − = − × × =⎜ ⎟×⎝ ⎠
 
(c) The rest energy of 0η is 2 28 8 2 13
1.00 MeV
(9.75 10 kg)(3.00 10 m s) 548 MeV.
1.60 10 J
mc − −
⎛ ⎞
= × × =⎜ ⎟×⎝ ⎠
 
(d) The kinetic energy lost by the protons is the energy that produces the 0 ,η 
548 MeV 2(274 MeV).= 
37.45. IDENTIFY: The relativistic expression for the kinetic energy is 2( 1)K mcγ= − , where 1
1 x
γ =
−
and 2 2/x v c= . 
The Newtonian expression for the kinetic energy is 2N
1
2
K mv= . 
SET UP: Solve for v such that N
3
2
K K= . 
Relativity 37-9 
EXECUTE: 2 2
3
( 1)
4
mc mvγ − = . 1 31
41
x
x
− =
−
. 
2
1 3
1
1 4
x
x
⎛ ⎞= +⎜ ⎟− ⎝ ⎠
. After a little algebra this becomes 
29 15 8 0x x+ − = . ( )21 15 (15) 4(9)(8)18x = − ± + . The positive root is 0.425x = . 2 2/x v c= , so 
 0.652v x c c= = . 
EVALUATE: The fractional increase of the relativistic expression above the nonrelativistic one increases as v increases. 
37.46. The fraction of the initial mass (a) that becomes energy is 3
(4.0015 u)
1 6.382 10 ,
2(2.0136 u)
−− = × and so the energy released 
per kilogram is 3 8 2 14(6.382 10 )(1.00 kg)(3.00 10 m s) 5.74 10 J.−× × = × 
(b) 
19
4
14
1.0 10 J
1.7 10 kg.
5.74 10 J kg
× = ×
×
 
37.47. (a) 2 2 26 8 2 9, (3.8 10 J) (2.998 10 m s) 4.2 10 kgE mc m E c= = = × × = × . 
1 kg is equivalent to 2.2 lbs, so 64.6 10m = × tons 
(b) The current mass of the sun is 301.99 10 kg,× so it would take it 
30 9 20 13(1.99 10 kg) (4.2 10 kg s) 4.7 10 s 1.5 10 years× × = × = × to use up all its mass. 
37.48. IDENTIFY: Since the final speed is close to the speed of light, there will be a considerable difference between the 
relativistic and nonrelativistic results. 
SET UP: The nonrelativistic work-energy theorem is 2 20
1 1
2 2
F x mv mvΔ = − , and the relativistic formula for a 
constant force is 2( 1)F x mcγΔ = − . 
(a) Using the classical work-energy theorem and solving for xΔ , we obtain 
2 2 9 8 2
0
6
( ) (0.100 10 kg)[(0.900)(3.00 10 m s)]
3.65 m.
2 2(1.00 10 N)
m v v
x
F
−− × ×Δ = = =
×
 
(b) Using the relativistic work-energy theorem for a constant force, we obtain 
2( 1)
.
mc
x
F
γ −Δ = 
For the given speed, 
2
1 2.29,
1 0.900
γ = =
−
 thus 
9 8 2
6
(2.29 1)(0.100 10 kg)(3.00 10 m s)
11.6 m.
(1.00 10 N)
x
−− × ×Δ = =
×
 
EVALUATE: (c) The distance obtained from the relativistic treatment is greater. As we have seen, more energy is 
required to accelerate an object to speeds close to c, so that force must act over a greater distance. 
37.49. (a) IDENTIFY and SET UP: 80 2.60 10 st
−Δ = × is the proper time, measured in the pion’s frame. The time 
measured in the lab must satisfy ,d c t= Δ where .u c≈ Calculate tΔ and then use Eq.(37.6) to calculate u. 
EXECUTE: 
3
6
8
1.20 10 m
4.003 10 s
2.998 10 m/s
d
t
c
−×Δ = = = ×
×
 
0
2 21 /
t
t
u c
ΔΔ =
−
 so 2 2 1/ 2 0(1 / )
t
u c
t
Δ− =
Δ
 and 
2
2 2 0(1 / )
t
u c
t
Δ⎛ ⎞− = ⎜ ⎟Δ⎝ ⎠
 
Write (1 )u c= − Δ so that 2 2 2( / ) (1 ) 1 2 1 2u c = − Δ = − Δ + Δ ≈ − Δ since Δ is small. 
Using this in the above gives 
2
01 (1 2 )
t
t
Δ⎛ ⎞− − Δ = ⎜ ⎟Δ⎝ ⎠
 
22 8
50
6
1 1 2.60 10 s
2.11 10
2 2 4.003 10 s
t
t
−
−
−
⎛ ⎞Δ ×⎛ ⎞Δ = = = ×⎜ ⎟⎜ ⎟Δ ×⎝ ⎠ ⎝ ⎠
 
EVALUATE: An alternative calculation is to say that the length of the tube must contract relative to the moving 
pion so that the pion travels that length before decaying. The contracted length must be 
8 8
0 (2.998 10 m/s)(2.60 10 s) 7.79 m.l c t
−= Δ = × × = 
2 2
0 1 /l l u c= − so 
2
2 2
0
1 /
l
u c
l
⎛ ⎞
− = ⎜ ⎟
⎝ ⎠
 
37-10 Chapter 37 
Then (1 )u c= − Δ gives 
2 2
5
3
0
1 1 7.79 m
2.11 10 ,
2 2 1.20 10 m
l
l
−⎛ ⎞ ⎛ ⎞Δ = = = ×⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠
 which checks. 
(b) IDENTIFY and SET UP: 2E mcγ= (Eq.(37.38). 
EXECUTE: 
2 2 5
1 1 1
154
21 / 2(2.11 10 )u c
γ
−
= = = =
Δ− ×
 
4154(139.6 MeV) 2.15 10 MeV 21.5 GeVE = = × = 
EVALUATE: The total energy is 154 times the rest energy. 
37.50. IDENTIFY and SET UP: The proper length of a side is 0l a= . The side along the direction of motion is shortened 
to 2 20 1 /l l v c= − . The sides in the two directions perpendicular to the motion are unaffected by the motion and 
still have a length a. 
EXECUTE: 2 3 2 21 /V a l a v c= = − 
37.51. IDENTIFY and SET UP: There must be a length contraction such that the length a becomes the same as b; 
0 0, . l a l b l= = is the distance measured by an observer at rest relative to the spacecraft. Use Eq.(37.16) and solve 
for u. 
EXECUTE: 2 2
0
1 /
l
u c
l
= − so 2 21 / ;b u c
a
= − 
1.40a b= gives 2 2/1.40 1 /b b u c= − and thus 2 2 21 / 1/(1.40)u c− = 
2 81 1/(1.40) 0.700 2.10 10 m/su c c= − = = × 
EVALUATE: A length on the spacecraft in the direction of the motion is shortened. A length perpendicular to the 
motion is unchanged. 
37.52. IDENTIFY and SET UP: The proper time 0tΔ is the time that elapses in the frame of the space probe. tΔ is the 
time that elapses in the frame of the earth. The distance traveled is 42.2 light years, as measured in the earth frame. 
EXECUTE: (a) Light travels 42.2 light years in 42.2 yr, so (42.2 yr) 42.6 yr
0.9910
c
t
c
⎛ ⎞Δ = =⎜ ⎟
⎝ ⎠
. 
2 2 2
0 1 / (42.6 yr) 1 (0.9910) 5.7 yrt t u cΔ = Δ − = − = . She measures her biological age to be 
19 yr 5.7 yr 24.7 yr.+ = 
(b) Her age measured by someone on earth is 19 yr 42.6 yr 61.6 yr+ = . 
37.53. (a) 
2
2
22
1 γ 1 99
 and 10 0.995.
γ 1001 ( )
v v
E mc
c cv c
γ γ −= = = ⇒ = ⇒ = =
−
 
(b) 
2
2 2 2 2 2 2 2 4 2( ) γ , γ 1
v
pc m v c E m c
c
⎛ ⎞⎛ ⎞= = +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
2 2
22 2
2
( ) 1 1
0.01 1%.
1 (10 (0.995))
1
E pc
E v
c
γ
−
⇒ = = = =
+⎛ ⎞+ ⎜ ⎟
⎝ ⎠
 
37.54. IDENTIFY and SET UP: The clock on the plane measures the proper time 0.tΔ 
44.00 h 4.00 h (3600 s/1 h) 1.44 10 s.tΔ = = = × 
0
2 21 /
t
t
u c
ΔΔ =
−
 and 2 20 1 /t t u cΔ = Δ − 
EXECUTE: 
u
c
 small so 
2
2 2 2 2 1/ 2
2
1
1 / (1 / ) 1 ;
2
u
u c u c
c
− = − ≈ − thus 
2
0 2
1
1
2
u
t t
c
⎛ ⎞
Δ = Δ −⎜ ⎟
⎝ ⎠
 
The difference in the clock readings is 
22
4 9
0 2 8
1 1 250 m/s
(1.44 10 s) 5.01 10 s.
2 2 2.998 10 m/s
u
t t t
c
−⎛ ⎞Δ − Δ = Δ = × = ×⎜ ⎟×⎝ ⎠
 The 
clock on the plane has the shorter elapsed time. 
EVALUATE: 0tΔ is always less than ;tΔ our results agree with this. The speed of the planeis much less than the 
speed of light, so the difference in the reading of the two clocks is very small. 
37.55. IDENTIFY: Since the speed is very close to the speed of light, we must use the relativistic formula for kinetic 
energy. 
Relativity 37-11 
SET UP: The relativistic formula for kinetic energy is 2
2 2
1
1
1
K mc
v c
⎛ ⎞
⎜ ⎟= −
⎜ ⎟−⎝ ⎠
 and the relativistic mass is 
rel 2 21
m
m
v c
=
−
. 
EXECUTE: (a) 12 67 10 eV 1.12 10 JK −= × = × . Using this value in the relativistic kinetic energy formula and 
substituting the mass of the proton for m, we get 2
2 2
1
1
1
K mc
v c
⎛ ⎞
⎜ ⎟= −
⎜ ⎟−⎝ ⎠
 
which gives 3
2 2
1
7.45 10
1 v c
= ×
−
and 
2
2 3 2
1
1
(7.45 10 )
v
c
− =
×
. Solving for v gives 
2
2 2
( )( ) 2( )
1
v c v c v c v
c c c
+ − −− = = , since c + v ≈ 2c. Substituting (1 )v c= − Δ , we have. 
[ ]2
2
2 (1 )2( )
1 2
c cv c v
c c c
− − Δ−− = = = Δ . Solving for Δ gives 
( )232 2 9
1
7.45 101 /
9 10
2 2
v c −×−Δ = = = × , to one 
significant digit. 
(b) Using the relativistic mass formula and the result that 3
2 2
1
7.45 10
1 v c
= ×
−
, we have 
3
rel 2 2 2 2
1
(7 10 )
1 1
m
m m m
v c v c
⎛ ⎞
⎜ ⎟= = = ×
⎜ ⎟− −⎝ ⎠
, to one significant digit. 
EVALUATE: At such high speeds, the proton’s mass is over 7000 times as great as its rest mass. 
37.56. IDENTIFY and SET UP: The energy released is 2( )E m c= Δ . 
4
1
(8.00 kg)
10
m
⎛ ⎞Δ = ⎜ ⎟
⎝ ⎠
. av
E
P
t
= . The change in 
gravitational potential energy is mg yΔ . 
EXECUTE: (a) 2 8 2 13
4
1
( ) (8.00 kg)(3.00 10 m/s) 7.20 10 J
10
E m c
⎛ ⎞= Δ = × = ×⎜ ⎟
⎝ ⎠
 
(b) 
13
19
av 6
7.20 10 J
1.80 10 W
4.00 10 s
E
P
t −
×= = = ×
×
 
(c) E U mg y= Δ = Δ . 
13
9
2 3
7.20 10 J
7.35 10 kg
(9.80 m/s )(1.00 10 m)
E
m
g y
×= = = ×
Δ ×
 
37.57. IDENTIFY and SET UP: In crown glass the speed of light is .
c
v
n
= Calculate the kinetic energy of an electron that 
has this speed. 
EXECUTE: 
8
82.998 10 m/s 1.972 10 m/s.
1.52
v
×= = × 
2 ( 1)K mc γ= − 
2 31 8 2 14 19(9.109 10 kg)(2.998 10 m/s) 8.187 10 J(1 eV/1.602 10 J) 0.5111 MeVmc − − −= × × = × × = 
2 2 8 8 2
1 1
1.328
1 / 1 ((1.972 10 m/s)/(2.998 10 m/s))v c
γ = = =
− − × ×
 
2 ( 1) (0.5111 MeV)(1.328 1) 0.168 MeVK mc γ= − = − = 
EVALUATE: No object can travel faster than the speed of light in vacuum but there is nothing that prohibits an 
object from traveling faster than the speed of light in some material. 
37.58. (a) 
( )
,
p E c E
v
m m mc
= = = where the atom and the photon have the same magnitude of momentum, .E c 
(b) ,
E
v c
mc
= so 2.E mc 
37.59. IDENTIFY and SET UP: Let S be the lab frame and S′ be the frame of the proton that is moving in the +x direction, 
so / 2u c= + . The reference frames and moving particles are shown in Figure 37.59. The other proton moves in 
37-12 Chapter 37 
the x− direction in the lab frame, so / 2v c= − . A proton has rest mass 27p 1.67 10 kgm
−= × and rest energy 
2
p 938 MeVm c = . 
EXECUTE: (a) 
2 2
/ 2 / 2 4
1 / 1 ( / 2)( / 2) / 5
v u c c c
v
uv c c c c
− − −′ = = = −
− − −
 
The speed of each proton relative to the other is 
4
5
c . 
(b) In nonrelativistic mechanics the speeds just add and the speed of each relative to the other is c. 
(c) 
2
2
2 21 /
mc
K mc
v c
= −
−
 
(i) Relative to the lab frame each proton has speed / 2v c= . The total kinetic energy of each proton is 
2
938 MeV
(938 MeV) 145 MeV
1
1
2
K = − =
⎛ ⎞− ⎜ ⎟
⎝ ⎠
. 
(ii) In its rest frame one proton has zero speed and zero kinetic energy and the other has speed 
4
5
c . In this frame 
the kinetic energy of the moving proton is 
2
938 MeV
(938 MeV) 625 MeV
4
1
5
K = − =
⎛ ⎞− ⎜ ⎟
⎝ ⎠
 
(d) (i) Each proton has speed / 2v c= and kinetic energy 
( )
2
221 1 938 MeV/ 2 117 MeV
2 2 8 8
mc
K mv m c
⎛ ⎞= = = = =⎜ ⎟
⎝ ⎠
 
(ii) One proton has speed 0v = and the other has speed c. The kinetic energy of the moving proton 
is 2
1 938 MeV
469 MeV
2 2
K mc= = = 
EVALUATE: The relativistic expression for K gives a larger value than the nonrelativistic expression. The kinetic 
energy of the system is different in different frames. 
 
Figure 37.59 
37.60. IDENTIFY and SET UP: Let S be the lab frame and let S′ the frame of the proton that is moving in the +x direction 
in the lab frame, as shown in Figure 37.60. In S′ the other proton moves in the x′− direction with speed / 2c , so 
/ 2v c′ = − . In the lab frame each proton has speed cα , where α is a constant that we need to solve for. 
EXECUTE: (a) 
21 /
v u
v
uv c
′ +=
′+
with v cα= − , u cα= + and 0.50v c′ = − gives 
2
0.50
1 ( )( 0.50 ) /
c c
c
c c c
αα
α
− +− =
+ −
and 
0.50
1 0.50
αα
α
− +− =
−
. 2 4 1 0α α− + = and 0.268α = or 3.73α = . Can’t have v c> , so only 0.268α = is physically 
allowed. The speed measured by the observer in the lab is 0.268c. 
(b) (i) 0.269v c= . 1.0380γ = . 2( 1) 35.6 MeVK mcγ= − = . 
Relativity 37-13 
(ii) 0.500v c= . 1.1547γ = . 2( 1) 145 MeVK mcγ= − = . 
 
Figure 37.60 
37.61. 2 2 2x c t′ ′= ( )22 2 2 2 2( )x ut c t ux cγ γ⇒ − = − 
2 2 2 21( ) 1 ( ) ( ) .
u
x ut c t ux c x x u c t u c x ct x c t
c c
⎛ ⎞⇒ − = − ⇒ + = + = + ⇒ = ⇒ =⎜ ⎟
⎝ ⎠
 
37.62. IDENTIFY and SET UP: Let S be the lab frame and let S′ be the frame of the nucleus. Let the +x direction be the 
direction the nucleus is moving. 0.7500u c= . 
EXECUTE: (a) 0.9995v c′ = + . 
2
0.9995 0.7500
0.999929
1 / 1 (0.7500)(0.9995)
v u c c
v c
uv c
′ + += = =
′+ +
 
(b) 0.9995v c′ = − . 0.9995 0.7500 0.9965
1 (0.7500)( 0.9995)
c c
v c
− += = −
+ −
 
(c) emitted in same direction: 
(i) 2
2 2 2
1 1
1 (0.511 MeV) 1 42.4 MeV
1 / 1 (0.999929)
K mc
v c
⎛ ⎞⎛ ⎞
⎜ ⎟= − = − =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
 
(ii) 2
2 2 2
1 1
1 (0.511 MeV) 1 15.7 MeV
1 / 1 (0.9995)
K mc
v c
⎛ ⎞⎛ ⎞
′ ⎜ ⎟= − = − =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
 
(d) emitted in opposite direction: 
(i) 2
2 2 2
1 1
1 (0.511 MeV) 1 5.60 MeV
1 / 1 (0.9965)
K mc
v c
⎛ ⎞⎛ ⎞
⎜ ⎟= − = − =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
 
(ii) 2
2 2 2
1 1
1 (0.511 MeV) 1 15.7 MeV
1 / 1 (0.9995)
K mc
v c
⎛ ⎞⎛ ⎞
′ ⎜ ⎟= − = − =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
 
37.63. IDENTIFY and SET UP: Use Eq.(37.30), with / ,a dv dt= to obtain an expression for / .dv dt Separate the 
variables v and t and integrate to obtain an expression for ( ).v t In this expression, let .t → ∞ 
EXECUTE: 2 2 3 / 2(1 / ) .
dv F
a v c
dt m
= = − (One-dimensional motion is assumed, and all the F, v, and a refer to x-
components.) 
2 2 3/ 2(1 / )
dv F
dt
v c m
⎛ ⎞= ⎜ ⎟− ⎝ ⎠
 
Integrate from 0,t = when 0,v = to time t, when the velocity is v. 
2 2 3 / 20 0(1 / )
v tdv F
dt
v c m
⎛ ⎞= ⎜ ⎟− ⎝ ⎠∫ ∫ 
Since F is constant, 
0
.
t F Ft
dt
m m
⎛ ⎞ =⎜ ⎟
⎝ ⎠∫ In the velocity integral make the change of variable / ;y v c= then / .dy dv c= 
/
/
2 2 3/ 2 2 3 / 2 2 1/ 2 2 20 0
0
(1 / ) (1 ) (1 ) 1 /
v c
v v cdv dy y v
c c
v c y y v c
⎡ ⎤
= = =⎢ ⎥− − − −⎣ ⎦
∫ ∫ 
Thus 
2 2
.
1 /
v Ft
mv c
=
−
 
37-14 Chapter 37 
Solve this equation for v: 
22
2 21 /
v Ft
v c m
⎛ ⎞= ⎜ ⎟− ⎝ ⎠
 and 
2
2 2 2(1 / )
Ft
v v c
m
⎛ ⎞= −⎜ ⎟
⎝ ⎠
 
2 2
2 1
Ft Ft
v
mc m
⎛ ⎞⎛ ⎞ ⎛ ⎞+ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
 so 
2 2 2 2 2
( / )
1 ( / )
Ft m Ft
v c
Ft mc m c F t
= =
+ +
 
As 
2 2 2 2 2 2
, 1,
Ft Ft
t
m c F t F t
→ ∞ → →
+
 so .v c→ 
EVALUATE: Note that 
2 2 2 2
Ft
m c F t+
 is always less than 1, so v c< always and v approaches c only when .t → ∞ 
37.64. Setting 0x = in Eq.(37.21), the first equation becomes x utγ′ = − and the last, upon multiplication by ,c becomes 
.ct ctγ′ = Squaring and subtracting gives 2 2 2 2 2 2 2 2 2 2γ ( ) ,c t x c t u t c t′ ′− = − = or 2 2 84.53 10 m.x c t t′ ′= − = × 
37.65. (a) IDENTIFY and SET UP: Use the Lorentz coordinate transformation (Eq.37.21) for 1 1( , )x t and 2 2( , ) :x t 
1 1
1 2 2
,
1 /
x ut
x
u c
−′ =
−
 2 22 2 21 /
x ut
x
u c
−′ =
−
 
2
1 1
1 2 2
/
,
1 /
t ux c
t
u c
−′ =
−
 
2
2 2
2 2 2
/
1 /
t ux c
t
u c
−′ =
−
 
Same point in S′ implies 1 2.x x′ ′=What then is 2 1 ?t t t′ ′ ′Δ = − 
EXECUTE: 1 2x x′ ′= implies 1 1 2 2x ut x ut− = − 
2 1 2 1( )u t t x x− = − and 2 1
2 1
x x x
u
t t t
− Δ= =
− Δ
 
From the time transformation equations, 
2
2 1 2 2
1
( / )
1 /
t t t t u x c
u c
′ ′ ′Δ = − = Δ − Δ
−
 
Using the result that 
x
u
t
Δ=
Δ
 gives 
2 2
2 2 2
1
( ( ) /(( ) ))
1 ( ) /(( ) )
t t x t c
x t c
′Δ = Δ − Δ Δ
− Δ Δ
 
2 2
2 2 2
( ( ) /(( ) ))
( ) ( ) /
t
t t x t c
t x c
Δ′Δ = Δ − Δ Δ
Δ − Δ
 
2 2 2
2 2
2 2 2
( ) ( ) /
( ) ( / ) ,
( ) ( ) /
t x c
t t x c
t x c
Δ − Δ′Δ = = Δ − Δ
Δ − Δ
 as was to be shown. 
This equation doesn’t have a physical solution (because of a negative square root) if 2 2( / ) ( )x c tΔ > Δ or .x c tΔ ≥ Δ 
(b) IDENTIFY and SET UP: Now require that 2 1t t′ ′= (the two events are simultaneous in S′ ) and use the Lorentz 
coordinate transformation equations. 
EXECUTE: 2 1t t′ ′= implies 
2 2
1 1 2 2/ /t ux c t ux c− = − 
2 1
2 1 2
x x
t t u
c
−⎛ ⎞− = ⎜ ⎟
⎝ ⎠
 so 
2
x
t u
c
Δ⎛ ⎞Δ = ⎜ ⎟
⎝ ⎠
 and 
2c t
u
x
Δ=
Δ
 
From the Lorentz transformation equations, 
2 1 2 2
1
( ).
1 /
x x x x u t
u c
⎛ ⎞
′ ′ ′Δ = − = Δ − Δ⎜ ⎟
−⎝ ⎠
 
Using the result that 2 /u c t x= Δ Δ gives 
2 2
2 2 2
1
( ( ) / )
1 ( ) /( )
x x c t x
c t x
′Δ = Δ − Δ Δ
− Δ Δ
 
2 2
2 2 2
( ( ) / )
( ) ( )
x
x x c t x
x c t
Δ′Δ = Δ − Δ Δ
Δ − Δ
 
2 2 2
2 2 2
2 2 2
( ) ( )
( ) ( )
( ) ( )
x c t
x x c t
x c t
Δ − Δ′Δ = = Δ − Δ
Δ − Δ
 
(c) IDENTIFY and SET UP: The result from part (b) is 2 2 2( ) ( )x x c t′Δ = Δ − Δ 
Relativity 37-15 
Solve for 2 2 2 2: ( ) ( ) ( )t x x c t′Δ Δ = Δ − Δ 
EXECUTE: 
2 2 2 2
8
8
( ) ( ) (5.00 m) (2.50 m)
1.44 10 s
2.998 10 m/s
x x
t
c
−′Δ − Δ −Δ = = = ×
×
 
EVALUATE: This provides another illustration of the concept of simultaneity (Section 37.2): events observed to 
be simultaneous in one frame are not simultaneous in another frame that is moving relative to the first. 
37.66. (a) 80.0 m s is non-relativistic, and 2
1
186 J.
2
K mv= = 
(b) 2 15( 1) 1.31 10 J.mcγ − = × 
(c) In Eq. (37.23), c) 8 8 72.20 10 m s, 1.80 10 m s,and so 7.14 10 m s.v u v′ = × = − × = × 
(d)
20.0 m
13.6 m.
γ
= 
(e) 8
8
20.0 m
9.09 10 s.
2.20 10 m s
−= ×
×
 
(f) 8 8
8
13.6 m
6.18 10 s, or 6.18 10 s.
2.20 10 m s
t
t t
γ
− −′ ′= = × = = ×
×
 
37.67. IDENTIFY and SET UP: An increase in wavelength corresponds to a decrease in frequency ( / ),f c λ= so the 
atoms are moving away from the earth. Receding, so use Eq.(37.26): 0
c u
f f
c u
−=
+
 
EXECUTE: Solve for u: 20( / ) ( )f f c u c u+ = − and 
2
0
2
0
1 ( / )
1 ( / )
f f
u c
f f
⎛ ⎞−= ⎜ ⎟+⎝ ⎠
 
0 0/ , /f c f cλ λ= = so 0 0/ /f f λ λ= 
2 2
80
2 2
0
1 ( / ) 1 (656.3/953.4)
0.357 1.07 10 m/s
1 ( / ) 1 (656.3/953.4)
u c c c
λ λ
λ λ
⎛ ⎞ ⎛ ⎞− −= = = = ×⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
 
EVALUATE: The relative speed is large, 36% of c. The cosmological implication of such observations will be 
discussed in Section 44.6. 
37.68. The baseball had better be moving non-relativistically, so the Doppler shift formula (Eq.(37.25)) becomes 
0 (1 ( )).f f u c≅ − In the baseball’s frame, this is the frequency with which the radar waves strike the baseball, and 
the baseball reradiates at f. But in the coach’s frame, the reflected waves are Doppler shifted again, so the detected 
frequency is 20 0 0(1 ( )) (1 ( )) (1 2( )), so 2 ( )f u c f u c f u c f f u c− = − ≈ − Δ = and the fractional frequency shift is 
0
2( ).
f
u c
f
Δ = In this case, 
7
8
0
(2.86 10 )
(3.00 10 m) 42.9 m s 154 km h 92.5 mi h.
2 2
f
u c
f
−Δ ×= = × = = = 
37.69. IDENTIFY and SET UP: 18500 light years 4.73 10 m= × . The proper distance l0 to the star is 500 light years. The 
energy needed is the kinetic energy of the rocket at its final speed. 
EXECUTE: (a) 0.50u c= . 
18
10
8
4.73 10 m
3.2 10 s 1000 yr
(0.50)(3.00 10 m/s)
d
t
u
×Δ = = = × =
×
 
The proper time is measured by the astronauts. 2 20 1 / 866 yrt t u cΔ = Δ − = 
2
2 8 2 19
2 2 2
1
(1000 kg)(3.00 10 m/s) 1 1.4 10 J
1 / 1 (0.500)
mc
K mc
v c
⎛ ⎞
⎜ ⎟= − = × − = ×
⎜ ⎟− −⎝ ⎠
 
This is 140% of the U.S. yearly use of energy. 
(b) 0.99u c= . 
18
10
8
4.73 10 m
1.6 10 s 505 yr
(0.99)(3.00 10 m/s)
d
t
u
×Δ = = = × =
×
, 0 71 yrtΔ = 
19 20
2
1
(9.00 10 J) 1 5.5 10 J
1 (0.99)
K
⎛ ⎞
⎜ ⎟= × − = ×
⎜ ⎟−⎝ ⎠
 
This is 55 times the U.S. yearly use. 
37-16 Chapter 37 
(c) 0.9999u c= . 
18
10
8
4.73 10 m
1.58 10 s 501 yr
(0.9999)(3.00 10 m/s)
d
t
u
×Δ = = = × =
×
, 0 7.1 yrtΔ = 
19 21
2
1
(9.00 10 J) 1 6.3 10 J
1 (0.9999)
K
⎛ ⎞
⎜ ⎟= × − = ×
⎜ ⎟−⎝ ⎠
 
This is 630 times the U.S. yearly use. 
The energy cost of accelerating a rocket to these speeds is immense. 
37.70. (a) As in the hint, both the sender and the receiver measure the same distance. However, in our frame, the ship has 
moved between emission of successive wavefronts, and we can use the time 1T f= as the proper time, with the 
result that 0 0.f f fγ= > 
(b) Toward: 
1/2
0
1 0.758
345 MHz 930 MHz
1 0.758
c u
f f
c u
+ +⎛ ⎞= = =⎜ ⎟− −⎝ ⎠
 
0 930 MHz 345 MHz 585 MHz.f f− = − = 
Away: 
1/2
0 0
1 0.758
345 MHz 128 MHz and 217 MHz.
1 0.758
c u
f f f f
c u
− −⎛ ⎞= = = − = −⎜ ⎟+ +⎝ ⎠
 
(c) 0 0 0γ 1.53 528 MHz, 183 MHz.f f f f= = − = The shift is still bigger than 0f , but not as large as the approaching 
frequency. 
37.71. The crux of this problem is the question of simultaneity. To be “in the barn at one time” for the runner is different 
than for a stationary observer in the barn. The diagram in Figure 37.71a shows the rod fitting into the barn at time 
0t = , according to the stationary observer. The diagram in Figure 37.71b is in the runner’s frame of reference. The 
front of the rod enters the barn at time 1t and leaves the back of the barn at time 2.t However, the back of the rod 
does not enter the front of the barn until the later time 3.t 
 
Figure 37.71 
37.72. In Eq.(37.23), , ( ),u V v c n′= = and so 
2
( / ) ( / )
.
1 ( / )1
c n V c n V
v
cV V nc
nc
+ += =
++
 For V non-relativistic, this is 
2 2
2
1
(( ) )(1 ( / )) ( / ) ( / ) ( / ) 1
c
v cn V V nc nc n V V n V nc V
n n
⎛ ⎞≈ + − = + − − ≈ + −⎜ ⎟
⎝ ⎠
 , so 
2
1
1 .k
n
⎛ ⎞= −⎜ ⎟
⎝ ⎠
 For water, 1.333n = 
and 0.437.k = 
Relativity 37-17 
37.73. (a) 
dv
a
dt
′ =
′
. 2( )dt dt udx cγ′ = − . 
2 2 2 2(1 ) (1 )
dv v u u
dv dv
uv c uv c c
−′ = +
− −
 
22 22
1
1 (1 )
dv v u u
dv uv c cuv c
′ − ⎛ ⎞= + ⎜ ⎟− − ⎝ ⎠
. 
2 2 2
2 2 2 2 2
1 ( ) 1
1 (1 ) (1 )
v u u c u c
dv dv dv
uv c uv c uv c
⎛ ⎞ ⎛ ⎞− −′ = + =⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠
 
2 2
2 22 2
2 2 2 2
(1 )
(1 ) 1(1 )
γ (1 ) γ(1 )
u c
dv
dv u cuv c
a
dt u dx c dt uv c uv cγ
−
−−′ = =
− − −
 
2 2 3 2 2 3(1 ) (1 ) .a u c uv c −= − − 
(b) Changing frames from S S′ → just involves changing ,a a v v a′ ′→ → − ⇒ =
3
2 2 3 2
2
(1 ) 1 .
uv
a u c
c
−′⎛ ⎞′ − +⎜ ⎟
⎝ ⎠
 
37.74. (a) The speed v′ is measured relative to the rocket, and so for the rocket and its occupant, 0.v′ = The acceleration 
as seen in the rocket is given to be ,a g′ = and so the acceleration as measured on the earth is 
3 22
2
1 .
du u
a g
dt c
⎛ ⎞
= = −⎜ ⎟
⎝ ⎠
 
(b) With 1 0v = when 0t = , 
1 1 1
12 2 3 2 2 2 3 2 2 20 0
1
1 1
. . .
(1 ) (1 ) 1
t vdu du v
dt dt t
g u c g u c g v c
= = =
− − −∫ ∫
 
(c) 2 2/ 1 ,dt dt dt u cγ′ = = − so the relation in part (b) between dt and du, expressed in terms of dt′ and du, is 
2 2 3 2 2 2 22 2
1 1
.
(1 ) (1 )1
du du
dt dt
g u c g u cu c
γ′ = = =
− −−
 
Integrating as above (perhaps using the substitution z u c= ) gives 11 arctanh .
c v
t
g c
⎛ ⎞′ = ⎜ ⎟
⎝ ⎠
 For those who wish to 
avoid inverse hyperbolic functions, the above integral may be done by the method of partial fractions; 
1
(1 )(1 ) 2 1 1
du du du
gdt
u c u c u c uc
⎡ ⎤′ = = +⎢ ⎥+ − + −⎣ ⎦
, which integrates to 11
1
c
1n
2
c v
t
g c v
⎛ ⎞+′ = ⎜ ⎟−⎝ ⎠
. 
(d) Solving the expression from part (c) for 1v in terms of 1 1 1, ( ) tanh( ),t v c gt c′= so that 
2
1 11 ( ) 1 cosh( ),v c gt c′− = using the appropriate indentities for hyperbolic functions. Using this in the expression 
found in part (b), 11 1
1
tanh( )
sinh( ),
1 cosh( )
c gt c c
t gt c
g gt c g
′ ′= =
′
 which may be rearranged slightly as 1 1sinh .
gt gt
c c
′⎛ ⎞= ⎜ ⎟
⎝ ⎠
 If 
hyperbolic functions are not used, 1v in terms of 1t′ is found to be 
1 1
1 1
/ /
1
/ /
gt c gt c
gt c gt c
v e e
c e e
′ ′−
′ ′−
−=
+
 which is the same as 
tanh( 1gt c′ ). Inserting this expression into the result of part (b) gives, after much algebra, 1 11 ( ),2
gt c gt cct e e
g
′ ′−= − 
which is equivalent to the expression found using hyperbolic functions. 
(e) After the first acceleration period (of 5 years by Stella’s clock), the elapsed time on earth is 
9
1 1sinh( ) 2.65 10 s 84.0 yr.
c
t gt c
g
′ ′= = × = 
The elapsed time will be the same for each of the four parts of the voyage, so when Stella has returned, Terra has 
aged 336 yr and the year is 2436. (Keeping more precision than is given in the problem gives February 7 of that 
year.) 
37.75. (a) 14 14 140 4.568110 10 Hz; 4.568910 10 Hz; 4.567710 10 Hzf f f+ −= × = × = × 
0 2 2
0
2 2
0
0
( )
 
( ) ( ( )) ( ( ))
( ( )) ( ( ))( )
 
( )
c u v
f f
c u v f c u v f c u v
f c u v f c u vc u v
f f
c u v
+
+
−
−
⎫+ += ⎪− + − + = + +⎪⇒⎬ − − = + −+ − ⎪= ⎪− − ⎭
 
37-18 Chapter 37 
where u is the velocity of the center of mass and v is the orbital velocity. 
2
0
2
0
( ) 1
( )
( ) 1
f f
u v c
f f
+
+
−
⇒ + =
+
 and 
2 2
0
2 2
0
( ) 1
( )
( ) 1
f f
u v c
f f
−
−
−− =
+
 
4 45.25 10 m s and 2.63 10 m su v u v⇒ + = × − = − × . 
This gives 41.31 10 m su = + × (moving toward at 13.1 km s) and 43.94 10 m/sv = × . 
(b) 43.94 10 m s; 11.0 days.v T= × = 2 R vtπ = ⇒ 
4
9(3.94 10 m s)(11.0 days)(24 hrs day)(3600 sec hr) 5.96 10 m
2
R
π
×= = × . This is about 
0.040 times the earth-sun distance. 
Also the gravitational force between them (a distance of 2R) must equal the centripetal force from the center of 
mass: 
2 2 2 9 4 2
29
sun2 11 2 2
( ) 4 4(5.96 10 m)(3.94 10 m s)
5.55 10 kg 0.279 m .
(2 ) 6.672 10 N m kg
Gm mv Rv
m
R R G −
× ×= ⇒ = = = × =
× ⋅
 
37.76. For any function ( , )f f x t= and ( , ), ( , ),x x x t t t x t′ ′ ′ ′= = let ( , ) ( ( , ), ( , ))F x t f x x t t x t′ ′ ′ ′ ′ ′= and use the standard 
(but mathematically improper) notation ( , ) ( , ).F x t f x t′ ′ ′ ′= The chain rule is then 
( , ) ( , ) ( , )
,
( , ) ( , ) ( , )
.
f x t f x t x f x t t
x x x t x
f x t f x t x f x t t
t x t t t
′ ′ ′ ′ ′ ′∂ ∂ ∂ ∂ ∂= +
′ ′∂ ∂ ∂ ∂ ∂
′ ′ ′ ′ ′ ′∂ ∂ ∂ ∂ ∂= +
′ ′∂ ∂ ∂ ∂ ∂
 
In this solution, the explicit dependence of the functions on the sets of dependent variables is suppressed, and the 
above relations are then 
f f x f t
x x x t x
′ ′∂ ∂ ∂ ∂ ∂= +
′ ′∂ ∂ ∂ ∂ ∂
, .
f f x f t
t x t t t
′ ′∂ ∂ ∂ ∂ ∂= +
′ ′∂ ∂ ∂ ∂ ∂
 
(a) 
2 2
2 2
1, , 0 and 1. Then, , and .
x x t t E E E E
v
x t x t x x x x
′ ′ ′ ′∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= = − = = = =
′ ′∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
 For the time derivative, 
.
E E E
v
t x t
∂ ∂ ∂= − +
′ ′∂ ∂ ∂
 To find the second time derivative, the chain rule must be applied to both terms; that is, 
2 2
2
2 2
2
,
.
E E E
v
t x x t x
E E E
v
t t x t t
∂ ∂ ∂ ∂= − +
′ ′ ′ ′∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂= − +
′ ′ ′ ′∂ ∂ ∂ ∂ ∂
 
Using these in
2
2
,
E
t
∂
∂
 collecting terms and equating the mixed partial derivatives gives 
2 2 2 2
2
2 2 2
2
E E E E
v v
t x x t t
∂ ∂ ∂ ∂= − +
′ ′ ′ ′∂ ∂ ∂ ∂ ∂
, and using this and the above expression for 
2
2
E
x
∂
′∂
gives the result. 
(b) For the Lorentz transformation, 2γ, , / and γ.
x x t t
v v c
x t x t
γ γ
′ ′ ′ ′∂ ∂ ∂ ∂= = = =
∂ ∂ ∂ ∂
 
The first partials are then 
2
γ γ , γ γ
E E v E E E E
v
x x c t t x t
∂ ∂ ∂ ∂ ∂ ∂= − = − +
′ ′ ′ ′∂ ∂ ∂ ∂ ∂ ∂
 
and the second partials are (again equating the mixed partials) 
2 2 2 2 2
2 2 2
2 2 4 2 2
2 2 2 2
2 2 2 2
2 2 2
γ γ 2γ
γ γ 2γ .
E E v E v E
x x c t c x t
E E E E
v v
t x t x t
∂ ∂ ∂ ∂= + −
′ ′ ′ ′∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂= + −
′ ′ ′ ′∂ ∂ ∂ ∂ ∂
 
Substituting into the wave equation and combining terms (note that the mixed partials cancel), 
2 2 2 2 2 2 2 2
2 2
2 2 2 2 2 4 2 2 2 2 2
1 1 1
γ 1 0.
E E v E v E E E
x c t c x c c t x c t
γ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂− = − + − = − =⎜ ⎟ ⎜ ⎟′ ′ ′ ′∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠
 
Relativity 37-19 
37.77. (a) In the center of momentum frame, the two protons approach each other with equal velocities (since the protons 
have the same mass). After the collision, the two protons are at rest─but now there are kaons as well. In this 
situation the kinetic energy of the protons must equal the total rest energy of the two kaons 2cm p2(γ 1)m c⇒ − = 
2
k2m c ⇒ 
k
cm
p
γ 1 1.526.
m
m
= + = The velocity of a proton in the center of momentum frame is then 
2
cm
cm 2
cm
γ 1
0.7554 .
γ
v c c
−= = 
To get the velocity of this proton in the lab frame, we must use the Lorentz velocity transformations. This is the 
same as “hopping” into the proton that will be our target and asking what the velocity of the projectile proton is. 
Taking the lab frame to be the unprimed frame moving to the left, cm cmandu v v v′= = (the velocity of the projectile 
proton in the center of momentum frame). 
2cm
lab lab lab lab p2 2
cm lab
2 2 2
2 1
0.9619 γ 3.658 (γ 1) 2494 MeV.
1 1 1
v u v
v c K m c
uv v v
c c c
′ += = = ⇒ = = ⇒ = − =′
+ + −
 
(b) lab
k
2494 MeV
2.526.
2 2(493.7 MeV)
K
m
= = 
(c) The center of momentum case considered in part (a) is the same as this situation. Thus, the kinetic energy 
required is just twice the rest mass energy of the kaons. cm 2(493.7 MeV) 987.4 MeV.K = = This offers a 
substantial advantage over the fixed target experiment in part (b). It takes less energy to create two kaons in the 
proton center of momentum frame.
38-1 
PHOTONS, ELECTRONS, AND ATOMS 
 38.1. IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by 0
h
V f
e e
φ= − . The 
slope of V0 versus f is h/e. The value fth of f when 0 0V = is related to φ by thhfφ = . 
EXECUTE: (a) From the graph, 15th 1.25 10 Hzf = × . Therefore, with the value of h from part (b), 
th 4.8 eVhfφ = = . 
(b) From the graph, the slope is 153.8 10 V s−× ⋅ . 16 15 34( )(slope) (1.60 10 C)(3.8 10 V s) 6.1 10 J sh e − − −= = × × ⋅ = × ⋅ 
(c) No photoelectrons are produced for thf f< . 
(d) For a different metal fth and φ are different. The slope is h/e so would be the same, but the graph would be 
shifted right or left so it has a different intercept with the horizontal axis. 
EVALUATE: As the frequency f of the light is increased above fth the energy of the photons in the light increases 
and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel. 
 38.2. IDENTIFY and SET UP: c f λ= relates frequency and wavelength and E hf= relates energy and frequency for a 
photon. 83.00 10 m/sc = × . 161 eV 1.60 10 J−= × . 
EXECUTE: (a) 
8
14
9
3.00 10 m/s
5.94 10 Hz
505 10 m
c
f
λ −
×= = = ×
×
 
(b) 34 14 19(6.626 10 J s)(5.94 10 Hz) 3.94 10 J 2.46 eVE hf − −= = × ⋅ × = × = 
(c) 212K mv= so 
19
15
2 2(3.94 10 J)
9.1 mm/s
9.5 10 kg
K
v
m
−
−
×= = =
×
 
 38.3. 
8
14
7
3.00 10 m s
5.77 10 Hz
λ 5.20 10 m
c
f −
×= = = ×
×
 
34
27
7
27 8 19
6.63 10 J s
1.28 10 kg m s
λ 5.20 10 m
(1.28 10 kg m s) (3.00 10 m s) 3.84 10 J 2.40 eV.
h
p
E pc
−
−
−
− −
× ⋅= = = × ⋅
×
= = × ⋅ × = × =
 
 38.4. IDENTIFY and SET UP: av
energy
P
t
= . 191 eV 1.60 10 J−= × . For a photon, hcE hf
λ
= = . 346.63 10 J sh −= × ⋅ . 
EXECUTE: (a) 3 2 16avenergy (0.600 W)(20.0 10 s) 1.20 10 J 7.5 10 eVP t
− −= = × = × = × 
(b) 
34 8
19
9
(6.63 10 J s)(3.00 10 m/s)
3.05 10 J 1.91 eV
652 10 m
hc
E
λ
−
−
−
× ⋅ ×= = = × =
×
 
(c) The number of photons is the total energy in a pulse divided by the energy of one photon: 
2
16
19
1.20 10 J
3.93 10 photons
3.05 10 J/photon
−
−
× = ×
×
. 
EVALUATE: The number of photonsin each pulse is very large. 
 38.5. IDENTIFY and SET UP: Eq.(38.2) relates the photon energy and wavelength. c f λ= relates speed, frequency and 
wavelength for an electromagnetic wave. 
EXECUTE: (a) E hf= so 
6 19
20
34
(2.45 10 eV)(1.602 10 J/1 eV)
5.92 10 Hz
6.626 10 J s
E
f
h
−
−
× ×= = = ×
× ⋅
 
(b) c f λ= so 
8
13
20
2.998 10 m/s
5.06 10 m
5.92 10 Hz
c
f
λ −×= = = ×
×
 
(c) EVALUATE: λ is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was 
converted to the SI unit of Joules. 
38
38-2 Chapter 38 
 38.6. IDENTIFY and SET UP: th 272 nmλ = . c f λ= . 
2
max
1
2
mv hf φ= − . At the threshold frequency, thf , max 0v → . 
154.136 10 eV sh −= × ⋅ . 
EXECUTE: (a) 
8
15
th 9
th
3.00 10 m/s
1.10 10 Hz
272 10 m
c
f
λ −
×= = = ×
×
. 
(b) 15 15th (4.136 10 eV s)(1.10 10 Hz) 4.55 eVhfφ
−= = × ⋅ × = . 
(c) 2 15 15max
1
(4.136 10 eV s)(1.45 10 Hz) 4.55 eV 6.00 eV 4.55 eV 1.45 eV
2
mv hf φ −= − = × ⋅ × − = − = 
EVALUATE: The threshold wavelength depends on the work function for the surface. 
 38.7. IDENTIFY and SET UP: Eq.(38.3): 2max
1
.
2
hc
mv hf φ φ
λ
= − = − Take the work function φ from Table 38.1. Solve 
for max .v Note that we wrote f as / .c λ 
EXECUTE: 
34 8
2 19
max 9
1 (6.626 10 J s)(2.998 10 m/s)
(5.1 eV)(1.602 10 J/1 eV)
2 235 10 m
mv
−
−
−
× ⋅ ×= − ×
×
 
2 19 19 20
max
1
8.453 10 J 8.170 10 J 2.83 10 J
2
mv − − −= × − × = × 
20
5
max 31
2(2.83 10 J)
2.49 10 m/s
9.109 10 kg
v
−
−
×= = ×
×
 
EVALUATE: The work function in eV was converted to joules for use in Eq.(38.3). A photon with 235 nmλ = 
has energy greater then the work function for the surface. 
 38.8. IDENTIFY and SET UP: th
th
hc
hfφ
λ
= = . The minimum φ corresponds to the minimum λ . 
EXECUTE: 
15 8
9
th
(4.136 10 eV s)(3.00 10 m/s)
1.77 eV
700 10 m
hcφ
λ
−
−
× ⋅ ×= = =
×
 
 38.9. IDENTIFY and SET UP: c f λ= . The source emits (0.05)(75 J) 3.75 J= of energy as visible light each second. 
E hf= , with 346.63 10 J sh −= × ⋅ . 
EXECUTE: (a) 
8
14
9
3.00 10 m/s
5.00 10 Hz
600 10 m
c
f
λ −
×= = = ×
×
 
(b) 34 14 19(6.63 10 J s)(5.00 10 Hz) 3.32 10 JE hf − −= = × ⋅ × = × . The number of photons emitted per second is 
19
19
3.75 J
1.13 10 photons
3.32 10 J/photon−
= ×
×
. 
(c) No. The frequency of the light depends on the energy of each photon. The number of photons emitted per 
second is proportional to the power output of the source. 
38.10. IDENTIFY: In the photoelectric effect, the energy of the photon is used to eject an electron from the surface, and 
any excess energy goes into kinetic energy of the electron. 
SET UP: The energy of a photon is E = hf, and the work function is given by φ = hf0, where f0 is the threshold frequency. 
EXECUTE: (a) From the graph, we see that Kmax = 0 when λ = 250 nm, so the threshold wavelength is 250 nm. 
Calling f0 the threshold frequency, we have 
f0 = c/λ0 = (3.00 × 108 m/s)/(250 nm) = 1.2 × 1015 Hz. 
(b) φ = hf0 = (4.136 × 10–15 eV s⋅ )(1.2 × 1015 Hz) = 4.96 eV = 5.0 eV 
(c) The graph (see Figure 38.10) is linear for λ < λ0 (1/λ > 1/λ0), and linear graphs are easier to interpret than curves. 
EVALUATE: If the wavelength of the light is longer than the threshold wavelength (that is, if 1/λ < 1/λ0), the 
kinetic energy of the electrons is really not defined since no photoelectrons are ejected from the metal. 
 
Figure 38.10 
Photons, Electrons, and Atoms 38-3 
38.11. IDENTIFY: Protons have mass and photons are massless. 
(a) SET UP: For a particle with mass, 2 / 2 .K p m= 
EXECUTE: 2 12p p= means 2 14 .K K= 
(b) SET UP: For a photon, .E pc= 
EXECUTE: 2 12p p= means 2 12 .E E= 
EVALUATE: The relation between E and p is different for particles with mass and particles without mass. 
38.12. IDENTIFY and SET UP: 20 max
1
2
eV mv= , where 0V is the stopping potential. The stopping potential in volts equals 
0eV in electron volts. 
2
max
1
2
mv hf φ= − . 
EXECUTE: (a) 20 max
1
2
eV mv= so 
15 8
0 9
(4.136 10 eV s)(3.00 10 m/s)
2.3 eV 4.96 eV 2.3 eV 2.7 eV
250 10 m
eV hf φ
−
−
× ⋅ ×= − = − = − =
×
. The stopping potential 
is 2.7 electron volts. 
(b) 2max
1
2.7 eV
2
mv = 
(c) 
19
5
max 31
2(2.7 eV)(1.60 10 J/eV)
9.7 10 m/s
9.11 10 kg
v
−
−
×= = ×
×
 
38.13. (a) IDENTIFY: First use Eq.(38.4) to find the work function .φ 
SET UP: 0eV hf φ= − so 0 0
hc
hf eV eVφ
λ
= − = − 
EXECUTE: 
34 8
19
9
(6.626 10 J s)(2.998 10 m/s)
(1.602 10 C)(0.181 V)
254 10 m
φ
−
−
−
× ⋅ ×= − ×
×
 
19 20 19 197.821 10 J 2.900 10 J 7.531 10 J(1 eV/1.602 10 J) 4.70 eVφ − − − −= × − × = × × = 
IDENTIFY and SET UP: The threshold frequency thf is the smallest frequency that still produces photoelectrons. 
It corresponds to max 0K = in Eq.(38.3), so th .hf φ= 
EXECUTE: 
c
f
λ
= says 
th
hc φ
λ
= 
34 8
7
th 19
(6.626 10 J s)(2.998 10 m/s)
2.64 10 m 264 nm
7.531 10 J
hcλ
φ
−
−
−
× ⋅ ×= = = × =
×
 
(b) EVALUATE: As calculated in part (a), 4.70 eV.φ = This is the value given in Table 38.1 for copper. 
38.14. IDENTIFY and SET UP: A photon has zero rest mass, so its energy and momentum are related by Eq.(37.40). 
Eq.(38.5) then relates its momentum and wavelength. 
EXECUTE: (a) 28 8 19(8.24 10 kg m/s)(2.998 10 m/s) 2.47 10 JE pc − −= = × ⋅ × = × = 
19 19(2.47 10 J)(1 eV/1.602 10 J)− −× × = 1.54 eV 
(b) 
h
p
λ
= so 
34
7
28
6.626 10 J s
8.04 10 m 804 nm
8.24 10 kg m/s
h
p
λ
−
−
−
× ⋅= = = × =
× ⋅
 
EVALUATE: This wavelength is longer than visible wavelengths; it is in the infrared region of the 
electromagnetic spectrum. To check our result we could verify that the same E is given by Eq.(38.2), using the λ 
we have calculated. 
38.15. IDENTIFY and SET UP: Balmer’s formula is 
2 2
1 1 1
.
2
R
nλ
⎛ ⎞= −⎜ ⎟
⎝ ⎠
 For the Hγ spectral line 5.n = Once we have ,λ 
calculate f from /f c λ= and E from Eq.(38.2). 
EXECUTE: (a) 
2 2
1 1 1 25 4 21
.
2 5 100 100
R R R
λ
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
 
Thus 7
7
100 100
 m 4.341 10 m 434.1 nm.
21 21(1.097 10 )R
λ −= = = × =
×
 
(b) 
8
14
7
2.998 10 m/s
6.906 10 Hz
4.341 10 m
c
f
λ −
×= = = ×
×
 
38-4 Chapter 38 
(c) 34 14 19(6.626 10 J s)(6.906 10 Hz) 4.576 10 J 2.856 eVE hf − −= = × ⋅ × = × = 
EVALUATE: Section 38.3 shows that the longest wavelength in the Balmer series (H )α is 656 nm and the 
shortest is 365 nm. Our result for Hγ falls within this range. The photon energies for hydrogen atom transitions are 
in the eV range, and our result is of this order. 
38.16. IDENTIFY and SET UP: For the Lyman series the final state is 1n = and the wavelengths are given by 
2 2
1 1 1
, 2,3,....
1
R n
nλ
⎛ ⎞= − =⎜ ⎟
⎝ ⎠
 For the Paschen series the final state is 3n = and the wavelengths are given by 
2 2
1 1 1
, 4,5,....
3
R n
nλ
⎛ ⎞= − =⎜ ⎟
⎝ ⎠
 7 11.097 10 mR −= × . The longest wavelength is for the smallest n and the shortest 
wavelength is for n → ∞ . 
EXECUTE: Lyman Longest: 
2 2
1 1 1 3
1 2 4
R
R
λ
⎛ ⎞= − =⎜ ⎟
⎝ ⎠
. 
7 1
4
121.5 nm
3(1.097 10 m )
λ −= =×
. 
Shortest: 
2 2
1 1 1
1
R R
λ
⎛ ⎞= − =⎜ ⎟∞⎝ ⎠
. 
7 1
1
91.16 nm
1.097 10 m
λ −= =×
 
Paschen Longest: 
2 2
1 1 1 7
3 4 144
R
R
λ
⎛ ⎞= − =⎜ ⎟
⎝ ⎠
. 
7 1
144
1875 nm
7(1.097 10 m )
λ −= =×
. 
Shortest: 
2 2
1 1 1
3 9
R
R
λ
⎛ ⎞= − =⎜ ⎟∞⎝ ⎠
. 
38.17. (a) 
34 8
19
7
(6.63 10 J s) (3.00 10 m s)
2.31 10 J 1.44 eV.
λ 8.60 10 m
hc
Eγ
−
−
−
× ⋅ ×= = = × =
×
 
So the internal energy of the atom increases by 1.44 eV to 6.52 eV 1.44 eVE = − + = 5.08 eV.− 
(b) 
34 8
19
7
(6.63 10 J s) (3.00 10 m s)
4.74 10 J 2.96 eV.
λ 4.20 10 m
hc
Eγ
−
−
−
× ⋅ ×= = = × =
×
 
So the final internal energy of the atom decreases to 2.68 eV 2.96 eV 5.64 eV.E = − − = − 
38.18. IDENTIFY and SET UP: The ionization threshold is at 0E = . The energy of an absorbed photon equals the 
energy gained by the atom and the energy of an emitted photon equals theenergy lost by the atom. 
EXECUTE: (a) 0 ( 20 eV) 20 eVEΔ = − − = 
(b) When the atom in the 1n = level absorbs a 18 eV photon, the final level of the atom is 4n = . The possible 
transitions from 4n = and corresponding photon energies are 4 3, 3 eVn n= → = ; 4 2, 8 eVn n= → = ; 
4 1, 18 eVn n= → = . Once the atom has gone to the 3n = level, the following transitions can occur: 
3 2, 5 eVn n= → = ; 3 1, 15 eVn n= → = . Once the atom has gone to the 2n = level, the following transition 
can occur: 2 1, 10 eVn n= → = . The possible energies of emitted photons are: 3 eV, 5 eV, 8 eV, 10 eV, 15 eV, 
and 18 eV. 
(c) There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed. 
(d) The photon energies for 3 2n n= → = and for 3 1n n= → = are 5 eV and 15 eV. The photon 
energy for 4 3n n= → = is 3 eV. The work function must have a value between 3 eV and 5 eV. 
38.19. IDENTIFY and SET UP: The wavelength of the photon is related to the transition energy i fE E− of the atom by 
i f
hc
E E
λ
− = where 61.240 10 eV mhc −= × ⋅ . 
EXECUTE: (a) The minimum energy to ionize an atom is when the upper state in the transition has 0E = , so 
1 17.50 eVE = − . For 5 1n n= → = , 73.86 nmλ = and 
6
5 1 9
1.240 10 eV m
16.79 eV
73.86 10 m
E E
−
−
× ⋅− = =
×
. 
5 17.50 eV 16.79 eV 0.71 eVE = − + = − . For 4 1n n= → = , 75.63 nmλ = and 4 1.10 eVE = − . For 
3 1n n= → = , 79.76 nmλ = and 3 1.95 eVE = − . For 2 1n n= → = , 94.54 nmλ = and 2 4.38 eVE = − . 
(b) i f 4 2 1.10 eV ( 4.38 eV) 3.28 eVE E E E− = − = − − − = and 
6
i f
1.240 10 eV m
378 nm
3.28 eV
hc
E E
λ
−× ⋅= = =
−
 
EVALUATE: The 4 2n n= → = transition energy is smaller than the 4 1n n= → = transition energy so the 
wavelength is longer. In fact, this wavelength is longer than for any transition that ends in the 1n = state. 
Photons, Electrons, and Atoms 38-5 
38.20. (a) Equating initial kinetic energy and final potential energy and solving for the separation radius r, 
19
14
6
0 0
1 (92 ) (2 ) 1 (184) (1.60 10 C)
5.54 10 m.
4 4 (4.78 10 J C)
e e
r
Kπ π
−
−×= = = ×
×P P
 
(b) The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force 
and the magnitude of the potential energy in a Coulombic field is 
6 19
14
(4.78 10 eV) (1.6 10 J ev)
13.8 N.
(5.54 10 m)
K
F
r
−
−
× ×= = =
×
 
38.21. (a) IDENTIFY: If the particles are treated as point charges, 1 2
0
1
.
4
q q
U
rπ
=
P
 
SET UP: 1 2q e= (alpha particle); 2 82q e= (gold nucleus); r is given so we can solve for U. 
EXECUTE: 
19 2
9 2 2 13
14
(2)(82)(1.602 10 C)
(8.987 10 N m /C ) 5.82 10 J
6.50 10 m
U
−
−
−
×= × ⋅ = ×
×
 
13 19 65.82 10 J(1 eV/1.602 10 J) 3.63 10 eV 3.63 MeVU − −= × × = × = 
(b) IDENTIFY: Apply conservation of energy: 1 1 2 2.K U K U+ = + 
SET UP: Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle 
momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies 1r ≈ ∞ and 1 0.U = Alpha 
particle stops implies 2 0.K = 
EXECUTE: Conservation of energy thus says 131 2 5.82 10 J 3.63 MeV.K U
−= = × = 
(c) 2
1
2
K mv= so 
13
7
27
2 2(5.82 10 J)
1.32 10 m/s
6.64 10 kg
K
v
m
−
−
×= = = ×
×
 
EVALUATE: / 0.044,v c = so it is ok to use the nonrelativistic expression to relate K and v. When the alpha 
particle stops, all its initial kinetic energy has been converted to electrostatic potential energy. 
38.22. (a), (b) For either atom, the magnitude of the angular momentum is 
2
h
π
= 34 21.05 10 kg m s.−× ⋅ 
38.23. IDENTIFY and SET UP: Use the energy to calculate n for this state. Then use the Bohr equation, Eq.(38.10), to 
calculate L. 
EXECUTE: 2(13.6 eV)/ ,nE n= − so this state has 13.6/1.51 3.n = = In the Bohr model. L n= U so for this state 
34 23 3.16 10 kg m /s.L −= × ⋅U = 
EVALUATE: We will find in Section 41.1 that the modern quantum mechanical description gives a different 
result. 
38.24. IDENTIFY and SET UP: For a hydrogen atom 
2
13.6 eV
nE n
= − . hcE
λ
Δ = , where EΔ is the magnitude of the 
energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted. 
EXECUTE: 4 1 2 2
1 1
(13.6 eV) 12.75 eV
4 1
E E E
⎛ ⎞Δ = − = − − = +⎜ ⎟
⎝ ⎠
. 
15 8(4.136 10 eV s)(3.00 10 m/s)
97.3 nm
12.75 eV
hc
E
λ
−× ⋅ ×= = =
Δ
. 153.08 10 Hz
c
f
λ
= = × . 
38.25. IDENTIFY: The force between the electron and the nucleus in 3+Be is 
2
2
0
1
,
4
Ze
F
rπ
=
P
 where 4Z = is the nuclear 
charge. All the equations for the hydrogen atom apply to 3+Be if we replace 2e by 2.Ze 
(a) SET UP: Modify Eq.(38.18). 
EXECUTE: 
4
2 2
0
1
8n
me
E
n h
= −
P
 (hydrogen) becomes 
2 2 4
2 2 3+
2 2 2 2 2
0 0
1 ( ) 1 13.60 eV
(for Be )
8 8n
m Ze me
E Z Z
n h n h n
⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠P P
 
The ground-level energy of 3+Be is 1 2
13.60 eV
16 218 eV.
1
E
⎛ ⎞= − = −⎜ ⎟
⎝ ⎠
 
EVALUATE: The ground-level energy of 3+Be is 2 16Z = times the ground-level energy of H. 
(b) SET UP: The ionization energy is the energy difference between the n → ∞ level energy and the 1n = level 
energy. 
38-6 Chapter 38 
EXECUTE: The n → ∞ level energy is zero, so the ionization energy of 3+Be is 218 eV. 
EVALUATE: This is 16 times the ionization energy of hydrogen. 
(c) SET UP: 
2 2
1 2
1 1 1
R
n nλ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
 just as for hydrogen but now R has a different value. 
EXECUTE: 
4
7 1
H 3
0
1.097 10 m
8
me
R
h c
−= = ×
P
 for hydrogen becomes 
4
2 7 1 8 1
Be 3
0
16(1.097 10 m ) 1.755 10 m
8
me
R Z
h c
− −= = × = ×
P
 for 3+Be . 
For 2n = to Be 2 2
1 1 1
1, 3 /4.
1 2
n R R
λ
⎛ ⎞= = − =⎜ ⎟
⎝ ⎠
 
8 1 94 /(3 ) 4 /(3(1.755 10 m )) 7.60 10 m 7.60 nm.Rλ − −= = × = × = 
EVALUATE: This wavelength is smaller by a factor of 16 compared to the wavelength for the corresponding 
transition in the hydrogen atom. 
(d) SET UP: Modify Eq.(38.12): 
2 2
0 2n
n h
r
meπ
= P (hydrogen). 
EXECUTE: 
2 2
0 2( )n
n h
r
m Zeπ
= P 3+(Be ). 
EVALUATE: For a given n the orbit radius for 3+Be is smaller by a factor of 4Z = compared to the 
corresponding radius for hydrogen. 
38.26. (a) We can find the photon’s energy from Eq. 38.8 
34 8 7 1 19
2 2 2 2
1 1 1 1
(6.63 10 J s) (3.00 10 m s) (1.097 10 m ) 4.58 10 J.
2 2 5
E hcR
n
− − −⎛ ⎞ ⎛ ⎞= − = × ⋅ × × − = ×⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
The 
corresponding wavelength is λ 434 nm.
E
hc
= = 
(b) In the Bohr model, the angular momentum of an electron with principal quantum number n is given by 
Eq. 38.10: .
2
h
L n
π
= Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the following 
loss in angular momentum (which we would assume is transferred to the photon): 
34
343(6.63 10 J s)(2 5) 3.17 10 J s.
2 2
h
L
ππ
−
−× ⋅Δ = − = − = − × ⋅ 
However, this prediction of the Bohr model is wrong (as shown in Chapter 41). 
38.27. (a) 
2 19 2
6
1 34
0 0
1 (1.60 10 C)
: 1 2.18 10 m/s
2 2 (6.63 10 J s)n
e
v n v
nh
−
−
×= = ⇒ = = ×
× ⋅P P
 
6 51 1
2 32 1.09 10 m s. 3 7.27 10 m s.2 3
v v
h v h v= ⇒ = = × = ⇒ = = × 
(b) Orbital period 
2 2 2 2 3 3
0 0
2 4
0
2 2 4
1 2
n
n
r n h me n h
v e nh me
π= = =
⋅
P P
P
 
2 34 3
160
1 31 19 4
3 15 3 15
2 1 3 1
4 (6.63 10 J s)
1 1.53 10 s
(9.11 10 kg) (1.60 10 C)
2 : (2) 1.22 10 s. 3 : (3) 4.13 10 s.
n T
n T T n T T
−
−
− −
− −
× ⋅= ⇒ = = ×
× ×
= = = × = = = ×
P
 
(c) number of orbits 
8
6
15
1.0 10 s
8.2 10 .
1.22 10 s
−
−
×= = ×
×
 
38.28. IDENTIFY and SET UP: 
2
13.6 eV
nE n
= − 
EXECUTE: (a) 
2
13.6 eV
nE n
= − and +1 2
13.6 eV
( 1)n
E
n
= −
+
 
2 2
1 2 2 2 2
1 1 ( 1)
( 13.6 eV) (13.6 eV)
( 1) ( )( 1)n n
n n
E E E
n n n n+
⎡ ⎤ − +Δ = − = − − = −⎢ ⎥+ +⎣ ⎦
 
2 2
2 1
(13.6 eV)
( )( 1)
n
E
n n
+Δ =
+
 As n becomes large, 
4 3
2 2
(13.6 eV) (13.6 eV)
n
E
n n
Δ → = 
Photons, Electrons, and Atoms 38-7 
Thus EΔ becomes small as n becomes large. 
(b) 2 1nr n r= so theorbits get farther apart in space as n increases. 
38.29. IDENTIFY and SET UP: The number of photons emitted each second is the total energy emitted divided by the 
energy of one photon. The energy of one photon is given by Eq.(38.2). E Pt= gives the energy emitted by the 
laser in time t. 
EXECUTE: In 1.00 s the energy emitted by the laser is 3 3(7.50 10 W)(1.00 s) 7.50 10 J.− −× = × 
The energy of each photon is 
34 8
20
6
(6.626 10 J s)(2.998 10 m/s)
1.874 10 J.
10.6 10 m
hc
E
λ
−
−
−
× ⋅ ×= = = ×
×
 
Therefore 
3
17
20
7.50 10 J/s
4.00 10 photons/s
1.874 10 J/photon
−
−
× = ×
×
 
EVALUATE: The number of photons emitted per second is extremely large. 
38.30. IDENTIFY and SET UP: Visible light has wavelengths from about 400 nm to about 700 nm. The energy of each 
photon is 
251.99 10 J mhc
E hf
λ λ
−× ⋅= = = . The power is the total energy per second and the total energy Etot is the 
number of photons N times the energy E of each photon. 
EXECUTE: (a) 193 nm is shorter than visible light so is in the ultraviolet. 
(b) 181.03 10 J 6.44 eV
hc
E
λ
−= = × = 
(c) tot
E NE
P
t t
= = so 
3 9
7
18
(1.50 10 W)(12.0 10 s)
1.75 10 photons
1.03 10 J
Pt
N
E
− −
−
× ×= = = ×
×
 
EVALUATE: A very small amount of energy is delivered to the lens in each pulse, but this still corresponds to a 
large number of photons. 
38.31. IDENTIFY: Apply Eq.(38.21): 5 3( ) /5
3
s pE E kTs
p
n
e
n
− −= 
SET UP: From Fig.38.24a in the textbook, 5 320.66 eV and 18.70 eVs pE E= = 
EXECUTE: 19 195 3 20.66 eV 18.70 eV 1.96 eV(1.602 10 J/1 eV) 3.140 10 Js pE E
− −− = − = × = × 
(a) 
19 23(3.140 10 J)/[(1.38 10 J/K)(300 K)] 75.79 335
3
1.2 10s
p
n
e e
n
− −− × × − −= = = × 
(b) 
19 23(3.140 10 J)/[(1.38 10 J/K)(600 K)] 37.90 175
3
3.5 10s
p
n
e e
n
− −− × × − −= = = × 
(c) 
19 23(3.140 10 J)/[(1.38 10 J/K)(1200 K)] 18.95 95
3
5.9 10s
p
n
e e
n
− −− × × − −= = = × 
(d) EVALUATE: At each of these temperatures the number of atoms in the 5s excited state, the initial state for the 
transition that emits 632.8 nm radiation, is quite small. The ratio increases as the temperature increases. 
38.32. 3 2 2 3 2 2 1 2
1/ 2
2 ( )
2
.P P
P E E KT
P
n
e
n
− −= 
From the diagram 
34 8
19
3/ 2 g 7
1
(6.626 10 J)(3.000 10 m s)
3.375 10 J.
λ 5.890 10 m
hc
E
−
−
− −
× ×Δ = = = ×
×
 
34 8
19 19 19
1 2 g 3/ 2 1/ 27
2
(6.626 10 J)(3.000 10 m s)
3.371 10 J. so 3.375 10 J 3.371 10 J
λ 5.896 10 m
hc
E E
−
− − −
− −−
× ×Δ = = = × Δ = × − × =
×
224.00 10 J.−×
22 23
3/ 2
1/ 2
2 (4.00 10 J) (1.38 10 J / K 500 K).
2
0.944.P
P
n
e
n
− −− × × ⋅= = So more atoms are in the 1 22 p state. 
38.33. AC max
minλ
hceV hf= = 
34 8
10
min 19
AC
(6.63 10 J s)(3.00 10 m s)
λ 3.11 10 m
(1.60 10 C)(4000 V)
hc
eV
−
−
−
× ⋅ ×
⇒ = = = ×
×
 
This is the same answer as would be obtained if electrons of this energy were used. Electron beams are much more 
easily produced and accelerated than proton beams. 
38-8 Chapter 38 
38.34. IDENTIFY and SET UP: 
hc
eV
λ
= , where λ is the wavelength of the x ray and V is the accelerating voltage. 
EXECUTE: (a) 
34 8
19 9
(6.63 10 J s)(3.00 10 m/s)
8.29 kV
(1.60 10 C)(0.150 10 m)
hc
V
eλ
−
− −
× ⋅ ×= = =
× ×
 
(b) 
34 8
11
19 3
(6.63 10 J s)(3.00 10 m/s)
4.14 10 m 0.0414 nm
(1.60 10 C)(30.0 10 V)
hc
eV
λ
−
−
−
× ⋅ ×= = = × =
× ×
 
(c) No. A proton has the same magnitude of charge as an electron and therefore gains the same amount of kinetic 
energy when accelerated by the same magnitude of potential difference. 
38.35. IDENTIFY: The initial electrical potential energy of the accelerated electrons is converted to kinetic energy which 
is then given to a photon. 
SET UP: The electrical potential energy of an electron is eVAC, where VAC
 is the accelerating potential, and the 
energy of a photon is hf. Since the energy of the electron is all given to a photon, we have eVAC = hf. For any wave, 
fλ = v. 
EXECUTE: (a) eVAC = hfmin gives 
fmin = eVAC/h = (1.60 × 10–19 C)(25,000 V)/(6.626 × 10–34 J s⋅ ) = 6.037 × 1018 Hz 
= 6.04 × 1018 Hz, rounded to three digits 
(b) λmin = c/fmax = (3.00 × 108 m/s)/(6.037 × 1018 Hz) = 4.97 × 10–11 m = 0.0497 nm 
(c) We assume that all the energy of the electron produces only one photon on impact with the screen. 
EVALUATE: These photons are in the x-ray and γ-ray part of the electromagnetic spectrum (see Figure 32.4 in the 
textbook) and would be harmful to the eyes without protective glass on the screen to absorb them. 
38.36. IDENTIFY and SET UP: The wavelength of the x rays produced by the tube is give by 
hc
eV
λ
= . 
(1 cos )
h
mc
λ λ φ′ = + − . 122.426 10 mh
mc
−= × . The energy of the scattered x ray is hc
λ′
. 
EXECUTE: (a) 
34 8
11
19 3
(6.63 10 J s)(3.00 10 m/s)
6.91 10 m 0.0691 nm
(1.60 10 C)(18.0 10 V)
hc
eV
λ
−
−
−
× ⋅ ×= = = × =
× ×
 
(b) 11 12(1 cos ) 6.91 10 m (2.426 10 m)(1 cos45.0 )
h
mc
λ λ φ − −′ = + − = × + × − ° . 
116.98 10 m 0.0698 nmλ −′ = × = . 
(c) 
15 8
11
(4.136 10 eV s)(3.00 10 m/s)
17.8 keV
6.98 10 m
hc
E
λ
−
−
× ⋅ ×= = =
′ ×
 
EVALUATE: The incident x ray has energy 18.0 keV. In the scattering event, the photon loses energy and its 
wavelength increases. 
38.37. IDENTIFY: Apply Eq.(38.23): C(1 cos ) (1 cos )
h
mc
λ λ φ λ φ′ − = − = − 
SET UP: Solve for C: (1 cos )λ λ λ λ φ′ ′ = + − 
The largest λ′ corresponds to 180 ,φ = ° so cos 1.φ = − 
EXECUTE: 9 12 11C2 0.0665 10 m 2(2.426 10 m) 7.135 10 m 0.0714 nm.λ λ λ
− − −′ = + = × + × = × = This wavelength 
occurs at a scattering angle of 180 .φ = ° 
EVALUATE: The incident photon transfers some of its energy and momentum to the electron from which it 
scatters. Since the photon loses energy its wavelength increases, .λ λ′ > 
38.38. (a) From Eq. (38.23), 
λ
cos 1 ,
( )h mc
φ Δ= − and so λ 0.0542 nm 0.0500 nm,Δ = − 
0.0042 nm
cos 1 0.731, and 137 .
0.002426 nm
φ φ= − = − = ° 
(b) 
0.0021 nm
λ 0.0521 nm 0.0500 nm. cos 1 0.134. 82.3 .
0.002426 nm
φ φΔ = − = − = = ° 
(c) λ 0,Δ = the photon is undeflected, cos 1φ = and 0.φ = 
38.39. IDENTIFY and SET UP: The shift in wavelength of the photon is (1 cos )
h
mc
λ λ φ′ − = − where λ′ is the 
wavelength after the scattering and 12c 2.426 10 m
h
mc
λ −= = × . The energy of a photon of wavelength λ is 
Photons, Electrons, and Atoms 38-9 
61.24 10 eV mhc
E
λ λ
−× ⋅= = . Conservation of energy applies to the collision, so the energy lost by the photon 
equals the energy gained by the electron. 
EXECUTE: (a) 12 13 4c (1 cos ) (2.426 10 m)(1 cos35.0 ) 4.39 10 m 4.39 10 nmλ λ λ φ
− − −′ − = − = × − = × = ×° 
(b) 4 44.39 10 nm 0.04250 nm 4.39 10 nm 0.04294 nmλ λ − −′ = + × = + × = 
(c) 42.918 10 eV
hc
Eλ λ
= = × and 42.888 10 eVhcEλ λ′
= = ×
′
 so the photon loses 300 eV of energy. 
(d) Energy conservation says the electron gains 300 eV of energy. 
38.40. The change in wavelength of the scattered photon is given by Eq. 38.23 
Δλ
(1 cos ) λ (1 cos ).
Δλλ λ
λ
h h
mc mc
φ φ= − ⇒ = −
⎛ ⎞
⎜ ⎟
⎝ ⎠
 
Thus, 
34
14
27 8
(6.63 10 J s)
λ (1 1) 2.65 10 m.
(1.67 10 kg)(3.00 10 m/s)(0.100)
−
−
−
× ⋅= + = ×
× ×
 
38.41. The derivation of Eq.(38.23) is explicitly shown in Equations (38.24) through (38.27) with the final substitution of 
λ and λ yielding λ λ (1 cos ).
h
p h p h
mc
φ′ ′ ′= = − = − 
38.42. From Eq. (38.30), (a) 
3
m
m
2.898 10 m K
λ 0.966 mm, and
3.00 K λ
c
f
−× ⋅= = = = 113.10 10 Hz.× Note that a more precise 
value of the Wien displacement law constant has been used. 
(b) A factor of 100 increase in the temperature lowers mλ by a factor of 100 to 9.66 mμ and raises the frequency 
by the same factor, to 133.10 10 Hz.× 
(c) Similarly, mλ 966 nm= 
14and 3.10 10 Hz.f = × 
38.43. (a) 4 2;H AeσT A r lπ= = 
1 41 4
3 8 2 4
100 W
2 (0.20 10 m)(0.30 m)(0.26)(5.671 10 W m K )
H
T
Aeσ π − −
⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟ × × ⋅⎝ ⎠ ⎝ ⎠
 
32.06 10 KT = × 
(b) 3m mλ 2.90 10 m K; λ 1410 nmT−= × ⋅ = 
Much of the emitted radiation is in the infrared. 
38.44. 
3 3
3
9
m
2.90 10 m K 2.90 10 m K
7.25 10 K.
400 10 m
T
λ
− −
−
× ⋅ × ⋅= = = ×
×
 
38.45. IDENTIFY and SET UP: The wavelength mλ where the Planck distribution peaks is given by Eq.(38.30). 
EXECUTE: 
3
3
m
2.90 10 m K
1.06 10 m 1.06 mm.
2.728 K
λ
−
−× ⋅= = × = 
EVALUATE: This wavelength is in the microwave portion of the electromagnetic spectrum. This radiation is often 
referred to as the “microwave background” (Section 44.7). Note that in Eq.(38.30), T must be in kelvins. 
38.46. IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law and Wien’s displacement 
law. 
SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power 
is P = σAT 4. Wien’s displacement law tells us that the peak-intensity wavelength is λm = (constant)/T. 
EXECUTE: (a) The hot and cool stars radiate the same total power, so the Stefan-Boltzmann law gives σAhTh4 = 
σAcTc4 ⇒ 4πRh2Th4 = 4πRc2Tc4 = 4π(3Rh)2Tc4 ⇒ Th4 = 9T 4 ⇒ h 3T T= = 1.7T, rounded to two significant digits. 
(b) Using Wien’s law, we take the ratio of the wavelengths, giving 
m c
m h
(hot) 1
(cool) 3 3
T T
T T
λ
λ
= = = = 0.58, rounded to two significant digits. 
EVALUATE: Although the hot star has only 1/9 the surface area of the cool star, its absolute temperature has to be 
only 1.7 times as great to radiate the same amount of energy. 
38.47. (a) Let / .hc kTα = To find the maximum in the Planck distribution: 
2 2 2 2
5 5 5 2
2 (2 ) 2 ( )
0 5
λ λ λ ( 1) λ ( 1) λ ( 1)λ
dI d hc hc hc λ
d d e e eα λ α λ α
π π π α⎛ ⎞ −= = = − −⎜ ⎟− − −⎝ ⎠
 
λ λ5( 1)λ 5 5 λ Solve 5 5 where .
λ λ
x hce e x e x
kT
α α αα α⇒ − − = ⇒ − + = ⇒ − = = = 
38-10 Chapter 38 
Its root is 4.965, so 4.965 λ .
λ (4.965)
α hc
kT
= ⇒ = 
(b) 
34 8
3
m 23
(6.63 10 J s)(3.00 10 m s)
λ 2.90 10 m K.
(4.965) (4.965)(1.38 10 J K)
hc
T
k
−
−
−
× ⋅ ×= = = × ⋅
×
 
38.48. IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law. 
SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power 
is P = σAT 4. 
EXECUTE: (a) I = σT 4 = (5.67 × 10–8 2 4W/m K⋅ )(24,000 K)4 = 1.9 × 1010 W/m2 
(b) Wien’s law gives λm = (0.00290 m K⋅ )/(24,000 K) = 1.2 × 10–7 m = 120 nm 
This is not visible since the wavelength is less than 400 nm. 
(c) P = AI ⇒ 4πR2 = P/I = (1.00 × 1025 W)/(1.9 × 1010 W/m2) 
which gives RSirius = 6.51 × 106 m = 6510 km. 
RSirius/Rsun = (6.51 × 106 m)/(6.96 × 109 m) = 0.0093, which gives 
RSirius = 0.0093 Rsun ≈ 1% Rsun 
(d) Using the Stefan-Boltzmann law, we have 
2 44 2 4
sun sun sun sun sun sun sun
4 2 4
Sirius Sirius Sirius Sirius Sirius Sirius Sirius
4
4
P A T R T R T
P A T R T R T
σ π
σ π
⎛ ⎞ ⎛ ⎞
= = = ⋅⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2 4
sun sun
Sirius sun
5800 K
39
0.00935 24,000 K
P R
P R
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 
EVALUATE: Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, it radiates 
only 1/39 times as much energy per second as our sun because it is so small. 
38.49. Eq. (38.32): 
2 2
5
2
( ) but 1 1 for
λ ( 1) 2
x
hc λkT
hc x
I λ e x x
e
π= = + + + ≈ +
−
 
2
5 4
2 2
1 ( ) Eq.
λ ( λ ) λ
hc ckT
x I λ
hc kT
π π
⇒ ≈ = =V (38.31), which is Rayleigh’s distribution. 
38.50. (a) Wien’s law: mλ
k
T
= . 
3
8
m
2.90 10 K m
λ 9.7 10 m 97 nm
30,000 K
−
−× ⋅= = × = 
This peak is in the ultraviolet region, which is not visible. The star is blue because the largest part of the visible 
light radiated is in the blue violet part of the visible spectrum 
(b) 4P σAT= (Stefan-Boltzmann law) 
26 8 2 4
2 4
9
W
(100, 000)(3.86 10 W) 5.67 10 (4 )(30,000 K)
m K
8.2 10 m
R
R
π−⎛ ⎞× = ×⎜ ⎟
⎝ ⎠
= ×
 
9
star sun 8
8.2 10 m
12
6.96 10 m
R R
×= =
×
 
(c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much of the power is 
radiated nonvisible wavelengths, which does not contribute to the visible luminosity. 
38.51. IDENTIFY and SET UP: Use c f λ= to relate frequency and wavelength and use E hf= to relate photon energy 
and frequency. 
EXECUTE: (a) One photon dissociates one AgBr molecule, so we need to find the energy required to dissociate a 
single molecule. The problem states that it requires 51.00 10 J× to dissociate one mole of AgBr, and one mole 
contains Avogadro’s number 23(6.02 10 )× of molecules, so the energy required to dissociate one AgBr is 
5
19
23
1.00 10 J/mol
1.66 10 J/molecule.
6.02 10 molecules/mol
−× = ×
×
 
The photon is to have this energy, so 19 191.66 10 J(1eV/1.602 10 J) 1.04 eV.E − −= × × = 
(b) 
hc
E
λ
= so 
34 8
6
19
(6.626 10 J s)(2.998 10 m/s)
1.20 10 m 1200 nm
1.66 10 J
hc
E
λ
−
−
−
× ⋅ ×= = = × =
×
 
(c) c f λ= so 
8
14
6
2.998 10 m/s
2.50 10 Hz
1.20 10 m
c
f
λ −
×= = = ×
×
 
(d) 34 6 26(6.626 10 J s)(100 10 Hz) 6.63 10 JE hf − −= = × ⋅ × = × 
26 19 76.63 10 J(1 eV/1.602 10 J) 4.14 10 eVE − − −= × × = × 
Photons, Electrons, and Atoms 38-11 
(e) EVALUATE: A photon with frequency 100 MHzf = has too little energy, by a large factor, to dissociate a 
AgBr molecule. The photons in the visible light from a firefly do individually have enough energy to dissociate 
AgBr. The huge number of 100 MHz photons can’t compensate for the fact that individually they have too little 
energy. 
38.52. (a) Assume a non-relativistic velocity and conserve momentum 
λ
h
mv⇒ = ⇒ .
λ
h
v
m
= 
(b) 
2 2
2
2
1 1
2 2 λ 2 λ
h h
K mv m
m m
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
. 
(c) 
2
2
λ
.
2 λ 2 λ
K h h
E m hc mc
= ⋅ = Recoil becomes an important concern for small m and small λ since this ratio 
becomes large in those limits. 
(d) 
34 8
7
19
(6.63 10 J s)(3.00 10 m s)
10.2 eV λ 1.22 10 m 122 nm.
(10.2 eV)(1.60 10 J eV)
hc
E
E
−
−
−
× ⋅ ×= ⇒ = = = × =
×
 
34 2
27 8
27 7 2
8
9
(6.63 10 J s)
8.84 10 J 5.53 10 eV.
2(1.67 10 kg)(1.22 10 m)
5.53 10 eV
5.42 10 . This is quite small so recoil can be neglected.
10.2 eV
K
K
E
−
− −
− −
−
−
× ⋅= = × = ×
× ×
×= = ×
 
38.53. IDENTIFY and SET UP: 
c
f
λ
= . The 0( , )f V values are: 
14(8.20 10 Hz, 1.48 V)× , 14(7.41 10 Hz, 1.15 V)× , 
14(6.88 10 Hz, 0.93 V)× , 14(6.10 10 Hz, 0.62 V)× , 14(5.49 10 Hz, 0.36 V)× , 14(5.18 10 Hz, 0.24 V)× . The graph 
of 0V versus f is given in Figure 38.53. 
EXECUTE: (a) The threshold frequency, thf , is f where 0 0V = . From the graph this is 
14
th 4.56 10 Hzf = × . 
(b) 
8
th 14
th
3.00 10 m/s
658 nm
4.56 10 Hz
c
f
λ ×= = =
×
 
(c) 15 14th (4.136 10 eV s)(4.56 10 Hz) 1.89 eVhfφ
−= = × ⋅ × = 
(d) 0eV hf φ= − so 0
h
V f
e
φ⎛ ⎞= −⎜ ⎟
⎝ ⎠
. The slope of the graph is 
h
e
. 
15
14 14
1.48 V 0.24 V
4.11 10 V/Hz
8.20 10 Hz 5.18 10 Hz
h
e
−−⎛ ⎞= = ×⎜ ⎟× − ×⎝ ⎠
 and 
15 19 34(4.11 10 V/Hz)(1.60 10 C) 6.58 10 J sh − − −= × × = × ⋅ . 
 
Figure 38.53 
38.54. (a) 
14
( ) (200 W)(0.10)
( ) (5.00 10 Hz)
dN dE dt P
dt dE dN hf h
= = = =
×
196.03 10 photons sec.× 
(b) Demand 11 2
2
( )
1.00 10 photons sec cm .
4
dN dt
rπ
= × ⋅ 
Therefore, 
1/ 219
11 2
6.03 10 photons sec
6930 cm 69.3 m.
4 (1.00 10 photons sec cm )
r
π
⎛ ⎞×= = =⎜ ⎟× ⋅⎝ ⎠
 
38.55. (a) IDENTIFY: Apply the photoelectric effect equation, Eq.(38.4). 
SET UP: 0 ( / ) .eV hf hcφ λ φ= − = − Call the stopping potential 01V for 1λ and 02V for 2.λ Thus 
01 1( / )eV hc λ φ= − and 02 2( / ) .eV hc λ φ= − Note that the work function φ is a property of the material and is 
independent of the wavelength of the light. 
EXECUTE: Subtracting one equation from the other gives 1 202 01
1 2
( ) .e V V hc
λ λ
λ λ
⎛ ⎞−− = ⎜ ⎟
⎝ ⎠
 
38-12 Chapter 38 
(b) 
34 8 9 9
0 19 9 9
(6.626 10 J s)(2.998 10 m/s) 295 10 m 265 10 m
0.476 V.
1.602 10 C (295 10 m)(265 10 m)
V
− − −
− − −
⎛ ⎞× ⋅ × × − ×Δ = =⎜ ⎟× × ×⎝ ⎠
 
EVALUATE: 0 ,e VΔ which is 0.476 eV, is the increase in photon energy from 295 nm to 265 nm. The stoppingpotential increases when λ deceases because the photon energy increases when the wavelength decreases. 
38.56. IDENTIFY: The photoelectric effect occurs, so the energy of the photon is used to eject an electron, with any 
excess energy going into kinetic energy of the electron. 
SET UP: Conservation of energy gives hf = hc/λ = Kmax + φ. 
EXECUTE: (a) Using hc/λ = Kmax + φ, we solve for the work function: 
φ = hc/λ – Kmax = (4.136 × 10–15 eV s⋅ )(3.00 × 108 m/s)/(124 nm) – 4.16 eV = 5.85 eV 
(b) The number N of photoelectrons per second is equal to the number of photons per second that strike the metal 
per second. N × (energy of a photon) = 2.50 W. N(hc/λ) = 2.50 W. 
N = (2.50 W)(124 nm)/[(6.626 × 10–34 J s⋅ )(3.00 × 108 m/s)] = 1.56 × 1018 electrons/s 
(c) N is proportional to the power, so if the power is cut in half, so is N, which gives 
N = (1.56 × 1018 el/s)/2 = 7.80 × 1017 el/s 
(d) If we cut the wavelength by half, the energy of each photon is doubled since E = hc/λ. To maintain the same 
power, the number of photons must be half of what they were in part (b), so N is cut in half to 7.80 × 1017 el/s. We 
could also see this from part (b), where N is proportional to λ. So if the wavelength is cut in half, so is N. 
EVALUATE: In part (c), reducing the power does not reduce the maximum kinetic energy of the photons; it only 
reduces the number of ejected electrons. In part (d), reducing the wavelength does change the maximum kinetic 
energy of the photoelectrons because we have increased the energy of each photon. 
38.57. IDENTIFY and SET UP: The energy added to mass m of the blood to heat it to f 100 CT = ° and to vaporize it is 
f i v( )Q mc T T mL= − + , with 4190 J/kg Kc = ⋅ and 
6
v 2.256 10 J/kgL = × . The energy of one photon is 
251.99 10 J mhc
E
λ λ
−× ⋅= = . 
EXECUTE: (a) 9 9 6(2.0 10 kg)(4190 J/kg K)(100 C 33 C) (2.0 10 kg)(2.256 10 J/kg)Q − −= × ⋅ − + × × =° ° 35.07 10 J−× 
The pulse must deliver 5.07 mJ of energy. 
(b) 
3
6
energy 5.07 10 J
11.3 W
450 10 s
P
t
−
−
×= = =
×
 
(c) One photon has energy 
25
19
9
1.99 10 J m
3.40 10 J
585 10 m
hc
E
λ
−
−
−
× ⋅= = = ×
×
. The number N of photons per pulse is the 
energy per pulse divided by the energy of one photon: 
3
16
19
5.07 10 J
1.49 10 photons
3.40 10 J/photon
N
−
−
×= = ×
×
 
38.58. (a) 0λ ,
hc
E
= and the wavelengths are: cesium: 590 nm, copper: 264 nm, potassium: 539 nm, zinc: 288 nm. 
b) The wavelengths of copper and zinc are in the ultraviolet, and visible light is not energetic enough to overcome 
the threshold energy of these metals. 
38.59. (a) IDENTIFY and SET UP: Apply Eq.(38.20): p1 2r
1 2 p
207
207
e
e
m mm m
m
m m m m
= =
+ +
 
EXECUTE: 
31 27
28
r 31 27
207(9.109 10 kg)(1.673 10 kg)
1.69 10 kg
207(9.109 10 kg) 1.673 10 kg
m
− −
−
− −
× ×= = ×
× + ×
 
We have used em to denote the electron mass. 
(b) IDENTIFY: In Eq.(38.18) replace em m= by 
4
r
r 2 2 2
0
1
: .
8n
m e
m E
n h
= −
P
 
SET UP: Write as 
4
r H
2 2 2
H 0
1
,
8n
m m e
E
m n h
⎛ ⎞⎛ ⎞
= −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠P
 since we know that 
4
H
2 2
0
1
13.60 eV.
8
m e
h
=
P
 Here Hm denotes the 
reduced mass for the hydrogen atom; 31 31H 0.99946(9.109 10 kg) 9.104 10 kg.m
− −= × = × 
EXECUTE: r
2
H
13.60 eV
n
m
E
m n
⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
 
28
1 31
1.69 10 kg
( 13.60 eV) 186( 13.60 eV) 2.53 keV
9.104 10 kg
E
−
−
×= − = − = −
×
 
Photons, Electrons, and Atoms 38-13 
(c) SET UP: From part (b), r H
2
H
,n
m R ch
E
m n
⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
 where 7 1H 1.097 10 mR
−= × is the Rydberg constant for the 
hydrogen atom. Use this result in i f
hc
E E
λ
= − to find an expression for 1/ .λ The initial level for the transition is 
the 2in = level and the final level is the 1fn = level. 
EXECUTE: r H H
2 2
H i f
hc m R ch R ch
m n nλ
⎛ ⎞⎛ ⎞
= − − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
r
H 2 2
H f i
1 1 1m
R
m n nλ
⎛ ⎞
= −⎜ ⎟
⎝ ⎠
 
28
7 1 9 1
31 2 2
1 1.69 10 kg 1 1
(1.097 10 m ) 1.527 10 m
9.104 10 kg 1 2λ
−
− −
−
× ⎛ ⎞= × − = ×⎜ ⎟× ⎝ ⎠
 
0.655 nmλ = 
EVALUATE: From Example 38.6 the wavelength of the radiation emitted in this transition in hydrogen is 122 nm. 
The wavelength for muonium is 3H
r
5.39 10
m
m
−= × times this. The reduced mass for hydrogen is very close to the 
electron mass because the electron mass is much less then the proton mass: p e/ 1836.m m = The muon mass is 
28
e207 1.886 10 kg.m
−= × The proton is only about 10 times more massive than the muon, so the reduced mass is 
somewhat smaller than the muon mass. The muon-proton atom has much more strongly bound energy levels and 
much shorter wavelengths in its spectrum than for hydrogen. 
38.60. (a) The change in wavelength of the scattered photon is given by Eq. 38.23 
34
9
31 8
λ λ (1 cos ) λ λ (1 cos )
(6.63 10 J s)
(0.0830 10 m) (1 1) 0.0781 nm.
(9.11 10 kg)(3.00 10 m s)
h h
mc mc
φ φ
−
−
−
′ ′− = − ⇒ = − − =
× ⋅× − + =
× ×
 
(b) Since the collision is one-dimensional, the magnitude of the electron’s momentum must be equal to the 
magnitude of the change in the photon’s momentum. Thus, 
34 9 1
e
23 23
1 1 1 1
(6.63 10 J s) (10 m )
λ λ 0.0781 0.0830
1.65 10 kg m s 2 10 kg m s.
p h − −
− −
−⎛ ⎞ ⎛ ⎞= − = × ⋅ +⎜ ⎟ ⎜ ⎟′⎝ ⎠ ⎝ ⎠
= × ⋅ ≈ × ⋅
 
(c) Since the electron is non relativistic ( 0.06),β =
2
16 16e
e 1.49 10 J 10 J.2
p
K
m
− −= = × ≈ 
38.61. IDENTIFY and SET UP: (1 cos )
h
mc
λ λ φ′ = + − 
180φ = ° so 2 0.09485 m.h
mc
λ λ′ = + = Use Eq.(38.5) to calculate the momentum of the scattered photon. Apply 
conservation of energy to the collision to calculate the kinetic energy of the electron after the scattering. The 
energy of the photon is given by Eq.(38.2), 
EXECUTE: (a) 24/ 6.99 10 kg m/s.p h λ −′ ′= = × ⋅ 
(b) e e; / /E E E hc hc Eλ λ′ ′= + = + 
16
e
1 1
( ) 1.129 10 J 705 eVE hc hc
λ λ
λ λ λλ
−′ −⎛ ⎞= − = = × =⎜ ⎟′ ′⎝ ⎠
 
EVALUATE: The energy of the incident photon is 13.8 keV, so only about 5% of its energy is transferred to the 
electron. This corresponds to a fractional shift in the photon’s wavelength that is also 5%. 
38.62. (a) 
2
180 so (1 cos ) 2 λ 0.0049 nm, so λ 0.1849 nm.
h
mc
φ φ ′= ° − = ⇒ Δ = = = 
(b) 17
1 1
2.93 10 J 183 eV.
λ λ
E hc −
⎛ ⎞Δ = − = × =⎜ ⎟′⎝ ⎠
 This will be the kinetic energy of the electron. 
(c) The kinetic energy is far less than the rest mass energy, so a non-relativistic calculation is adequate; 
62 8.02 10 m s.v K m= = × 
38-14 Chapter 38 
38.63. IDENTIFY and SET UP: The Hα line in the Balmer series corresponds to the 3n = to 2n = transition. 
2
13.6 eV
nE n
= − . hc E
λ
= Δ . 
EXECUTE: (a) The atom must be given an amount of energy 3 1 2 2
1 1
(13.6 eV) 12.1 eV
3 1
E E
⎛ ⎞− = − − =⎜ ⎟
⎝ ⎠
. 
(b) There are three possible transitions. 3 1n n= → = : 12.1 eVEΔ = and 103 nmhc
E
λ = =
Δ
; 
3 2n n= → = : 
2 2
1 1
(13.6 eV) 1.89 eV
3 2
E
⎛ ⎞Δ = − − =⎜ ⎟
⎝ ⎠
and 657 nmλ = ; 2 1n n= → = : 
2 2
1 1
(13.6 eV) 10.2 eV
2 1
E
⎛ ⎞Δ = − − =⎜ ⎟
⎝ ⎠
and 122 nmλ = . 
38.64. 
( )
ex g( ) ex g2
1 2 1
( )
.
ln /
E E kT E En e T
n k n n
− − − −= ⇒ = 
18
ex 2 g ex g
13.6 eV
3.4 eV. 13.6 eV. 10.2 eV 1.63 10 J.
4
E E E E E −
−= = = − = − − = = × 
(a) 122
1
10 .
n
n
−= 
18
23 12
(1.63 10 J)
4275 K.
(1.38 10 J K) ln(10 )
T
−
− −
− ×= =
×
 
(b) 82
1
10 .
n
n
−= 
18
23 8
(1.63 10 J)
6412 K.
(1.38 10 J K ) ln(10 )
T
−
− −
− ×= =
×
 
(c) 42
1
10 .
n
n
−= 
18
23 4
(1.63 10 J)
12824 K.
(1.38 10 J K ) ln(10 )
T
−
− −
− ×= =
×
 
(d) For absorption to take place in the Balmer series, hydrogen must start in the 2n = state. From part (a), colder 
stars have fewer atoms in this state leading to weaker absorption lines. 
38.65. (a) IDENTIFY and SET UP: The photon energy is given to the electron in the atom. Some of this energy 
overcomes the binding energy of the atom and what is left appears as kinetic energy of the free electron. Apply 
f i ,hf E E= − the energy given to the electron in the atom when a photonis absorbed. 
EXECUTE: The energy of one photon is 
34 8
9
(6.626 10 J s)(2.998 10 m/s)
85.5 10 m
hc
λ
−
−
× ⋅ ×=
×
 
18 192.323 10 J(1 eV/1.602 10 J) 14.50 eV.
hc
λ
− −= × × = 
The final energy of the electron is f i .E E hf= + In the ground state of the hydrogen atom the energy of the electron 
is i 13.60 eV.E = − Thus f 13.60 eV 14.50 eV 0.90 eV.E = − + = 
(b) EVALUATE: At thermal equilibrium a few atoms will be in the 2n = excited levels, which have an energy of 
13.6 eV/4 3.40 eV, 10.2 eV− = − greater than the energy of the ground state. If an electron with 3.40 eVE = − 
gains 14.5 eV from the absorbed photon, it will end up with 14.5 eV 3.4 eV 11.1 eV− = of kinetic energy. 
38.66. IDENTIFY: The diffraction grating allows us to determine the peak-intensity wavelength of the light. Then 
Wien’s displacement law allows us to calculate the temperature of the blackbody, and the Stefan-Boltzmann law 
allows us to calculate the rate at which it radiates energy. 
SET UP: The bright spots for a diffraction grating occur when d sin θ = mλ. Wien’s displacement law is 
3
peak
2.90 10 m K
T
λ
−× ⋅= , and the Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the 
total radiated power is P = σAT 4. 
EXECUTE: (a) First find the wavelength of the light: 
λ = d sin θ = [1/(385,000 lines/m)] sin(11.6°) = 5.22 × 10–7 m 
Now use Wien’s law to find the temperature: T = (2.90 × 10–3 m K⋅ )/(5.22 × 10–7 m) = 5550 K. 
(b) The energy radiated by the blackbody is equal to the power times the time, giving 
U = Pt = IAt = σAT 4t, which gives 
t = U/(σAT 4) = (12.0 × 106 J)/[(5.67 × 10–8 2 4W/m K⋅ )(4π)(0.0750 m)2(5550 K)4] = 3.16 s. 
EVALUATE: By ordinary standards, this blackbody is very hot, so it does not take long to radiate 12.0 MJ of 
energy. 
Photons, Electrons, and Atoms 38-15 
38.67. IDENTIFY: Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the Stefan-
Boltzmann law apply to its radiation. 
SET UP: Wien’s displacement law is 
3
peak
2.90 10 m K
T
λ
−× ⋅= , and the Stefan-Boltzmann law says that the 
intensity of the radiation is I = σT 4, so the total radiated power is P = σAT 4. 
EXECUTE: (a) First use Wien’s law to find the peak wavelength: 
λm = (2.90 × 10–3 m K⋅ )/(3000 K) = 9.667 × 10–7 m 
Call N the number osf photons/second radiated. N × (energy per photon) = IA = σAT 4. 
N (hc/λm) = σAT 4. 
4
m ATN
hc
λ σ= . 
7 8 2 4 8 2 4
34 8
(9.667 10 m)(5.67 10 W/m K )(4 )(600 6.96 10 m) (3000 K)
(6.626 10 J s)(3.00 10 m/s)
N
π− −
−
× × ⋅ × ×=
× ⋅ ×
. 
N = 5 × 1049 photons/s. 
(b) 
2 44 2 4
B B B B B B S
4 2 4
S S S S S S S
4 600 3000 K
4 5800 K
I A A T R T R
I A A T R T R
σ π
σ π
⎛ ⎞ ⎛ ⎞= = = ⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
 = 3 × 104 
EVALUATE: Betelgeuse radiates 30,000 times as much energy per second as does our sun! 
38.68. IDENTIFY: The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody. 
The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a 
thermophile in the water measures the wavelength and frequency of the light in the water. 
SET UP: By the Stefan-Boltzman law, the net power radiated by the blackbody is ( )4 4sphere waterdQ A T Tdt σ= − . Since 
this heat evaporates water, the rate at which water evaporates is v
dQ dm
L
dt dt
= . Wien’s displacement law is 
3
m
2.90 10 m K
T
λ
−× ⋅= , and the wavelength in the water is λw = λ0/n. 
EXECUTE: (a) The net radiated heat is ( )4 4sphere waterdQ A T Tdt σ= − and the evaporation rate is v
dQ dm
L
dt dt
= , where 
dm is the mass of water that evaporates in time dt. Equating these two rates gives ( )4 4v sphere waterdmL A T Tdt σ= − . 
( )( )2 4 4sphere water
v
4 R T Tdm
dt L
σ π −
= . 
( )( )8 2 4 2 4 4 4
3
5.67 10 W/m K 4 (0.120 m) (498 K) (373 K)
1.92 10 kg/s 0.193 g/s
2256 10 J/Kg
dm
dt
π−
−
⎡ ⎤× ⋅ −⎣ ⎦= = × =
×
 
(b) (i) Wien’s law gives λm = (0.00290 m K⋅ )/(498 K) = 5.82 × 10–6 m 
But this would be the wavelength in vacuum. In the water the thermophile organism would measure λw = λ0/n = 
(5.82 × 10–6 m)/1.333 = 4.37 × 10–6 m = 4.37 µm 
(ii) The frequency is the same as if the wave were in air, so 
f = c/λ0 = (3.00 ×108 m/s)/(5.82 × 10–6 m) = 5.15 × 1013 Hz 
EVALUATE: An alternative way is to use the quantities in the water:
0
/
/
c n
f
nλ
= = c/λ0, which gives the same 
answer for the frequency. An organism in the water would measure the light coming to it through the water, so the 
wavelength it would measure would be reduced by a factor of 1/n. 
38.69. IDENTIFY: The energy of the peak-intensity photons must be equal to the energy difference between the n = 1 
and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to 
radiate with its peak intensity at this wavelength. 
SET UP: In the Bohr model, the energy of an electron in shell n is 
2
13.6 eV
nE n
= − , and Wien’s displacement law 
is 
3
m
2.90 10 m K
T
λ
−× ⋅= . The energy of a photon is E = hf = hc/λ. 
EXECUTE: First find the energy (ΔE) that a photon would need to excite the atom. The ground state of the atom 
is n = 1 and the third excited state is n = 4. This energy is the difference between the two energy levels. Therefore 
38-16 Chapter 38 
ΔE = ( ) 2 2
1 1
13.6 eV
4 1
⎛ ⎞− −⎜ ⎟
⎝ ⎠
 = 12.8 eV. Now find the wavelength of the photon having this amount of energy. 
hc/λ = 12.8 eV and 
λ = (4.136 × 10–15 eV s⋅ )(3.00 × 108 m/s)/(12.8 eV) = 9.73 ×10–8 m 
Now use Wien’s law to find the temperature. T = (0.00290 m K⋅ )/(9.73 × 10–8 m) = 2.98 × 104 K. 
EVALUATE: This temperature is well above ordinary room temperatures, which is why hydrogen atoms are not in 
excited states during everyday conditions. 
38.70. IDENTIFY and SET UP: Electrical power is VI. Q mc T= Δ . 
EXECUTE: (a) 3 3(0.010) (0.010)(18.0 10 V)(60.0 10 A) 10.8 W 10.8 J/sVI −= × × = = 
(b) The energy in the electron beam that isn’t converted to x rays stays in the target and appears as thermal energy. 
For 1.00 st = , 3(0.990) (1.00 s) 1.07 10 JQ VI= = × and 
31.07 10 J
32.9 K
(0.250 kg)(130 J/kg K)
Q
T
mc
×Δ = = =
⋅
. The 
temperature rises at a rate of 32.9 K/s. 
EVALUATE: The target must be made of a material that has a high melting point. 
38.71. IDENTIFY: Apply conservation of energy and conservation of linear momentum to the system of atom plus 
photon. 
(a) SET UP: Let trE be the transition energy, phE be the energy of the photon with wavelength ,λ′ and rE be 
the kinetic energy of the recoiling atom. Conservation of energy gives ph r tr .E E E+ = 
ph
hc
E
λ
=
′
 so tr r
hc
E E
λ
= −
′
 and 
tr r
.
hc
E E
λ′ =
−
 
EXECUTE: If the recoil energy is neglected then the photon wavelength is tr/ .hc Eλ = 
tr r tr tr r tr
1 1 1
1
1 /
hc
hc
E E E E E E
λ λ λ
⎛ ⎞ ⎛ ⎞⎛ ⎞′Δ = − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠ ⎝ ⎠⎝ ⎠
 
1
r r
r tr tr tr
1
1 1
1 / E
E E
E E E
−
⎛ ⎞
= − ≈ +⎜ ⎟− ⎝ ⎠
 since r
tr
1
E
E
V 
(We have used the binomial theorem, Appendix B.) 
Thus r
tr tr
,
hc E
E E
λ
⎛ ⎞
Δ = ⎜ ⎟
⎝ ⎠
 or since 2rtr / , .
E
E hc
hc
λ λ λ⎛ ⎞= Δ = ⎜ ⎟
⎝ ⎠
 
SET UP: Use conservation of linear momentum to find r :E Assuming that the atom is initially at rest, the 
momentum rp of the recoiling atom must be equal in magnitude and opposite in direction to the momentum 
ph /p h λ= of the emitted photon: r/ .h pλ = 
EXECUTE: 
2
r
r ,2
p
E
m
= where m is the mass of the atom, so 
2
r 2
.
2
h
E
mλ
= 
Use this result in the above equation: 
2 2
2r
2
;
2 2
E h h
hc m hc mc
λλ λ
λ
⎛ ⎞⎛ ⎞⎛ ⎞Δ = = =⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
 
note that this result for λΔ is independent of the atomic transition energy. 
(b) For a hydrogen atom pm m= and 
34
16
27 8
p
6.626 10 J s
6.61 10 m
2 2(1.673 10 kg)(2.998 10 m/s)
h
m c
λ
−
−
−
× ⋅Δ = = = ×
× ×
 
EVALUATE: The correction is independent of n. The wavelengths of photons emitted in hydrogen atom 
transitionsare on the order of 7100 nm 10 m,−= so the recoil correction is exceedingly small. 
38.72. (a) 1 1 2 2λ ( )(1 cos ), λ ( )(1 cos ),h mc θ h mc θΔ = − Δ = − and so the overall wavelength shift is 
1 2λ ( )(2 cos cos ).h mc θ θΔ = − − 
(b) For a single scattering through angle s, λ ( )(1 cos ).θ h mc θΔ = − For two successive scatterings through an 
angle of 2θ for each scattering, 
tλ 2( )(1 cos 2).h mc θΔ = − 
2 2
s
2
s t
1 cos 2(1 cos ( 2)) and λ ( )2(1 cos ( 2))
cos( 2) 1so1 cos ( 2) (1 cos( 2)) and λ λ
θ θ h mc θ
θ θ θ
− = − Δ = −
≤ − ≥ − Δ ≥ Δ
 
Photons, Electrons, and Atoms 38-17 
Equality holds only when 180 .θ = ° 
(c) ( )2(1 cos30.0 ) 0.268( ).h mc h mc− ° = 
(d) ( )(1 cos60 ) 0.500( ),h mc h mc− ° = which is indeed greater than the shift found in part (c). 
38.73. IDENTIFY and SET UP: Find the average change in wavelength for one scattering and use that in λΔ in 
Eq.(38.23) to calculate the average scattering angle .φ 
EXECUTE: (a) The wavelength of a 1 MeV photon is 
15 8
12
6
(4.136 10 eV s)(2.998 10 m/s)
1 10 m
1 10 eV
hc
E
λ
−
−× ⋅ ×= = = ×
×
 
The total change in wavelength therefore is 9 12 9500 10 m 1 10 m 500 10 m.− − −× − × = × 
If this shift is produced in 2610 Compton scattering events, the wavelength shift in each scattering event is 
9
33
26
500 10 m
5 10 m.
1 10
λ
−
−×Δ = = ×
×
 
(b) Use this λΔ in (1 cos )h
mc
λ φΔ = − and solve for .φ We anticipate that φ will be very small, since λΔ is 
much less than / ,h mc so we can use 2cos 1 / 2.φ φ≈ − 
2 2(1 (1 / 2))
2
h h
mc mc
λ φ φΔ = − − = 
33
11 9
12
2 2(5 10 m)
6.4 10 rad (4 10 )
( / ) 2.426 10 mh mc
λφ
−
− −
−
Δ ×= = = × = × °
×
 
φ in radians is much less than 1 so the approximation we used is valid. 
(c) IDENTIFY and SET UP: We know the total transit time and the total number of scatterings, so we can calculate 
the average time between scatterings. 
EXECUTE: The total time to travel from the core to the surface is 6 7 13(10 y)(3.156 10 s/y) 3.2 10 s.× = × There are 
2610 scatterings during this time, so the average time between scatterings is 
13
13
26
3.2 10 s
3.2 10 s.
10
t −
×= = × 
The distance light travels in this time is 8 13(3.0 10 m/s)(3.2 10 s) 0.1 mmd ct −= = × × = 
EVALUATE: The photons are on the average scattered through a very small angle in each scattering event. The 
average distance a photon travels between scatterings is very small. 
38.74. (a) The final energy of the photon is ,and ,
λ
hc
E E E K′ ′= = +
′
 where K is the kinetic energy of the electron after 
the collision. Then, 
2
2 2 1 2
λ
λ .
( λ ) ( λ ) ( 1) λ 1
1 1
(1 )
hc hc hc
E K hc K hc mc mc
h v c
γ
′
= = = =
′ ′ ′ ′+ + + − ⎡ ⎤
+ −⎢ ⎥−⎣ ⎦
 
2( ( 1)K mc γ= − since the relativistic expression must be used for three-figure accuracy). 
(b) arccos(1 λ ( )).h mcφ = − Δ 
(c) 
( )( )
12
1 221.80
3.00
1
1 1 1.25 1 0.250, 2.43 10 m
1
h
mc
γ −− = − = − = = ×
−
 
3
3
12 31 8
34
5.10 10 mm
λ 3.34 10 nm
(5.10 10 m)(9.11 10 kg)(3.00 10 m s)(0.250)
1
(6.63 10 J s)
−
−
− −
−
×
⇒ = = ×
× × ×+
× ⋅
. 
12 12
12
(5.10 10 m 3.34 10 m)
arccos 1 74.0 .
2.43 10 m
φ
− −
−
⎛ ⎞× − ×= − = °⎜ ⎟×⎝ ⎠
 
38.75. (a) IDENTIFY and SET UP: Conservation of energy applied to the collision gives e ,E E Eλ λ′= + where eE is the 
kinetic energy of the electron after the collision and Eλ and Eλ′ are the energies of the photon before and after the 
collision. The energy of a photon is related to its wavelength according to Eq.(38.2). 
38-18 Chapter 38 
EXECUTE: e
1 1
E hc hc
λ λ
λ λ λλ
′ −⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟′ ′⎝ ⎠ ⎝ ⎠
 
9
34 8
e 9 9
0.0032 10 m
(6.626 10 J s)(2.998 10 m/s)
(0.1100 10 m)(0.1132 10 m)
E
−
−
− −
⎛ ⎞×= × ⋅ × ⎜ ⎟× ×⎝ ⎠
 
17
e 5.105 10 J 319 eVE
−= × = 
2
e
1
2
E mv= so 
17
7e
31
2 2(5.105 10 J)
1.06 10 m/s
9.109 10 kg
E
v
m
−
−
×= = = ×
×
 
(b) The wavelength λ of a photon with energy eE is given by e /E hc λ= so 
34 8
17
e
(6.626 10 J s)(2.998 10 m/s)
3.89 nm
5.105 10 J
hc
E
λ
−
−
× ⋅ ×= = =
×
 
EVALUATE: Only a small portion of the incident photon’s energy is transferred to the struck electron; this is why 
the wavelength calculated in part (b) is much larger than the wavelength of the incident photon in the Compton 
scattering. 
38.76. IDENTIFY: Apply the Compton scattering formula c(1 cos ) (1 cos )
h
mc
λ λ λ φ λ φ′ − = Δ = − = − 
(a) SET UP: Largest λΔ is for 180 .φ = ° 
EXECUTE: For c180 , 2 2(2.426 pm) 4.85 pm.φ λ λ= ° Δ = = = 
(b) SET UP: c (1 cos )λ λ λ φ′ − = − 
Wavelength doubles implies 2λ λ′ = so .λ λ λ′ − = Thus C (1 cos ). λ λ φ λ= − is related to E by Eq.(38.2). 
EXECUTE: / ,E hc λ= so smallest energy photon means largest wavelength photon, so 180φ = ° and 
c2 4.85 pm.λ λ= = Then 
34 8
14 19
12
(6.626 10 J s)(2.998 10 m/s)
4.096 10 J(1 eV/1.602 10 J)
4.85 10 m
hc
E
λ
−
− −
−
× ⋅ ×= = = × × =
×
 
0.256 MeV. 
EVALUATE: Any photon Compton scattered at 180φ = ° has a wavelength increase of c2 4.85 pm.λ = 4.85 pm is 
near the short-wavelength end of the range of x-ray wavelengths. 
38.77. (a) 
2
5 λ
2
(λ) but λ
λ ( 1)hc kT
hc c
I
e f
π= =
−
 
2 5
5 3
2 2
( )
( ) ( 1) ( 1)hf kT hf kT
hc hf
I f
c f e c e
π π
⇒ = =
− −
 
(b) 
0
20
(λ) ( )
c
I d I f df
f
λ
∞
∞
⎛ ⎞−= ⎜ ⎟
⎝ ⎠
∫ ∫ 
3 4 3 4 5 4 5 4 4
4
2 2 3 2 3 3 2 2 30 0
2 2 ( ) 2 ( ) 1 (2 ) ( ) 2
(2 )
( 1) 1 240 240 15hf kT x
hf df kT x kT kT k T
dx
c e c h e c h h c c h
π π π π ππ
∞ ∞
= = = = =
− −∫ ∫ 
(c) The expression 
5 4 4
3 2
2
15
k T
h c
π σ= as shown in Eq. (38.36). Plugging in the values for the constants we get 
8 2 45.67 10 W m Kσ −= × ⋅ . 
38.78. 4, , and ; combining,I σT P IA E Pt= = Δ = 
3
4 6 2 8 2 4 4
(100 J)
8.81 10 s 2.45 hrs.
(4.00 10 m )(5.67 10 W m K )(473 K)
E
t
A Tσ − −
Δ= = = × =
× × ⋅
 
38.79. (a) The period was found in Exercise 38.27b: 
2 3 3
0
4
4 n h
T
me
= P and frequency is just 
4
2 3 3
0
1
.
4
me
f
T n h
= =
P
 
(b) Eq. (38.6) tells us that 2 1
1
( ).f E E
h
= − So 
4
2 3 2 2
0 2 1
1 1
8
me
f
h n n
⎛ ⎞
= −⎜ ⎟
⎝ ⎠P
 (from Eq. (38.18)). 
If 2 1 2 2 2 2
2 1
1 1 1 1
and 1, then
( 1)
n n n n
n n n n
= = + − = −
+
 
4
2 2 2 3 2 3 3
0
1 1 1 2 2
1 1 1 for large .
(1 1 ) 4
me
n f
n n n n n n h
⎛ ⎞ ⎛ ⎞⎛ ⎞= − ≈ − − + = ⇒ ≈⎜ ⎟ ⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠⎝ ⎠ P
 
Photons, Electrons, and Atoms 38-19 
38.80. Each photon has momentum ,
λ
h
p = and if the rate at which the photons strike the surface is ( ),dN dt the force 
on the surface is ( λ)( ),h dN dt and the pressure is ( λ)( ) .h dN dt A The intensity is 
( )( ) ( )( λ) ,I dN dt E A dN dt hc A= = and comparison of the two expressions gives the pressure as ( ).I c 
38.81. Momentum: ( )p P p P p P p P′ ′ ′ ′ ′ ′⇒ − = − − ⇒ = − +p + P = p + P 
energy: pc E p c E′ ′+ = + 2 2 2( ) ( )p c P c mc′ ′= + + 
2 2 2 2( ) ( ) ( )pc p c E P c mc′ ′⇒ − + = + 2 2 2 2 2( ) (( ) ) 2 ( ) ( )Pc p p c P p p c mc′ ′= + + − + + 
2 2 2 2 2 2( ) ( ) 2( )( ) 2 ( ) 4 2 ( )pc p c E E pc p c Pc p p Ec p p pp c Ec p p′ ′ ′ ′ ′ ′− + = + + − + + − − + − 
22( )( ) 0Pc p p′+ + = 
2 2 2
2
2 2
( 2 ) ( )
2 2 ( )
2 λ ( ) 2
λ λ λ
(λ( ) 2 )
λ
p Pc pc Ec p Ec Pc
Ec Pc E Pc
p p p
pc Ec Pc pc E Pc
hc E Pc E Pc hc
E Pc E Pc E Pc
E Pc hc
E Pc
′⇒ − − = − −
+ +′⇒ = =
+ − + −
+ − −⎛ ⎞ ⎛ ⎞′⇒ = = +⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠
− +′⇒ =
+
 
If
22
2 2 2 2, ( ) 1
mc
E mc Pc E mc E
E
⎛ ⎞
= − = − ⎜ ⎟
⎝ ⎠
W 
221
1
2
mc
E
E
⎛ ⎞⎛ ⎞⎜ ⎟≈ − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
2 21 ( )
2
mc
E Pc
E
⇒ − ≈
2 2 2 4
1
( ) λ
λ 1
2 (2 ) 4
λ mc hc hc m c
E E E E hcE
⎛ ⎞
⇒ ≈ + = +⎜ ⎟
⎝ ⎠
 
(b) If 6 0 9λ 10.6 10 m, 1.00 10 eV 1.60 10 JE− 1 −= × = × = × 
31 2 4 6
9 9
(9.11 10 kg) (10.6 10 m)
λ 1
1.60 10 J 4 (1.6 10 J)
hc c
hc
− −
− −
⎛ ⎞× ×′⇒ ≈ +⎜ ⎟× ×⎝ ⎠
 
16 15(1.24 10 m)(1 56.0) 7.08 10 m.− −= × + = × 
(c) These photons are gamma rays. We have taken infrared radiation and converted it into gamma rays! Perhaps 
useful in nuclear medicine, nuclear spectroscopy, or highenergy physics: wherever controlled gamma ray sources 
might be useful. 
THE WAVE NATURE OF PARTICLES 
 39.1. IDENTIFY and SET UP: 
h h
p mv
λ = = . For an electron, 319.11 10 kgm −= × . For a proton, 271.67 10 kgm −= × . 
EXECUTE: (a) 
34
10
31 6
6.63 10 J s
1.55 10 m 0.155 nm
(9.11 10 kg)(4.70 10 m/s)
λ
−
−
−
× ⋅= = × =
× ×
 
(b) λ is proportional to 1
m
, so 
31
10 14e
p e 27
p
9.11 10 kg
(1.55 10 m) 8.46 10 m
1.67 10 kg
m
m
λ λ
−
− −
−
⎛ ⎞ ⎛ ⎞×= = × = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠
. 
 39.2. IDENTIFY and SET UP: For a photon, 
hc
E
λ
= . For an electron or proton, hp
λ
= and 
2
2
p
E
m
= , so 
2
22
h
E
mλ
= . 
EXECUTE: (a) 
15 8
9
(4.136 10 eV s)(3.00 10 m/s)
6.2 keV
0.20 10 m
hc
E
λ
−
−
× ⋅ ×= = =
×
 
(b) 
22 34
18
2 9 31
6.63 10 J s 1
6.03 10 J 38 eV
2 0.20 10 m 2(9.11 10 kg)
h
E
mλ
−
−
− −
⎛ ⎞× ⋅= = = × =⎜ ⎟× ×⎝ ⎠
 
(c)
31
e
p e 27
p
9.11 10 kg
(38 eV) 0.021 eV
1.67 10 kg
m
E E
m
−
−
⎛ ⎞ ⎛ ⎞×= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠
 
EVALUATE: For a given wavelength a photon has much more energy than an electron, which in turn has more 
energy than a proton. 
 39.3. (a) 
34
24
10
(6.63 10 J s)
λ 2.37 10 kg m s.
λ (2.80 10 m)
h h
p
p
−
−
−
× ⋅= ⇒ = = = × ⋅
×
 
(b) 
2 24 2
18
31
(2.37 10 kg m s)
3.08 10 J 19.3 eV.
2 2(9.11 10 kg)
p
K
m
−
−
−
× ⋅= = = × =
×
 
 39.4. λ
2
h h
p mE
= = 
34
15
27 6 19
(6.63 10 J s)
7.02 10 m.
2(6.64 10 kg) (4.20 10 eV) (1.60 10 J eV)
−
−
− −
× ⋅= = ×
× × ×
 
 39.5. IDENTIFY and SET UP: The de Broglie wavelength is .
h h
p mv
λ = = In the Bohr model, ( / 2 ), nmvr n h π= 
so /(2 ).nmv nh rπ= Combine these two expressions and obtain an equation for λ in terms of n. Then 
2 2
.n n
r r
h
nh n
π πλ ⎛ ⎞= =⎜ ⎟
⎝ ⎠
 
EXECUTE: (a) For 101 1 01, 2 with 0.529 10 m, son r r aλ π
−= = = = × 10 102 (0.529 10 m) 3.32 10 mλ π − −= × = × 
12 ;rλ π= the de Broglie wavelength equals the circumference of the orbit. 
(b) For 44, 2 / 4.n rλ π= = 
2
0 4 0 so 16 .nr n a r a= = 
10 9
0 02 (16 ) / 4 4(2 ) 4(3.32 10 m) 1.33 10 ma aλ π π
− −= = = × = × 
42 / 4;rλ π= the de Broglie wavelength is 
1 1
4n
= times the circumference of the orbit. 
EVALUATE: As n increases the momentum of the electron increases and its de Broglie wavelength decreases. For 
any n, the circumference of the orbits equals an integer number of de Broglie wavelengths. 
39
39-2 Chapter 39 
 39.6. (a) For a nonrelativistic particle, 
2
, so
2
p
K
m
= .
2
h h
p Km
λ = = 
(b) 34 -19 -31 11(6.63 10 J s) 2(800 eV)(1.60 10 J/eV)(9.11 10 kg) 4.34 10 m.− −× ⋅ × × = × 
 39.7. IDENTIFY: A person walking through a door is like a particle going through a slit and hence should exhibit wave 
properties. 
SET UP: The de Broglie wavelength of the person is λ = h/mv. 
EXECUTE: (a) Assume m = 75 kg and v = 1.0 m/s. 
λ = h/mv = (6.626 × 10–34 J ⋅ s)/[(75 kg)(1.0 m/s)] = 8.8 × 10–36 m 
EVALUATE: (b) A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too small 
to show wave behavior through a “slit” that is about 1035 times as wide as the wavelength. Hence ordinary objects 
do not show wave behavior in everyday life. 
 39.8. Combining Equations 37.38 and 37.39 gives 2 1.p mc γ= − 
(a) 2 12( ) 1 4.43 10 m.
h
h mc
p
λ γ −= = − = × (The incorrect nonrelativistic calculation gives 125.05 10 m.)−× 
(b) 2 13( ) 1 7.07 10 m.h mc γ −− = × 
 39.9. IDENTIFY and SET UP: A photon has zero mass and its energy and wavelength are related by Eq.(38.2). An 
electron has mass. Its energy is related to its momentum by 2 / 2E p m= and its wavelength is related to its 
momentum by Eq.(39.1). 
EXECUTE: (a) photon 
34 8
19
(6.626 10 J s)(2.998 10 m/s)
 so 62.0 nm
(20.0 eV)(1.602 10 J/eV)
hc hc
E
E
λ
λ
−
−
× ⋅ ×= = = =
×
 
electron 
2 /(2 ) so 2E p m p mE= = = 31 19 242(9.109 10 kg)(20.0 eV)(1.602 10 J/eV) 2.416 10 kg m/s− − −× × = × ⋅ 
/ 0.274 nmh pλ = = 
(b) photon 19/ 7.946 10 J 4.96 eVE hc λ −= = × = 
electron 27/ so / 2.650 10 kg m/sh p p hλ λ −= = = × ⋅ 
2 24 5/(2 ) 3.856 10 J 2.41 10 eVE p m − −= = × = × 
(c) EVALUATE: You should use a probe of wavelength approximately 250 nm. An electron with 250 nmλ = has 
much less energy than a photon with 250 nm,λ = so is less likely to damage the molecule. Note that /h pλ = 
applies to all particles, those with mass and those with zero mass. /E hf hc λ= = applies only to photons and 
2 / 2E p m= applies only to particles with mass. 
39.10. IDENTIFY: Any moving particle has a de Broglie wavelength. The speed of a molecule, and hence its de Broglie 
wavelength, depends on the temperature of the gas. 
SET UP: The average kinetic energy of the molecule is Kav = 3/2 kT, and the de Broglie wavelength is λ = 
h/mv = h/p. 
EXECUTE: (a) Combining Kav = 3/2 kT and K = p
2/2m gives 3/2 kT = pav
2/2m and pav = 3mkT . The de Broglie 
wavelength is 
3
h h
p mkT
λ = = =
( )( )( )
34
10
27 23
6.626 10 J s
1.08 10 m
3 2 1.67 10 kg 1.38 10 J/K 273 K
−
−
− −
× ⋅ = ×
× × ×
. 
(b) For an electron, λ = h/p = h/mv gives 
( )( )
34
31 10
6.626 10 J s
9.11 10 kg 1.08 10 m
h
v
mλ
−
− −
× ⋅= =
× ×
 = 6.75 × 106 m/s 
This is about 2% the speed of light, so we do not need to use relativity. 
(c) For photon: 
E = hc/λ = (6.626 × 10–34 J ⋅ s)(3.00 × 108 m/s)/(1.08 × 10–10 m) = 1.84 × 10–15 J 
For the H2 molecule: Kav = (3/2)kT = 3/2 (1.38 × 10–23 J/K)(273 K) = 5.65 × 10–21 J 
For the electron: K = ½ mv2 = ½ (9.11 × 10–31 kg)(6.73 × 106 m/s)2 = 2.06 × 10–17 J 
EVALUATE: The photon has about 100 times more energy than the electron and 300,000 times more energy than 
the H2 molecule. This shows that photons of a given wavelength will have much more energy than particles of the 
same wavelength. 
The Wave Nature of Particles 39-3 
39.11. IDENTIFY and SET UP: Use Eq.(39.1). 
EXECUTE: 
34
34
3
6.626 10 J s
3.90 10 m
(5.00 10 kg)(340 m/s)
h h
p mv
λ
−
−
−
× ⋅= = = = ×
×
 
EVALUATE: This wavelength is extremely short; the bullet will not exhibit wavelike properties. 
39.12. (a) h mv v h mλ λ= → = 
Energy conservation: 2
1
2
e V mvΔ = 
2
2 2 34 2
2 19 31 9 2
(6.626 10 J s)
66.9 V
2 2 2 2(1.60 10 C) (9.11 10 kg) (0.15 10 m)
h
m
mv hm
V
e e em
λ
λ
−
− − −
⎛ ⎞
⎜ ⎟ × ⋅⎝ ⎠Δ = = = = =
× × ×
 
(b) 
34 8
15
photon 9
(6.626 10 J s) (3.0 10 m s)
1.33 10 J
0.15 10 m
hc
E hf
λ
−
−
−
× ⋅ ×= = = = ×
×
 
photone V K EΔ = = and 
15
photon
19
1.33 10 J
8310 V
1.6 10 C
E
V
e
−
−
×Δ = = =
×
 
39.13. (a) 0.10 nmλ = . 6so ( ) 7.3 10 m sp mv h v h mλ λ= = = = × . 
(b) 2
1
150 eV
2
E mv= = 
(c) E / 12 KeVhc λ= = 
(d) The electron is a better probe because for the same λ it has less energy and is less damaging to the structure 
being probed. 
39.14. IDENTIFY: The electrons behave like waves and are diffracted by the slit. 
SET UP: We use conservation of energy to find the speed of the electrons, and then use this speed to find their de 
Broglie wavelength, which is λ = h/mv. Finally we know that the first dark fringe for single-slit diffraction occurs 
when a sin θ = λ. 
EXECUTE: (a) Use energy conservation to find the speed of the electron: ½ mv2 = eV. 
v = 
( )19
31
2 1.60 10 C (100 V)2
9.11 10 kg
eV
m
−
−
×
=
×
 = 5.93 × 106 m/s 
which is about 2% the speed of light, so we can ignore relativity. 
(b) First find the de Broglie wavelength: 
( )( )
34
31 6
6.626 10 J s
9.11 10 kg 5.93 10 m/s
h
mv
λ
−
−
× ⋅= =
× ×
 = 1.23 × 10–10 m = 0.123 nm 
For the first single slit dark fringe, we have a sin θ = λ, which gives 
101.23 10 m
sin sin(11.5 )
a
λ
θ
−×= =
°
 = 6.16 × 10–10 m = 0.616 nm 
EVALUATE: The slit width is around 5 times the de Broglie wavelength of the electron, and both are much 
smaller than the wavelength of visible light. 
39.15. For m =1, sin
2
h
d θ
mE
λ = = . 
2 34 2
20
2 2 27 11 2 2
(6.63 10 J s)
6.91 10 J 0.432 eV.
2 sin 2(1.675 10 kg) (9.10 10 m) sin (28.6 )
h
E
md θ
−
−
− −
× ⋅= = = × =
× × °39.16. Intensity maxima occur when sin λ.d θ m= λ so sin .
2 2
h h mh
d θ
p ME ME
= = = (Careful! Here, m is the order 
of the maxima, whereas M is the mass of the incoming particle.) 
(a) 
34
31 19
(2)(6.63 10 J s)
2 sin 2(9.11 10 kg)(188 eV)(1.60 10 J/eV) sin(60.6 )
mh
d
ME θ
−
− −
× ⋅= =
× × °
 
 102.06 10 m 0.206 nm.−= × = 
(b) m = 1 also gives a maximum. 
34
31 19 10
(1) (6.63 10 J s)
arcsin 25.8
2(9.11 10 kg) (188 eV) (1.60 10 J eV) (2.06 10 m)
θ
−
− − −
⎛ ⎞× ⋅⎜ ⎟= = °
⎜ ⎟× × ×⎝ ⎠
. 
39-4 Chapter 39 
This is the only other one. If we let 3,m ≥ then there are no more maxima. 
(c) 
2 2 2 34 2
2 2 31 10 2 2
18
(1) (6.63 10 J s)
2 sin 2(9.11 10 kg) (2.60 10 m) sin (60.6 )
7.49 10 J 46.8 eV.
m h
E
Md θ
−
− −
−
× ⋅= =
× × °
= × =
 
Using this energy, if we let 2, then sin 1. Thus, there is no 2m θ m= > = maximum in this case. 
39.17. The condition for a maximum is sin . , so arcsin .
h h mh
d m
p Mv dMv
θ λ λ θ ⎛ ⎞= = = = ⎜ ⎟
⎝ ⎠
 (Careful! Here, m is the order of 
the maximum, whereas M is the incoming particle mass.) 
(a) 11 arcsin
h
m θ
dMv
⎛ ⎞= ⇒ = ⎜ ⎟
⎝ ⎠
 
 
34
6 31 4
6.63 10 J s
arcsin 2.07 .
(1.60 10 m) (9.11 10 kg) (1.26 10 m s)
−
− −
⎛ ⎞× ⋅= = °⎜ ⎟× × ×⎝ ⎠
 
34
2 6 31 4
(2) (6.63 10 J s)
2 arcsin 4.14 .
(1.60 10 m) (9.11 10 kg) (1.26 10 m s)
m θ
−
− −
⎛ ⎞× ⋅= ⇒ = =⎜ ⎟× × ×⎝ ⎠
° 
(b) For small angles (in radians!) , soy Dθ≅ 1
radians
(50.0 cm) (2.07 ) 1.81cm
180
π
y
⎛ ⎞≈ ° =⎜ ⎟°⎝ ⎠
, 
2
radians
(50.0 cm) (4.14 ) 3.61 cm
180
π
y
⎛ ⎞≈ ° =⎜ ⎟°⎝ ⎠
 and 2 1 3.61 cm 1.81cm 1.81 cm.y y− = − = 
39.18. IDENTIFY: Since we know only that the mosquito is somewhere in the room, there is an uncertainty in its 
position. The Heisenberg uncertainty principle tells us that there is an uncertainty in its momentum. 
SET UP: The uncertainty principle is xx pΔ Δ ≥ . 
EXECUTE: (a) You know the mosquito is somewhere in the room, so the maximum uncertainty in its horizontal 
position is Δ x = 5.0 m. 
(b) The uncertainty principle gives xx pΔ Δ ≥ , and Δpx = mΔvx since we know the mosquito’s mass. This gives 
 xx m vΔ Δ ≥ , which we can solve for Δvx to get the minimum uncertainty in vx. 
34
-6
1.055 10 J s
(1.5 10 kg)(5.0 m)x
v
m x
−× ⋅Δ = =
Δ ×
 = 1.4 × 10–29 m/s 
which is hardly a serious impediment! 
EVALUATE: For something as “large” as a mosquito, the uncertainty principle places a negligible limitation on 
our ability to measure its speed. 
39.19. (a) IDENTIFY and SET UP: Use / 2xx p h πΔ Δ ≥ to calculate xΔ and obtain xvΔ from this. 
EXECUTE: 
34
28
6
6.626 10 J s
1.055 10 kg m/s
2 2 (1.00 10 m)x
h
p
xπ π
−
−
−
× ⋅Δ ≥ = = × ⋅
Δ ×
 
28
321.055 10 kg m/s 8.79 10 m/s
1200 kg
x
x
p
v
m
−
−Δ × ⋅Δ = = = × 
(b) EVALUATE: Even for this very small xΔ the minimum xvΔ required by the Heisenberg uncertainty principle 
is very small. The uncertainty principle does not impose any practical limit on the simultaneous measurements of 
the positions and velocities of ordinary objects. 
39.20. IDENTIFY: Since we know that the marble is somewhere on the table, there is an uncertainty in its position. The 
Heisenberg uncertainty principle tells us that there is therefore an uncertainty in its momentum. 
SET UP: The uncertainty principle is xx pΔ Δ ≥ . 
EXECUTE: (a) Since the marble is somewhere on the table, the maximum uncertainty in its horizontal position is 
Δ x = 1.75 m. 
(b) Following the same procedure as in part (b) of problem 39.18, the minimum uncertainty in the horizontal 
velocity of the marble is 
( )
341.055 10 J s
0.0100 kg (1.75 m)x
v
m x
−× ⋅Δ = =
Δ
 = 6.03 × 10–33 m/s 
(c) The uncertainty principle tells us that we cannot know that the marble’s horizontal velocity is exactly zero, so 
the smallest we could measure it to be is 6.03 × 10–33 m/s, from part (b). The longest time it could remain on the 
The Wave Nature of Particles 39-5 
table is the time to travel the full width of the table (1.75 m), so t = x/vx = (1.75 m)/(6.03 × 10–33 m/s) = 2.90 × 1032 
s = 9.20 × 1024 years 
Since the universe is about 14 × 109 years old, this time is about 
24
14
9
9.0 10 yr
6 10
14 10 yr
× ≈ ×
×
 times the age of the universe! Don’t hold your breath! 
EVALUATE: For household objects, the uncertainty principle places a negligible limitation on our ability to 
measure their speed. 
39.21. Heisenberg’s Uncertainty Principles tells us that .
2x
h
x p
π
Δ Δ ≥ We can treat the standard deviation as a direct 
measure of uncertainty. 10 25 35 34Here (1.2 10 m) (3.0 10 kg m s) 3.6 10 J s but 1.05 10 J s
2x
h
x p
π
− − − −Δ Δ = × × ⋅ = × ⋅ = × ⋅ 
Therefore so the claim is .
2x
h
x p not valid
π
Δ Δ < 
39.22. (a) ( ) ( ) 2 ,xx m v h πΔ Δ ≥ and setting (0.010)x xv vΔ = and the product of the uncertainties equal to / 2h π (for the 
minimum uncertainty) gives (2 (0.010) ) 57.9 m s.xv h πm x= Δ = 
(b) Repeating with the proton mass gives 31.6 mm s. 
39.23. 
34
32 13
3
(6.63 10 J s)
2.03 10 J 1.27 10 eV.
2 2 (5.2 10 s)
h
E
π t π
−
− −
−
× ⋅Δ > = = × = ×
Δ ×
 
39.24. IDENTIFY and SET UP: The Heisenberg Uncertainty Principle says 
2x
h
x p
π
Δ Δ ≥ . The minimum allowed 
xx pΔ Δ is / 2h π . x xp m vΔ = Δ . 
EXECUTE: (a) 
2x
h
m x v
π
Δ Δ = . 
34
4
27 12
6.63 10 J s
3.2 10 m/s
2 2 (1.67 10 kg)(2.0 10 m)x
h
v
m xπ π
−
− −
× ⋅Δ = = = ×
Δ × ×
 
(b) 
34
4
31
6.63 10 J s
4.6 10 m
2 2 (9.11 10 kg)(0.250 m/s)x
h
x
m vπ π
−
−
−
× ⋅Δ = = = ×
Δ ×
 
39.25. 
2
h
E t
π
Δ Δ = . 
34
14 4
21
(6.63 10 J s)
1.39 10 J 8.69 10 eV 0.0869 MeV.
2 2 (7.6 10 s)
h
E
π t π
−
−
−
× ⋅Δ = = = × = × =
Δ ×
 
2
5
2
0.0869 MeV
2.81 10 .
3097 MeV
E c
E c
−Δ = = × 
39.26. 2 9 2 10 2. . 2.06 10 eV 3.30 10 J .
2
h
E t E mc m c c
π
−Δ Δ = Δ = Δ Δ = × = × 
34
25
2 10
6.63 10 J s
3.20 10 s.
2 2 (3.30 10 J)
h
t
π mc π
−
−
−
× ⋅Δ = = = ×
Δ ×
 
39.27. IDENTIFY and SET UP: For a photon 
25
ph
1.99 10 J mhc
E
λ λ
−× ⋅= = . For an electron 
22 2
e 2
1
2 2 2
p h h
E
m m mλ λ
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
. 
EXECUTE: (a) photon 
25
17
ph 9
1.99 10 J m
1.99 10 J
10.0 10 m
E
−
−
−
× ⋅= = ×
×
 
electron 
34 2
21
e 31 9 2
(6.63 10 J s)
2.41 10 J
2(9.11 10 kg)(10.0 10 m)
E
−
−
− −
× ⋅= = ×
× ×
 
17
ph 3
21
e
1.99 10 J
8.26 10
2.41 10 J
E
E
−
−
×= = ×
×
 
(b) The electron has much less energy so would be less damaging. 
EVALUATE: For a particle with mass, such as an electron, 2~E λ− . For a massless photon 1~E λ− . 
39.28. (a) 
2 2 2( ) ( )
, so 419 V.
2 2 2
p h h
eV K V
m m me
λ λ= = = = = 
(b) The voltage is reduced by the ratio of the particle masses, 
31
27
9.11 10 kg
(419 V) 0.229 V.
1.67 10 kg
−
−
× =
×
 
39.29. IDENTIFY and SET UP: ( ) sin .x A kxψ = The position probability density is given by 2 2 2( ) sin .x A kxψ = 
EXECUTE: (a) The probability is highest where sin 1 so 2 / / 2, 1, 3, 5,kx kx x n nπ λ π= = = = … 
/ 4, 1, 3, 5, so / 4, 3 / 4, 5 /4,x n n xλ λ λ λ= = =… … 
39-6 Chapter 39 
(b) The probability of finding the particle is zero where 
2
0,ψ = which occurs where sin 0kx = and 
2 / , 0, 1, 2,kx x n nπ λ π= = = … 
/ 2, 0, 1, 2, so 0, / 2, , 3 / 2,x n n xλ λ λ λ= = =… … 
EVALUATE: The situation is analogous to a standing wave, with the probability analogous to the square of the 
amplitude of the standing wave. 
39.30. sin ,ψ ωt∗ ∗Ψ = so 2 2* * 2 2sin sinψ ψ ωt ψ ωtΨ = Ψ Ψ = = . 2Ψ is not time-independent, so Ψ is not the 
wavefunction for a stationary state. 
39.31. IDENTIFY: To describe a real situation, a wave function must be normalizable. 
SET UP: |ψ |2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere, 
ψ must have the property that ∫|ψ |2 dV = 1 when the integral is taken over all space. 
EXECUTE: (a) In one dimension, as we have here, the integral discussed above is of the form 2| ( ) | 1x dxψ
∞
−∞
=∫ . 
(b) Using the result from part (a), we have ( )
2
2 2
2
ax
ax ax ee dxe dx
a
∞
∞ ∞
−∞ −∞
−∞
= = = ∞∫ ∫ . Hence this wave function cannot 
be normalized and therefore cannot be a valid wave function. 
(c) We only need to integrate this wave function of 0 to ∞ because it is zero for x < 0. For normalization we have 
21 | | dxψ
∞
−∞
= ∫ = ( )
2 2 2
2 2 2
0 0
0
2 2
bx
bx bx A e AAe dx A e dx
b b
∞−∞ ∞− −= = =
−∫ ∫ , which gives 
2
1
2
A
b
= , so 2A b= . 
EVALUATE: If b were positive, the given wave function could not be normalized, so it would not be allowable. 
39.32. (a) The uncertainty in the particle position is proportional to the width of ( )ψ x , and is inversely proportional to 
α . This can be seen by either plotting the function for different values of α , finding the expectation 
value 2 2 2x ψ x dx= ∫ for the normalized wave function or by finding the full width at half-maximum. The 
particle’s uncertainty in position decreases with increasing α . The dependence of the expectation value 2x〈 〉 on α 
may be found by considering 
2
2
2 2
2
2
x
x
x e dx
x
e dx
α
α
∞
−
−∞
∞
−
−∞
〈 〉 =
∫
∫
 =
221 ln
2
xe dxα
α
∞
−
−∞
⎡ ⎤∂− ⎢ ⎥∂ ⎣ ⎦
∫
21 1 1
ln ,
2 42
ue du
α αα
∞
−
−∞
⎡ ⎤∂= − =⎢ ⎥∂ ⎣ ⎦
∫ 
where the substitution u xα= has been made. 
(b) Since the uncertainty in position decreases, the uncertainty in momentum must increase. 
39.33. *( , ) and ( , )
x iy x iy
f x y f x y
x iy x iy
⎛ ⎞ ⎛ ⎞− += =⎜ ⎟ ⎜ ⎟+ −⎝ ⎠ ⎝ ⎠
 
2 * 1.
x iy x iy
f f f
x iy x iy
⎛ ⎞ ⎛ ⎞− +
⇒ = = ⋅ =⎜ ⎟ ⎜ ⎟+ −⎝ ⎠ ⎝ ⎠
 
39.34. The same. 
2 *( , , ) ( , , ) ( , , )ψ x y z ψ x y z ψ x y z= 
2 *( , , ) ( ( , , ) )( ( , , ) )i i iψ x y z e ψ x y z e ψ x y z eφ φ φ− += *( , , ) ( , , ).ψ x y z ψ x y z= 
The complex conjugate means convert all i’s to–i’s and vice-versa. 1.i ie eφ φ−⋅ = 
39.35. IDENTIFY: To describe a real situation, a wave function must be normalizable. 
SET UP: |ψ |2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere, 
ψ must have the property that ∫|ψ |2 dV = 1 when the integral is taken over all space. 
EXECUTE: (a) For normalization of the one-dimensional wave function, we have 
21 | | dxψ
∞
−∞
= ∫ = ( ) ( )
0 02 2 2 2 2 2
0 0
bx bx bx bxAe dx Ae dx A e dx A e dx
∞ ∞− −
−∞ −∞
+ = +∫ ∫ ∫ ∫ . 
02 2 2
2
0
1
2 2
bx bxe e A
A
b b b
∞−
−∞
⎧ ⎫⎪ ⎪= + =⎨ ⎬−⎪ ⎪⎩ ⎭
, which gives 12.00 mA b −= = = 1.41 m–1/2 
(b) The graph of the wavefunction versus x is given in Figure 39.35. 
(c) (i) 
5.00 m 2
0.500 m
| |P dxψ
+
−
= ∫ = 
5.00 m 2 2
0
2 bxA e dx
+ −∫ , where we have used the fact that the wave function is an even 
function of x. Evaluating the integral gives 
P = ( ) ( )
2 1
2 (0.500 m) 2.00
1
(2.00 m )
1 1 0.865
2.00 m
bA e e
b
−
− −
−
− −− = − = 
There is a little more than an 86% probability that the particle will be found within 50 cm of the origin. 
The Wave Nature of Particles 39-7 
(ii) P = ( )
2 1
0 02 2 2
1
2.00 m 1
2 2(2.00 m ) 2
bx bx AAe dx A e dx
b
−
−−∞ −∞
= = = =∫ ∫ = 0.500 
There is a 50-50 chance that the particle will be found to the left of the origin, which agrees with the fact that the 
wave function is symmetric about the y-axis. 
(iii) P = 
1.00 m 2 2
0.500 m
bxA e dx−∫ 
= ( ) ( )-1 -1
2
2(2.00 m )(1.00 m) 2(2.00 m )(0.500 m) 4 21 0.0585
2 2
A
e e e e
b
− − − −− = − − =
−
 
EVALUATE: There is little chance of finding the particle in regions where the wave function is small. 
 
Figure 39.35 
39.36. Eq. (39.18): 
2 2
22
d ψ
Uψ Eψ
m dx
− + = . Let 1 2Aψ Bψψ = + 
2 2
1 2 1 2 1 22
( ) ( ) ( )
2
d
Aψ Bψ U Aψ Bψ E Aψ Bψ
m dx
−
⇒ + + + = +
2 2 2 2
1 2
1 1 2 22 2
0.
2 2
d ψ d ψ
A Uψ Eψ B Uψ Eψ
m dx m dx
⎛ ⎞ ⎛ ⎞
⇒ − + − + − + − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
But each of 1ψ and 2ψ satisfy Schrödinger’s equation separately so the equation still holds true, for any A or B. 
39.37. 
2 2
1 1 2 22
.
2
d ψ
Uψ BEψ CE ψ
m dx
− + = + If ψ were a solution with energy E, then 1 1 2 2 1 2BEψ CE ψ BEψ CEψ+ = + or 
1 1 2 2( ) ( ) .B E E ψ C E E ψ− = − This would mean that 1ψ is a constant multiple of 2 1 2, and andψ ψ ψ would be wave 
functions with the same energy. However, 1 2E E≠ , so this is not possible, and ψ cannot be a solution to Eq. (39.18). 
39.38. (a) 
34
31 19
(6.63 10 J s)
2 2(9.11 10 kg)(40 eV)(1.60 10 J eV)
h
mK
λ
−
− −
× ⋅= =
× ×
101.94 10 m.−= × 
(b) 
31 1 2
7
19
(2.5 m)(9.11 10 kg)
6.67 10 s.
2 2(40 eV)(1.6 10 J eV)
R R
v E m
−
−
−
×= = = ×
×
 
(c) The width
λ
is 2 ' and ,y yw w R w v t p t ma
= = Δ = Δ where t is the time found in part (b) and a is the slit width. 
Combining the expressions for 28
2
, 2.65 10 kg m s.y
m R
w p
at
λ −Δ = = × ⋅ 
(d) 0.40 m,
2 y
h
y μ
π p
Δ = =
Δ
 which is the same order of magnitude. 
39.39. (a) 12 eVE hc λ= = 
(b) Find E for an electron with 60.10 10 m.λ −= × 27so 6.626 10 kg m sh p p hλ λ −= = = × ⋅ . 
2 4(2 ) 1.5 10 eVE p m −= = × . 4so 1.5 10 VE q V V −= Δ Δ = × 
27 31 3(6.626 10 kg m s) (9.109 10 kg) 7.3 10 m sv p m − −= = × ⋅ × = × 
(c) Same λ so same p. 2 27 8/(2 ) but now 1.673 10 kg so 8.2 10 eV andE p m m E− −= = × = × 88.2 10 V.V −Δ = × 
27 27(6.626 10 kg m s) (1.673 10 kg) 4.0 m sv p m − −= = × ⋅ × = 
39.40. (a) Single slit diffraction: sina θ mλ= . 9 8sin (150 10 m)sin20 5.13 10 ma θλ − −= = × ° = × 
h mv v h mλ λ= → = . 
34
4
31 8
6.626 10 J s
1.42 10 m s
(9.11 10 kg)(5.13 10 m)
v
−
− −
× ⋅= = ×
× ×
 
(b) 2sin 2a θ λ= . 
8
2 9
5.13 10 m
sin 2 2 0.684
150 10 m
θ
a
λ −
−
⎛ ⎞×= ± = ± = ±⎜ ⎟×⎝ ⎠
. 2 43.2θ = ± ° 
39-8 Chapter 39 
39.41. IDENTIFY: The electrons behave like waves and produce a double-slit interference pattern after passing through 
the slits. 
SET UP: The first angle at which destructive interference occurs is given by d sin θ = λ/2. The de Broglie 
wavelength of each of the electrons is λ = h/mv. 
EXECUTE: (a) First find the wavelength of the electrons. For the first dark fringe, we have d sin θ = λ/2, which gives 
(1.25 nm)(sin 18.0°) = λ/2 , and λ = 0.7725 nm. Now solve the de Broglie wavelength equation for the speed of the 
electron: 
34
31 9
6.626 10 J s
(9.11 10 kg)(0.7725 10 m)
h
v
mλ
−
− −
× ⋅= =
× ×
 = 9.42 × 105 m/s 
which is about 0.3% the speed of light, so they are nonrelativistic. 
(b) Energy conservation gives eV = ½ mv2 and 
V = mv2/2e = (9.11 × 10–31 kg)(9.42 × 105 m)2/[2(1.60 × 10–19 C)] = 2.52 V 
EVALUATE: The hole must be much smaller than the wavelength of visible light for the electrons to show diffraction. 
39.42. IDENTIFY: The alpha particles and protons behave as waves and exhibit circular-aperture diffraction after passing 
through the hole. 
SET UP: For a round hole, the first dark ring occurs at the angle θ for which sinθ = 1.22λ /D, where D is the 
diameter of the hole. The de Broglie wavelength for a particle is λ = h/p = h/mv. 
EXECUTE: Taking the ratio of the sines for the alpha particle and proton gives 
p p p
sin 1.22
sin 1.22
α α αθ λ λ
θ λ λ
= = 
The de Broglie wavelength gives λp = h/pp and λα = h/pα, so p
p p
sin /
sin /
ph p
h p p
α α
α
θ
θ
= = . Using K = p2/2m, we have 
2p mK= . Since the alpha particle has twice the charge of the proton and both are accelerated through the same 
potential difference, Kα = 2Kp. Therefore p p p2p m K= and p p2 2 (2 ) 4p m K m K m Kα α α α α= = = . 
Substituting these quantities into the ratio of the sines gives 
p pp p
p p
2sin
sin 24
m Kp m
p mm K
α
α αα
θ
θ
= = = 
Solving for sin θα gives 
27
27
1.67 10 kg
sin sin1 5.0
2(6.64 10 kg)α
θ
−
−
×= °
×
 and θα = 5.3°. 
EVALUATE: Since sin θ is inversely proportional to the mass of the particle, the larger-mass alpha particles form 
their first dark ring at a smaller angle than the ring for the lighter protons. 
39.43. IDENTIFY: Both the electrons and photons behave like waves and exhibit single-slit diffraction after passing 
through their respective slits. 
SET UP: The energy of the photon is E = hc/λ and the de Broglie wavelength of the electron is λ = h/mv= h/p. 
Destructive interference for a single slit first occurs when a sin θ = λ. 
EXECUTE: (a) For the photon: λ = hc/E and a sinθ = λ. Since the a and θ are the same for the photons and 
electrons, they must both have the same wavelength. Equating these two expressions for λ gives a sin θ = hc/E. 
For the electron, λ = h/p = 
2
h
mK
 and a sin θ = λ. Equating these two expressions for λ gives a sin θ = 
2
h
mK
. 
Equating the two expressions for asinθ gives hc/E = 
2
h
mK
, which gives 7 1/22 (4.05 10 J )E c mK K−= = × 
(b) 
22 2E c mK mc
K K K
= = . Since v << c, mc2 > K, so the square root is >1. Therefore E/K > 1, meaning that the 
photon has more energy than the electron. 
EVALUATE: As we have seen in Problem 39.10, when a photon and a particle have the same wavelength, the 
photon has more energy than the particle. 
39.44. According to Eq.(35.4) 
6sin (40.0 10 m)sin(0.0300 rad)
600 nm.
2
d θ
m
λ
−×= = = The velocity of an electron with 
this wavelength is given by Eq.(39.1) 
34
3
31 9
(6.63 10 J s)
1.21 10 m s.
λ (9.11 10 kg)(600 10 m)
p h
v
m m
−
− −
× ⋅= = = = ×
× ×
 
The Wave Nature of Particles 39-9 
Since this velocity is much smaller than c we can calculate the energy of the electron classically 
2 31 3 2 251 1 (9.11 10 kg)(1.21 10 m s) 6.70 10 J 4.19 eV.
2 2
K mv μ− −= = × × = × = 
39.45. The de Broglie wavelength of the blood cell is 
34
17
14 3
(6.63 10 J s)
λ 1.66 10 m.
(1.00 10 kg)(4.00 10 m s)
h
mv
−
−
− −
× ⋅= = = ×
× ×
 
We need not be concerned about wave behavior. 
39.46. (a) 
1 22
21
v
h
ch
p mv
λ
⎛ ⎞
−⎜ ⎟
⎝ ⎠= =
2 2 2
2 2 2 2 2
2 2
1
v h v
m v h h
c c
λ ⎛ ⎞⇒ = − = −⎜ ⎟
⎝ ⎠
 
2
2 2 2 2 2
2
v
m v h h
c
λ⇒ + = 
2 2
2
2 2 2 2
2 2
2 2
1
h c
v
h m c
m
c h
λλ
⇒ = =
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
 1 22
.
1
c
v
mc
h
λ
⇒ =
⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
(b) 
2
1 22
1
1 (1 ) .
2
1
( )
c mc
v c c
h
h mc
λ
λ
⎛ ⎞⎛ ⎞= ≈ − = − Δ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞ ⎝ ⎠⎛ ⎞
⎜ ⎟+ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
2 2 2
2
.
2
m c
h
λΔ = 
(c) 151.00 10 m .
h
mc
λ −= × << 
31 2 8 2 15 2
8
34 2
(9.11 10 kg) (3.00 10 m s) (1.00 10 m)
8.50 10
2(6.63 10 J s)
− −
−
−
× × ×Δ = = ×
× ⋅
 
8(1 Δ) (1 8.50 10 ) .v c c−⇒ = − = − × 
39.47. (a) Recall .
2 2
h h h
p mE mq V
λ = = =
Δ
So for an electron: 
34
10
31 19
6.63 10 J s
λ 1.10 10 m.
2(9.11 10 kg)(1.60 10 C)(125 V)
λ
−
−
− −
× ⋅= ⇒ = ×
× ×
 
(b) For an alpha particle: 
34
13
27 19
6.63 10 J s
9.10 10 m.
2(6.64 10 kg)2(1.60 10 C)(125 V)
λ
−
−
− −
× ⋅= = ×
× ×
 
39.48. IDENTIFY and SET UP: The minimum uncertainty product is 
2x
h
x p
π
Δ Δ = . 1x rΔ = , where 1r is the radius of the 
1n = Bohr orbit. In the 1n = Bohr orbit, 1 1 2
h
mv r
π
= and 1 1
12
h
p mv
rπ
= = . 
EXECUTE: 
34
24
10
1
6.63 10 J s
2.0 10 kg m/s
2 2 2 (0.529 10 m)x
h h
p
x rπ π π
−
−
−
× ⋅Δ = = = = × ⋅
Δ ×
. This is the same as the magnitude of 
the momentum of the electron in the 1n = Bohr orbit. 
EVALUATE: Since the momentum is the same order of magnitude as the uncertainty in the momentum, the 
uncertainty principle plays a large role in the structure of atoms. 
39.49. IDENTIFY and SET UP: Combining the two equations in the hint gives 2( 2 )PC K K mc= + and 
2
.
( 2 )
hc
K K mc
λ =
+
 
EXECUTE: (a) With 23K mc= this becomes 
2 2 2
.
153 (3 2 )
hc h
mcmc mc mc
λ = =
+
 
(b) (i) 2 31 8 2 133 3(9.109 10 kg)(2.998 10 m/s) 2.456 10 J 1.53 MeVK mc − −= = × × = × = 
34
13
31 8
6.626 10 J s
6.26 10 m
15 15(9.109 10 kg)(2.998 10 m/s)
h
mc
λ
−
−
−
× ⋅= = = ×
× ×
 
(ii) K is proportional to m, so for a proton p e( / )(1.53 MeV) 1836(1.53 MeV) 2810 MeVK m m= = = 
λ is proportional to 1/m, so for a proton 13 13 16e p( / )(6.26 10 m) (1/1836)(6.626 10 m) 3.41 10 mm mλ
− − −= × = × = × 
EVALUATE: The proton has a larger rest mass energy so its kinetic energy is larger when 23 .K mc= The proton 
also has larger momentum so has a smaller .λ 
39-10 Chapter 39 
39.50. (a) 
34
20
15
(6.626 10 J s)
2.1 10 kg m s.
2 (5.0 10 m)π
−
−
−
× ⋅ = × ⋅
×
 
(b) 2 2 2 2 13( ) ( ) 1.3 10 J 0.82 MeV.K pc mc mc −= + − = × = 
(c) The result of part (b), about 61 MeV 1 10 eV= × , is many orders of magnitude larger than the potential energy 
of an electron in a hydrogen atom. 
39.51. (a) IDENTIFY and SET UP: / 2xx p h πΔ Δ ≥ 
Estimate 15 as 5.0 10 m.x x −Δ Δ ≈ × 
EXECUTE: Then the minimum allowed xpΔ is 
34
20
15
6.626 10 J s
2.1 10 kg m/s
2 2 (5.0 10 m)x
h
p
xπ π
−
−
−
× ⋅Δ ≈ = = × ⋅
Δ ×
 
(b) IDENTIFY and SET UP: Assume 202.1 10 kg m/s.p −≈ × ⋅ Use Eq.(37.39) to calculate E, and then 2.K E mc= − 
EXECUTE: 2 2 2( ) ( )E mc pc= + 
2 31 8 2 14(9.109 10 kg)(2.998 10 m/s) 8.187 10 Jmc − −= × × = × 
20 8 12(2.1 10 kg m/s)(2.998 10 m/s) 6.296 10 Jpc − −= × ⋅ × = × 
14 2 12 2 12(8.187 10 J) (6.296 10 J) 6.297 10 JE − − −= × + × = × 
2 12 14 12 196.297 10 J 8.187 10 J 6.215 10 J(1 eV/1.602 10 J) 39 MeVK E mc − − − −= − = × − × = × × = 
(c) IDENTIFY and SET UP: The Coulomb potential energy for a pair of point charges is given by Eq.(23.9). The 
proton has charge +e and the electron has charge –e. 
EXECUTE: 
2 9 2 2 19 2
14
15
(8.988 10 N m / C )(1.602 10 C)
4.6 10 J 0.29 MeV
5.0 10 m
ke
U
r
−
−
−
× ⋅ ×= − = − = − × = −
×
 
EVALUATE: The kinetic energy of the electron required by the uncertainty principle would be much larger than 
the magnitude of the negative Coulomb potential energy. The total energy of the electron would be large and 
positive and the electron could not be bound within the nucleus. 
39.52. (a) Take the direction of the electron beam to be the x-direction and the direction of motion perpendicular to the 
beam to be the y-direction. 
34
31 3
6.626 10 J s
0.23 m/s
2 2 (9.11 10 kg)(0.50 10 m)
y
y
p h
v
m m yπ π
−
− −
Δ × ⋅Δ = = = =
Δ × ×
 
(b) The uncertainty rΔ in the position of the point where the electrons strike the screen is 
109.56 10 m,
2 2
y
y
x
p x h x
r v t
m v πm y K m
−ΔΔ = Δ = = = ×
Δ
 
(c) This is far too small to affect the clarity of the picture. 
39.53. IDENTIFY and SET UP: 
2
h
E t
π
Δ Δ ≥ . Take the minimum uncertainty product, so 
2
h
E
tπ
Δ =
Δ
, with 
178.4 10 st −Δ = × . e264m m= . 2
E
m
c
ΔΔ = . 
EXECUTE: 
34
18
17
6.63 10 J s
1.26 10 J
2 (8.4 10 s)
E
π
−
−
−
× ⋅Δ = = ×
×
. 
18
35
8 2
1.26 10 J
1.4 10 kg
(3.00 10 m/s)
m
−
−×Δ = = ×
×
. 
35
8
31
1.4 10 kg
5.8 10
(264)(9.11 10 kg)
m
m
−
−
−
Δ ×= = ×
×
 
39.54. IDENTIFY: The insect behaves like a wave as it passes through the hole in the screen. 
SET UP: (a) For wave behavior to show up, the wavelength of the insect must be of the order of the diameter of 
the hole. The de Broglie wavelength is λ = h/mv. 
EXECUTE: The de Broglie wavelength of the insect must be of the order of the diameter of the hole in the screen, 
so λ ≈ 5.00 mm. The de Broglie wavelength gives 
( )( )
34
6
6.626 10 J s
1.25 10 kg 0.00400 m
h
v
mλ
−
−
× ⋅= =
×
 = 1.33 × 10–25 m/s 
(b) t = x/v = (0.000500 m)/(1.33 × 10–25 m/s) = 3.77 × 1021 s = 1.4 × 1010 yr 
The universe is about 14 billion years old (1.4 × 1010 yr), so this time would be about 85,000 times the age of the 
universe. 
EVALUATE: Don’t expect to see a diffracting insect! Wave behavior of particles occurs only at the very small scale. 
39.55. IDENTIFY and SET UP: Use Eq.(39.1) to relate your wavelength and speed. 
EXECUTE: (a) 
34
356.626 10 J s, so 1.1 10 m/s
(60.0 kg)(1.0 m)
h h
v
mv m
λ
λ
−
−× ⋅= = = = × 
The Wave Nature of Particles 39-11 
(b) 34 7 27
35
distance 0.80 m
7.3 10 s(1 y/3.156 10 s) 2.3 10 y
velocity 1.1 10 m/s
t −= = = × × = ××
 
Since you walk through doorways much more quickly than this, you will not experience diffraction effects. 
EVALUATE: A 1 kg object moving at 1 m/s has a de Broglie wavelength 346.6 10 m,λ −= × which is exceedingly 
small. An object like you has a very, very small λ at ordinary speeds and does not exhibit wavelikeproperties. 
39.56. (a) 19 72.58 eV 4.13 10 J, with a wavelength of 4.82 10 m 482 nm
hc
E
E
λ− −= = × = = × = 
(b) 
34
28 9
7
(6.63 10 J s)
6.43 10 J 4.02 10 eV.
2 2 (1.64 10 s)
h
E
π t π
−
− −
−
× ⋅Δ = = = × = ×
Δ ×
 
(c) , so ( ) 0, andE hc E E E Eλ λ λ λ λ= Δ + Δ = Δ = Δ , so 
28
7 16 7
19
6.43 10 J
(4.82 10 m) 7.50 10 m 7.50 10 nm.
4.13 10 J
E Eλ λ
−
− − −
−
⎛ ⎞×Δ = Δ = × = × = ×⎜ ⎟×⎝ ⎠
 
39.57. IDENTIFY: The electrons behave as waves whose wavelength is equal to the de Broglie wavelength. 
SET UP: The de Broglie wavelength is λ = h/mv, and the energy of a photon is E = hf = hc/λ. 
EXECUTE: (a) Use the de Broglie wavelength to find the speed of the electron. 
( )( )
34
31 9
6.626 10 J s
9.11 10 kg 1.00 10 m
h
v
mλ
−
− −
× ⋅= =
× ×
 = 7.27 × 105 m/s 
which is much less than the speed of light, so it is nonrelativistic. 
(b) Energy conservation gives eV = ½ mv2. 
V = mv2/2e = (9.11 × 10–31 kg)(7.27 × 105 m/s)2/[2(1.60 × 10–19 C)] = 1.51 V 
(c) K = eV = e(1.51 V) = 1.51 eV, which is about ¼ the potential energy of the NaCl crystal, so the electron would 
not be too damaging. 
(d) E = hc/λ = (4.136 × 10–15 eV s)(3.00 × 108 m/s)/(1.00 × 10–9 m) = 1240 eV 
which would certainly destroy the molecules under study. 
EVALUATE: As we have seen in Problems 39.10 and 39.43, when a particle and a photon have the same 
wavelength, the photon has much more energy. 
39.58. sin sin , and ( ) ( 2 ), and soθ θ h p h mEλ λ
λ
′′ ′ ′ ′= = = arcsin sin
2
hθ θ
mEλ
⎛ ⎞′ = ⎜ ⎟′⎝ ⎠
. 
34
11 31 3 19
(6.63 10 J s)sin 35.8
arcsin 20.9
(3.00 10 m) 2(9.11 10 kg)(4.50 10 )(1.60 10 J eV)
θ
−
− − + −
⎛ ⎞× ⋅ °′ ⎜ ⎟= = °
⎜ ⎟× × × ×⎝ ⎠
 
39.59. (a) The maxima occur when 2 sind θ mλ= as described in Section 38.7. 
(b) 
2
h h
p mE
λ = = . 
( )
34
10
37 19
(6.63 10 J s)
1.46 10 m 0.146 nm
2(9.11 10 kg)(71.0 eV) 1.60 10 J/eV
λ
−
−
− −
× ⋅= = × =
× ×
. 
1sin (Note: This
2
mθ m
d
λ− ⎛ ⎞= ⎜ ⎟
⎝ ⎠
 is the order of the maximum, not the mass.) 
10
1
11
(1)(1.46 10 m)
sin 53.3 .
2(9.10 10 m)
−
−
−
⎛ ⎞×
⇒ = °⎜ ⎟×⎝ ⎠
 
(c) The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming 
electrons by .eφ An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength. A 
smaller wavelength gives a smaller angle θ (see part (b)). 
39.60. (a) Using the given approximation, ( )2 21 ( ) , ( )
2
E h x m kx dE dx kx= + = − 2 3( ),h mx and the minimum energy 
occurs when 2 3 2( ), or .
h
kx h mx x
mk
= = The minimum energy is then .h k m 
(b) They are the same. 
39.61. (a) IDENTIFY and SET UP: .U A x= Eq.(7.17) relates force and potential. The slope of the function A x is not 
continuous at 0x = so we must consider the regions x > 0 and x < 0 separately. 
EXECUTE: For 
( )
0, so and .
d Ax
x x x U Ax F A
dx
> = = = − = − For 0, so andx x x U Ax< = − = − 
( )
 .
d Ax
F A
dx
−= − = + We can write this result as / ,F A x x= − valid for all x except for x = 0. 
39-12 Chapter 39 
(b) IDENTIFY and SET UP: Use the uncertainty principle, expressed as ,p x hΔ Δ ≈ and as in Problem 39.50 
estimate pΔ by p and xΔ by x. Use this to write the energy E of the particle as a function of x. Find the value of x 
that gives the minimum E and then find the minimum E. 
EXECUTE: 
2
2
p
E K U A x
m
= + = + 
, so /px h p h x≈ ≈ 
Then 
2
2
.
2
h
E A x
mx
≈ + 
For 
2
2
0, .
2
h
x E Ax
mx
> = + 
To find the value of x that gives minimum E set 0.
dE
dx
= 
2
3
2
0
2
h
A
mx
−= + 
1/ 32 2
3 and 
h h
x x
mA mA
⎛ ⎞
= = ⎜ ⎟
⎝ ⎠
 
With this x the minimum E is 
1/ 32 / 32 2
2 / 3 1/ 3 2 / 3 2 / 3 1/ 3 2 / 3
2
1
2 2
h mA h
E A h m A h m A
m h mA
− −⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
 
1/ 32 2
3
2
h A
E
m
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
 
EVALUATE: The potential well is shaped like a V. The larger A is the steeper the slope of U and the smaller the 
region to which the particle is confined and the greater is its energy. Note that for the x that minimizes E, 2K = U. 
39.62. For this wave function, 1 21 2 , so
iω t iω tψ e ψ e∗ ∗ ∗Ψ = + 
1 2 1 2 1 2 2 1( ) ( )2
1 2 1 2 1 1 2 2 1 2 2 1( )( )
iω t iω t iω t iω t i ω ω t i ω ω t*ψ e ψ e ψ e ψ e ψ ψ ψ ψ ψ ψ e ψ ψ e .− − − −∗ ∗ ∗ ∗ ∗ ∗Ψ = Ψ Ψ = + + = + + + 
The frequencies 1 2andω ω are given as not being the same, so 
2Ψ is not time-independent, and Ψ is not the 
wave function for a stationary state. 
39.63. The time-dependent equation, with the separated form for ( , )x tΨ as given becomes 
2 2
2
( ) ( ) .
2
d ψ
i ψ iω U x ψ
m dx
⎛ ⎞
− = − +⎜ ⎟
⎝ ⎠
 
Since ψ is a solution of the time-independent solution with energy ,E the term in parenthesis is ,Eψ and so 
, and ( ).ω E ω E= = 
39.64. (a) 
2
2
π E Eω π f .
h
= = = 2 2π π pk p .
hλ
= = = 
2 2 2( )
2 2 2
p k kω E K ω .
m m m
= = = = ⇒ = 
(b) From Problem 39.63 the time-dependent Schrödinger’s equation is 
2 2
2
( , )
2
ψ x t
m x
∂− +
∂
 
( , )
( ) ( , ) . ( ) 0 for a free particle, so
ψ x t
U x ψ x t i U x
t
∂= =
∂
2
2
( , ) 2 ( , )
.
ψ x t mi ψ x t
x t
∂ ∂= −
∂ ∂
 
Try ( , ) cos( )ψ x t kx ωt= − : 
 
2
2
2
( , ) sin( )
( , )
sin( ) and cos( ).
ψ
x t Aω kx ωt
t
ψ x t ψ
Ak kx ωt Ak kx ωt
x x
∂ = −
∂
∂ ∂= − − = −
∂ ∂
 
Putting this into the Schrödinger’s equation, 2
2
cos( ) sin( ).
mi
Ak kx ωt Aω kx ωt⎛ ⎞− = − −⎜ ⎟
⎝ ⎠
 
This is not generally true for all andx t so is not a solution. 
The Wave Nature of Particles 39-13 
(c) Try ( , ) sin( )ψ x t A kx ωt= − : 
2
2
2
( , )
cos( )
( , ) ( , )
cos( ) and sin( ).
ψ x t
Aω kx ωt
t
ψ x t ψ x t
Ak kx ωt Ak kx ωt
x x
∂ = − −
∂
∂ ∂= − = − −
∂ ∂
 
Again, 2
2
sin( ) cos( ) is not generally true for
mi
Ak kx ωt Aω kx ωt all⎛ ⎞− − = − −⎜ ⎟
⎝ ⎠
andx t so is not a solution. 
(d) Try ( , ) cos( ) sin( )ψ x t A kx ωt B kx ωt= − + − : 
2
2 2
2
( , )
sin( ) cos( )
( , ) ( , )
sin( ) cos( ) and cos(( ) sin( ).
ψ x t
Aω kx ωt Bω kx ωt
t
ψ x t ψ x t
Ak kx ωt Bk kx ωt Ak kx ωt Bk kx ωt
x x
∂ = + − − −
∂
∂ ∂= − − + − = − − − −
∂ ∂
 
Putting this into the Schrödinger’s equation, 
2 2 2cos( ) sin( ) ( sin( ) cos( )).
mi
Ak kx ωt Bk kx ωt Aω kx ωt Bω kx ωt− − − − = − + − − − 
Recall that 
2
.
2
kω
m
= Collect sin and cos terms. 
2 2( ) cos( ) ( ) sin (A iB k kx ωt iA B k kx+ − + − − ) 0.ωt = This is only true if B = iA. 
39.65. (a) IDENTIFY and SET UP: Let the y-direction be from the thrower to the catcher, and let the x-direction be 
horizontal and perpendicular to the y-direction. A cube with volume 3 3 3125 cm 0.125 10 mV −= = × has side length 
1/ 3 3 3 1/ 3(0.125 10 m ) 0.050 m.l V −= = × = Thus estimate as 0.050 m.x xΔ Δ ≈ Use the uncertainty principle to 
estimate .xpΔ 
EXECUTE: / 2xx p h πΔ Δ ≥ then gives 
0.0663 J s
0.21 kg m/s
2 2 (0.050 m)x
h
p
xπ π
⋅Δ ≈ = = ⋅
Δ
 
(The value of h in this other universe has been used.) 
(b) IDENTIFY and SET UP: ( )xx v tΔ = Δ is the uncertainty in the x-coordinate of the ball when it reaches the 
catcher, where t is the time it takes the ball to reach the second student. Obtain xvΔ from .xpΔ 
EXECUTE: The uncertainty in the ball’s horizontal velocity is 
0.21 kg m/s
0.84 m/s
0.25 kg
x
x
p
v
m
Δ ⋅Δ = = = 
The time it takes the ball to travel to the second student is 
12 m
2.0 s.
6.0 m/s
t = = The uncertainty in the x-coordinate 
of the ball when it reaches the second student that is introduced by is ( ) (0.84 m/s)(2.0 s) 1.7 m.x xv x v tΔ Δ = Δ = = 
The ball could miss the second student by about 1.7 m. 
EVALUATE: A game of catch would be very different in this universe. We don’t notice the effects of the 
uncertainty principle in everyday life because h is so small. 
39.66. (a) 
2 2 22 2 2 2( ).αx βy γzψ A x e− + += To save some algebra, let 2,u x= so that 2ψ = 2 ( , )uue f y zα− . 
2 2
0 0
1 1
(1 2 ) ; the maximum occurs at , .
2 2
ψ u ψ u x
u
α
α α
∂ = − = = ±
∂
 
(b) ψ vanishes at 0,x = so the probability of finding the particle in the 0x = plane is zero. The wave