Ch31 ISM

Prévia do material em texto

Chapter 31 
Properties of Light 
 
Conceptual Problems 
 
1 • [SSM] A ray of light reflects from a plane mirror. The angle 
between the incoming ray and the reflected ray is 70°. What is the angle of 
reflection? (a) 70°, (b) 140°, (c) 35°, (d) Not enough information is given to 
determine the reflection angle. 
 
Determine the Concept Because the angles of incidence and reflection are equal, 
their sum is 70° and the angle of reflection is 35°. ( )c is correct. 
 
2 • A ray of light passes in air is incident on the surface of a piece of 
glass. The angle between the normal to the surface and the incident ray is 40°, and 
the angle between the normal and the refracted ray 28°. What is the angle between 
the incident ray and the refracted ray? (a) 12°, (b) 28°, (c) 40° ,(d) 68° 
 
Determine the Concept The angle between the incident ray and the refracted ray 
is the difference between the angle of incidence and the angle of refraction. 
( )a is correct. 
 
3 • During a physics experiment, you are measuring refractive indices of 
different transparent materials using a red helium-neon laser beam. For a given 
angle of incidence, the beam has an angle of refraction equal to 28° in material A 
and an angle of refraction equal to 26° in material B. Which material has the 
larger index of refraction? (a) A, (b) B, (c) The indices of refraction are the same. 
(d) You cannot determine the relative magnitudes of the indices of refraction from 
the data given. 
 
Determine the Concept The refractive index is a measure of the extent to which 
a material refracts light that passes through it. Because the angles of incidence are 
the same for materials A and B and the angle of refraction is smaller (more 
bending of the light) for material B, its index of refraction is larger than that of A. 
( )b is correct. 
 
4 • A ray of light passes from air into water, striking the surface of the 
water at an angle of incidence of 45º. Which, if any, of the following four 
quantities change as the light enters the water: (a) wavelength, (b) frequency, 
(c) speed of propagation, (d) direction of propagation, (e) none of the above? 
 
 
 
 
2889 
Chapter 31 
 
2890 
 
Determine the Concept When light passes from air into water its wavelength 
changes ( waterairwater nλλ = ), its speed changes ( waterwater ncv = ), and the 
direction of its propagation changes in accordance with Snell’s law. Its frequency 
does not change, so ( )a , ( )c and ( )d change. 
 
5 • Earth’s atmosphere decreases in density as the altitude increases. As a 
consequence, the index of refraction of the atmosphere also decreases as altitude 
increases. Explain how one can see the Sun when it is below the horizon. (The 
horizon is the extension of a plane that is tangent to Earth’s surface.) Why does 
the setting Sun appear flattened? 
 
Determine the Concept The decrease in the index of refraction n of the 
atmosphere with altitude results in refraction of the light from the Sun, bending it 
toward the normal to the surface of Earth. Consequently, the Sun can be seen even 
after it is just below the horizon. 
Earth
Atmosphere
 
 
6 • A physics student playing pocket billiards wants to strike her cue ball 
so that it hits a cushion and then hits the eight ball squarely. She chooses several 
points on the cushion and then measures the distances from each point to the cue 
ball and to the eight ball. She aims at the point for which the sum of these 
distances is least. (a) Will her cue ball hit the eight ball? (b) How is her method 
related to Fermat’s principle? Neglect any effects due to ball rotation. 
 
Determine the Concept (a) Yes. (b) Her procedure is based on Fermat’s 
principle of least time. The ball presumably bounces off the cushion with an angle 
of reflection equal to the angle of incidence, just as a light ray would do if the 
cushion were a mirror. The least time would also be the shortest distance of travel 
for the light ray. 
 
7 • [SSM] A swimmer at point S in Figure 31-53 develops a leg cramp 
while swimming near the shore of a calm lake and calls for help. A lifeguard at 
point L hears the call. The lifeguard can run 9.0 m/s and swim 3.0 m/s. She knows 
physics and chooses a path that will take the least time to reach the swimmer. 
Which of the paths shown in the figure does the lifeguard take? 
 
Determine the Concept The path through point D is the path of least time. In 
analogy to the refraction of light, the ratio of the sine of the angle of incidence to 
the sine of the angle of refraction equals the ratio of the speeds of the lifeguard in 
Properties of Light 
 
 
2891
each medium. Careful measurements from the figure show that path LDS is the 
path that best satisfies this criterion. 
 
8 • Material A has a higher index of refraction than material B. Which 
material has the larger critical angle for total internal reflection when the material 
is in air? (a) A, (b) B, (c) The angles are the same. (d) You cannot compare the 
angles based on the data given. 
 
Determine the Concept Because the product of the index of refraction on the 
incident side of the interface and the sine of the critical angle is equal to one (the 
index of refraction of air), the material with the smaller index of refraction will 
have the larger critical angle. ( )b is correct. 
 
9 • [SSM] A human eye perceives color using a structure which is 
called a cone that is is located on the retina. Three types of molecules compose 
these cones and each type of molecule absorbs either red, green, or blue light by 
resonance absorption. Use this fact to explain why the color of an object that 
appears blue in air appears blue underwater, in spite of the fact that the 
wavelength of the light is shortened in accordance with Equation 31-6. 
 
Determine the Concept In resonance absorption, the molecules respond to the 
frequency of the light through the Einstein photon relation E = hf. Neither the 
wavelength nor the frequency of the light within the eyeball depend on the index 
of refraction of the medium outside the eyeball. Thus, the color appears to be the 
same in spite of the fact that the wavelength has changed. 
 
10 • Let θ be the angle between the transmission axes of two polarizing 
sheets. Unpolarized light of intensity I is incident upon the first sheet. What is the 
intensity of the light transmitted through both sheets? (a) I cos2 θ, 
(b) (I cos2 θ)/2, (c) (I cos2 θ)/4, (d) I cos θ, (e) (I cos θ)/4, (f) None of the above 
 
Picture the Problem The intensity of the light transmitted by the second 
polarizer is given by where ,II θ20trans cos= .210 II = Therefore, θ221trans cosII = 
and )(b is correct. 
 
11 •• [SSM] Draw a diagram to explain how Polaroid sunglasses reduce 
glare from sunlight reflected from a smooth horizontal surface, such as the surface 
found on a pool of water. Your diagram should clearly indicate the direction of 
polarization of the light as it propagates from the Sun to the reflecting surface and 
then through the sunglasses into the eye. 
 
Determine the Concept The following diagram shows unpolarized light from the 
sun incident on the smooth surface at the polarizing angle for that particular 
Chapter 31 
 
2892 
 
surface. The reflected light is polarized perpendicular to the plane of incidence, 
i.e., in the horizontal direction. The sunglasses are shown in the correct 
orientation to pass vertically polarized light and block the reflected sunlight. 
Smooth surface 
Light from
the sun
Polaroid
sunglasses
θ θ
θr
PP
 
 
12 • Use the photon model of light to explain why, biologically, it is far 
less dangerous to stand in front of an intense beam of red light than a very weak 
beam of gamma radiation. HINT: Ionization of molecules in tissue can cause 
biological damage. Molecules absorb light in what form, wave or particle? 
 
Determine the Concept Molecules require that a certain minimum energy be 
absorbed before they ionize. The red light photons contain considerablyless 
energy than the gamma photons so, even though there are likely to be fewer 
photons in the gamma beam, each one is potentially dangerous. 
 
13 • Three energy states of an atom are A, B and C. State B is 2.0 eV above 
state A and state C is 3.00 eV above state B. Which atomic transition results in the 
emission of the shortest wavelength of light? (a) B → A, (b) C → B, (c) C → A, 
(d) A→ C 
 
Determine the Concept Because the wavelength of the light emitted in an atomic 
transition is inversely proportional to the energy difference between the energy 
levels, the highest energy difference produces the shortest wavelength light. 
( )c is correct. 
 
14 • In Problem 13, if the atom is initially in state A, which transition 
results in the emission of the longest wavelength light? (a) A → B, (b) B → C, (c) 
A → C, (d) B→ A 
 
 
 
 
 
Properties of Light 
 
 
2893
Determine the Concept Because the energy required to induce an atomic 
transition varies inversely with the wavelength of the light that must be absorbed 
to induce the transition and the transition from A to B is the lowest energy 
transition, the transition B→ A results in the longest wavelength light. ( )d is 
correct. 
 
15 • [SSM] What role does the helium play in a helium–neon laser? 
 
Determine the Concept The population inversion between the state E2,Ne and the 
state 1.96 eV below it (see Figure 31-51) is achieved by inelastic collisions 
between neon atoms and helium atoms excited to the state E2,He. 
 
16 • When a beam of visible white light that passes through a gas of atomic 
hydrogen at room temperature is viewed with a spectroscope, dark lines are 
observed at the wavelengths of the hydrogen atom emission series. The atoms that 
participate in the resonance absorption then emit light of the same wavelength as 
they return to the ground state. Explain why the observed spectrum nevertheless 
exhibits pronounced dark lines. 
 
Determine the Concept Although the excited atoms emit light of the same 
frequency on returning to the ground state, the light is emitted in a random 
direction, not exclusively in the direction of the incident beam. Consequently, the 
beam intensity is greatly diminished at this frequency. 
 
17 • [SSM] Which of the following types of light would have the highest 
energy photons? (a) red (b) infrared (c) blue (d) ultraviolet 
 
Determine the Concept The energy of a photon is directly proportional to the 
frequency of the light and inversely proportional to its wavelength. Of the 
portions of the electromagnetic spectrum include in the list of answers, ultraviolet 
light has the highest frequency. ( )d is correct. 
 
Estimation and Approximation 
 
18 • Estimate the time required for light to make the round trip during 
Galileo’s experiment to measure the speed of light. Compare the time of the round 
trip to typical human response times. How accurate do you think this experiment 
is? 
 
Picture the Problem We can use the distance, rate, and time relationship to 
estimate the time required to travel 6 km. 
 
 
Chapter 31 
 
2894 
 
Express the distance d the light in 
Galileo’s experiment traveled in 
terms of its speed c and the 
elapsed time Δt: 
 
tcd Δ= ⇒
c
dt =Δ 
Substitute numerical values and 
evaluate Δt: s 102m/s10998.2
km6
Δ 58
−×=×=t 
Because human reaction is 
approximately 0.3 s: 
4
5
reaction 102
s102
s 3.0
Δ
Δ ×≈×= −t
t 
or ( ) tt Δ102Δ 4reaction ×≈ 
 
Because human reaction time is so much longer than the travel time for the light, 
there was no way that Galileo’s experiment could demonstrate that the speed of 
light was not infinite. 
 
19 • Estimate the time delay in receiving a light on your retina when you 
are wearing eyeglasses compared to when you are not wearing your eyeglasses. 
 
Determine the Concept We’ll assume that the source of the photon is a distance 
L from your retina and express the difference in the photon’s travel time when 
you are wearing your glasses. Let the thickness of your glasses by 2 mm and the 
index of refraction of the material from which they are constructed by 1.5. 
 
When you are not wearing your 
glasses, the time required for a light 
photon, originating a distance L 
away, to reach your retina is given 
by: 
 
c
Lt =0Δ 
If glass of thickness d and index of 
refraction n is inserted in the path of 
the photon, its travel time becomes: 
 ( ) ( )
c
dnt
c
dnL
nc
d
c
dLttt
1
Δ
1
Δ
0
glassin airin 
−+=−+=
+−=+=
 
 
The time delay is the difference 
between Δt and Δt0: 
 
( )
c
dnttt 1ΔΔ 0delay
−=−= 
Substitute numerical values and 
evaluate tdelay: 
( )( ) ps 3
m/s10998.2
mm 215.1
8delay ≈×
−=t 
 
20 •• Estimate the number of photons that enter your eye if you look for a 
tenth of a second at the Sun. What energy is absorbed by your eye during that 
Properties of Light 
 
 
2895
time, assuming that all the photons are absorbed? The total power output of the 
Sun is 4.2 × 1026 W. 
 
Picture the Problem The rate at which photons enter your eye is the ratio of 
power incident on your pupil to the energy per photon. We’ll assume that the 
electromagnetic radiation from the Sun is at 550 nm and that, therefore, its 
photons have energy (given by λhchfE == ) of 2.25 eV. 
 
The rate at which photons enter your 
eye is the ratio of the rate at which 
energy is incident on your pupil to 
the energy carried by each photon: 
 
photonper 
pupilon 
incident 
E
P
dt
dN = (1) 
 
Sun thefrom distance
sEarth'at sphere
Sun
pupil
pupilon 
incident 
Sun A
P
A
P
I == 
 
The intensity of the radiation from 
the Sun is given by: 
Sun
Sun thefrom distance
sEarth'at sphere
pupil
pupilon 
incident PA
A
P = (2) 
 
Solving for yields: 
pupilon 
incident P
Substituting in equation (1) yields: 
 
photonper 
Sun
Sun thefrom distance
sEarth'at sphere
pupil
E
P
A
A
dt
dN = 
 
Substitute for the two areas and 
simplify to obtain: 
photonper 
Sun
2
-SunEarth
pupil
photonper 
Sun
2
-SunEarth
2
pupil4
1
4
4
E
P
R
d
E
P
R
d
dt
dN
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
= π
π
 
 
Substitute numerical values and evaluate dN/dt: 
 
( ) 11519
262
11 s 10237.3
eV
J 101.602eV 25.2
 W102.4
m 1050.14
mm 1 −
− ×=××
×⎟⎟⎠
⎞
⎜⎜⎝
⎛
×=dt
dN 
or, separating variables and integrating this expression, 
 ( )tN 115 s 10237.3 −×= 
 
Chapter 31 
 
2896 
 
Evaluating N for t = 0.1 s yields: 
 
( ) ( )( )
photons 103
s 1.0s 10237.3s 1.0
14
115
×≈
×= −N
 
 
The energy deposited, assuming all 
the photons are absorbed, is the 
product of the rate at which energy is 
incident on the pupil and the time 
during which it is delivered: 
 
tPE
pupilon 
incident = 
Substituting for from 
equation (2) yields: 
pupilon 
incident P tP
A
A
E Sun
Sun thefrom distance
sEarth'at sphere
pupil= 
 
Substitute for the two areas and 
simplify to obtain: 
tP
R
d
tP
R
d
E
Sun
2
-SunEarth
pupil
Sun2
-SunEarth
2
pupil4
1
4
4
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
= π
π
 
 
Substitute numerical values and evaluate E for t = 0.1 s: 
 
( ) ( ) ( )( ) mJ 0.1mJ 1167.0s 1.0 W102.4m 1050.14 mm 1s 1.0 26
2
11 ≈=×⎟⎟⎠
⎞
⎜⎜⎝
⎛
×=E 
 
21 •• Römer was observing the eclipses of Jupiter’s moon Io with the hope 
that they would serve as a highly accurate clock that would be independent of 
longitude. (Prior to GPS, such a clock was needed for accurate navigation.) Io 
eclipses (enters the umbra of Jupiter’s shadow) every 42.5 h. Assuming an eclipse 
of Io is observed on Earth on June 1 at midnight when Earth is at location A (as 
shown in Figure 31-54), predict the expected time of observation of an eclipse 
one-quarter of a year later when Earth is at location B, assuming (a) the speed of 
light is infinite and (b) the speed of light is 2.998 × 108 m/s. 
 
Picture the Problem We can use the period of Io’s motion and the position of the 
earth at B to find the number of eclipses of Io during Earth’s movement and then 
use this information to find the number of days before a night-time eclipse.During the 42.5 h between eclipses of Jupiter’s moon, Earth moves from A to B, 
increasing the distance from Jupiter by approximately the distance from Earth to 
the Sun, making the path for the light longer and introducing a delay in the onset 
of the eclipse. 
 
Properties of Light 
 
 
2897
(a) Find the time it takes Earth to 
travel from point A to point B: 
h4.2191
d
h24
4
d24.365
4
earth
=
×==→ Tt BA 
 
Because there are 42.5 h between 
eclipses of Io, the number of eclipses 
N occurring in the time it takes for 
the earth to move from A to B is: 
56.51
h42.5
h4.2191
Io
=== →
T
tN BA 
 
Hence, in one-fourth of a year, there will be 51.56 eclipses. Because we want to 
find the next occurrence that happens in the evening hours, we’ll use 52 as the 
number of eclipses. We’ll also assume that Jupiter is visible so that the eclipse of 
Io can be observed at the time we determine. 
 
Relate the time t(N) at which the 
Nth eclipse occurs to N and the 
period TIo of Io: 
 
( ) IoNTNt = 
Evaluate t(52) to obtain: 
 ( ) ( )
d083.92
h24
d1h5.425252
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×=t
 
 
Subtract the number of whole days to 
find the clock time t: 
 
( )
a.m.00:2
h992.1
d
h24
d0.083
d92d92.083d9252
≈
=×=
−=−= tt
 
 
Because June, July, and August have 
30, 31, and 31 d, respectively, the 
date is: 
 
1September 
(b) Express the time delay Δt in the 
arrival of light from Io due to Earth’s 
location at B: 
 
c
rt sun-earth=Δ 
Substitute numerical values and 
evaluate Δt: 
 min34.8
s 60
min 1s500
m/s10998.2
m105.1
8
11
=
×=×
×=Δt
 
 
Hence, the eclipse will actually occur at 2:08 a.m., September 1 
Chapter 31 
 
2898 
 
22 •• If the angle of incidence is small enough, the small angle 
approximation sin θ ≈ θ may be used to simplify Snell’s law of refraction. 
Determine the maximum value of the angle that would make the value for the 
angle differ by no more than one percent from the value for the sine of the angle. 
(This approximation will be used in connection with image formation by spherical 
surfaces in Chapter 32.) 
 
Picture the Problem We can express the relative error in using the small angle 
approximation and then either use 1) trial-and-error methods, 2) a spreadsheet 
program, or 3) the Solver capability of a scientific calculator to solve the 
transcendental equation that results from setting the error function equal to 0.01. 
 
Express the relative error δ in using 
the small angle approximation: 
 
( ) 1
sinsin
sin −=−= θ
θ
θ
θθθδ 
A spreadsheet program was used to plot the following graph of δ (θ ). 
0.000
0.002
0.004
0.006
0.008
0.010
0.012
0.014
0.016
0.00 0.05 0.10 0.15 0.20 0.25 0.30
theta (radians)
de
lta
(th
et
a)
 
From the graph, we can see that δ (θ ) < 1% for θ ≤ 0.24 radians. In degree 
measure, °≤ 14θ 
 
Remarks: Using the Solver program on a TI-85 gave θ = 0.244 radians. 
 
The Speed of Light 
 
23 • Mission Control sends a brief wake-up call to astronauts in a spaceship 
that is far from Earth. 5.0 s after the call is sent, Mission Control can hear the 
groans of the astronauts. How far from Earth is the spaceship? (a) 7.5 × 108 m, 
(b) 15 × 108 m, (c) 30 × 108 m, (d) 45 × 108 m, (e) The spaceship is on the moon. 
 
 
Properties of Light 
 
 
2899
Picture the Problem We can use the distance, rate, and time relationship to find 
the distance to the spaceship. 
 
Relate the distance d to the spaceship 
to the speed of electromagnetic 
radiation in a vacuum and to the time 
for the message to reach the 
astronauts: 
tcd Δ= 
( )( )
m105.7
s5.2m/s10998.2
8
8
×=
×=d
 
and )(a is correct. 
Noting that the time for the message 
to reach the astronauts is half the 
time for Mission Control to hear 
their response, substitute numerical 
values and evaluate d: 
 
24 • The distance from a point on the surface of Earth to a point on the 
surface of the moon is measured by aiming a laser light beam at a reflector on the 
surface of the moon and measuring the time required for the light to make a round 
trip. The uncertainty in the measured distance Δx is related to the uncertainty in 
the measured time Δt by Δx = 12 cΔt. If the time intervals can be measured to 
±1.00 ns, (a) find the uncertainty of the distance. (b) Estimate the percentage 
uncertainty in the distance. 
 
Picture the Problem We can use the given information that the uncertainty in the 
measured distance Δx is related to the uncertainty in the time Δt by Δx = cΔt to 
evaluate Δx. 
 
(a) The uncertainty in the distance is: 
 
tcx ΔΔ 21= 
 
Substitute numerical values and 
evaluate Δx: 
( )( )
cm0.15
ns00.1m/s10998.2Δ 821
±=
±×=x
 
 
(b)The percent uncertainty in the 
distance to the Moon is: 
 %10
m 103.84
cm 0.15Δ
8
8
Moon Earth to
Moon Earth to
−≈
×=x
x
 
 
25 •• [SSM] Ole Römer discovered the finiteness of the speed of light by 
observing Jupiter’s moons. Approximately how sensitive would the timing 
apparatus need to be in order to detect a shift in the predicted time of the moon’s 
eclipses that occur when the moon happens to be at perigee (3.63×105 km ) and 
those that occur when the moon is at apogee ( 4.06 ×105 km )? Assume that an 
Chapter 31 
 
2900 
 
instrument should be able to measure to at least one-tenth the magnitude of the 
effect it is to measure. 
 
Picture the Problem His timing apparatus would need to be sensitive enough to 
measure the difference in times for light to travel to Earth when the moon is at 
perigee and at apogee. 
 
The sensitivity of the timing 
apparatus would need to be one-tenth 
of the difference in time for light to 
reach Earth from the two positions of 
the moon of Jupiter: 
 
tΔy Sensitivit 101= 
where Δt is the time required for light 
to travel between the two positions of 
the moon. 
The time required for light to travel 
between the two positions of the 
moon is given by: 
 
c
dd
t perigeeat 
moon
apogeeat 
moon 
Δ
−
= 
 
c
dd
10
y Sensitivit perigeeat 
moon 
apogeeat 
moon −
= 
 
Substituting for Δt yields: 
Substitute numerical values and 
evaluate the required sensitivity: 
 
( )( )
ms 14
m/s 10998.210
km 1063.3km 1006.4
Δ 8
55
=
×
×−×=t
 
 
Remarks: Instruments with this sensitivity did not exist in the 17th century. 
 
Reflection and Refraction 
 
26 • Calculate the fraction of light energy reflected from an air–water 
interface at normal incidence. 
 
Picture the Problem Use the equation relating the intensity of reflected light at 
normal incidence to the intensity of the incident light and the indices of refraction 
of the media on either side of the interface. 
 
Express the intensity I of the light 
reflected from an air-water interface 
at normal incidence in terms of the 
indices of refraction and the intensity 
I0 of the incident light: 
 
0
2
waterair
waterair I
nn
nnI ⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−= 
Solve for the ratio I/I0: 
 
2
waterair
waterair
0
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−=
nn
nn
I
I 
Properties of Light 
 
 
2901
Substitute numerical values and 
evaluate I/I0: %0.233.100.1
33.100.1 2
0
=⎟⎠
⎞⎜⎝
⎛
+
−=
I
I 
 
27 • A ray of light is incident on one of a pair of mirrors set at right angles 
to each A ray of light is incident on one of two mirrors that are set at right angles 
to each other. The plane of incidence is perpendicular to both mirrors. Show that 
after reflecting from each mirror, the ray will emerge traveling in the direction 
opposite to the incident direction, regardless of the angle of incidence. 
 
Picture the Problem The diagram 
shows ray 1 incident on the vertical 
surface at an angle θ1, reflected as ray 
2, and incident on the horizontal 
surface at an angle of incidence θ3. 
We’ll prove that rays 1 and 3 are 
parallel by showing that θ1 = θ4, i.e., 
by showing that they make equal 
angles with the horizontal. Note that 
the law of reflection has been used in 
identifying equal angles of incidence 
and reflection. 
1
2
3
θ
θ
θ
θ θ
θ
1
1
2
33
4
 
 
We know that the angles of the righttriangle formed by ray 2 and the two 
mirror surfaces add up to 180°: 
 
( ) °=−°+°+ 1809090 12 θθ 
or 
21 θθ = 
The sum of θ2 and θ3 is 90°: 
 
23 90 θθ −°= 
Because 21 θθ = : 13 90 θθ −°= 
 
The sum of θ4 and θ3 is 90°: °=+ 9043 θθ 
 
Substitute for θ3 to obtain: ( ) °=+−° 9090 41 θθ ⇒ 41 θθ = 
 
28 •• (a) A ray of light in air is incident on an air–water interface. Using a 
spreadsheet or graphing program, plot the angle of refraction as a function of the 
angle of incidence from 0º to 90º. (b) Repeat Part (a), but for a ray of light in 
water that is incident on a water–air interface. [For Part (b), there is no reflected 
ray for angles of incidence that are greater than the critical angle.] 
 
 
 
Chapter 31 
 
2902 
 
Picture the Problem Diagrams showing the light rays for the two cases are 
shown below. In (a) the light travels from air into water and in (b) it travels from 
water into air. 
 
(a) 
Air
Water
θ
θ
1
2
1n
2n
 
(b) 
Air
Water
1n
2n
θ
θ
1
2
 
 
2211 sinsin θθ nn = 
where the angles of incidence and 
refraction are θ1 and θ2, respectively. 
 
(a) Apply Snell’s law to the air-
water interface to obtain: 
Solving for θ2 yields: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 1
2
11
2 sinsin θθ n
n 
 
A spreadsheet program to graph θ2 as a function of θ1 is shown below. The 
formulas used to calculate the quantities in the columns are as follows: 
 
Cell Content/Formula Algebraic Form 
n1B1 1 
n2B2 1.33333 
A6 0 θ1 (deg) 
A7 A6 + 5 θ1 + Δθ 
B6 A6*PI()/180 π 
1801
θ ×
C6 ASIN(($B$1/$B$2)*SIN(B6))
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
1
2
11 sinsin θ
n
n 
D6 C6*180/PI() 
πθ
180
2 × 
 
 
 
 A B C D 
1 n1= 1 
2 n2= 1.33333 
Properties of Light 
 
 
2903
3 
4 θ1 θ1 θ2 θ2
5 (deg) (rad) (rad) (deg) 
6 0 0.00 0.000 0.00 
7 1 0.02 0.013 0.75 
8 2 0.03 0.026 1.50 
9 3 0.05 0.039 2.25 
 
21 87 1.52 0.847 48.50 
22 88 1.54 0.847 48.55 
23 89 1.55 0.848 48.58 
24 90 1.57 0.848 48.59 
 
A graph of θ2 as a function of θ1 follows: 
0
5
10
15
20
25
30
35
40
45
50
0 10 20 30 40 50 60 70 80 90
Angle of incidence, deg
A
ng
le
 o
f r
ef
ra
ct
io
n,
 d
eg
 
 
(b) Change the contents of cell B1 to 1.33333 and the contents of cell B2 to 1 to 
obtain the following graph: 
0
10
20
30
40
50
60
70
80
90
0 10 20 30 40 50
Angle of incidence, deg
A
ng
le
 o
f r
ef
ra
ct
io
n,
 d
eg
 
Note that as the angle of incidence approaches the critical angle for a water-air 
interface (48.6°), the angle of refraction approaches 90°. 
 
Chapter 31 
 
2904 
 
29 •• The red light from a helium-neon laser has a wavelength of 632.8 nm 
in air. Find the (a) speed, (b) wavelength, and (c) frequency of helium-neon laser 
light in air, water, and glass. (The glass has an index of refraction equal to 1.50.) 
 
Picture the Problem We can use the definition of the index of refraction to find 
the speed of light in the three media. The wavelength of the light in each medium 
is its wavelength in air divided by the index of refraction of the medium. The 
frequency of the helium-neon laser light is the same in all media and is equal to its 
value in air. The wavelength of helium-neon laser light in air is 632.8 nm. 
 
The speed of light in a medium 
whose index of refraction is n is 
given by: 
 
n
cv = (1) 
The wavelength of light in a medium 
whose index of refraction is n is 
given by: 
 
n
air
n
λλ = (2) 
The frequency of the light is equal to 
its frequency in air independently of 
the medium in which the light is 
propagating: 
 
Hz 1074.4
nm 632.8
m/s 10998.2
14
8
×=
×== λ
cf
 
Substitute numerical values in 
equation (1) and evaluate vwater: 
m/s1025.2
33.1
m/s10998.2
8
8
water
×=
×=v
 
 
Substitute numerical values in 
equation (2) and evaluate λwater: nm 47633.1
nm 8.632
water ==λ 
 
The other speeds and wavelengths are found similarly and are summarized in the 
following table: 
 
 
 (a) speed (b) wavelength (c) frequency 
(m/s) (nm) (Hz) 
Air 81000.3 × 141074.4 ×633 
Water 81025.2 × 141074.4 ×476 
glass 81000.2 × 141074.4 ×422 
 
 
Properties of Light 
 
 
2905
30 •• The index of refraction for silicate flint glass is 1.66 for violet light 
that has a wavelength in air equal to 400 nm and 1.61 for red light that has a 
wavelength in air equal to 700 nm. A ray of 700-nm–wavelength red light and a 
ray of 400-nm-wavelength violet light both have angles of refraction equal to 30º 
upon entering the glass from air. (a) Which is greater, the angle of incidence of 
the ray of red light or the angle of incidence of the ray of violet light? Explain 
your answer. (b) What is the difference between the angles of incidence of the 
two rays? 
 
Picture the Problem Let the subscript 1 refer to the air and the subscript 2 to the 
silicate glass and apply Snell’s law to the air-glass interface. 
 
(a) Because the index of refraction for violet light is larger than that of red light, 
for a given incident angle violet light would refract more than red light. Thus to 
exhibit the same refraction angle, violet light would require an angle of incidence 
larger than that of red light. 
 
(b) Express the difference in their 
angles of incidence: 
 
red1,violet1,Δ θθθ −= (1) 
Apply Snell’s law to the air-glass 
interface to obtain: 
 
2211 sinsin θθ nn = 
Solving for θ1 yields: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 2
1
21
1 sinsin θθ n
n
 
 
Substitute for violet1,θ and red1,θ in equation (1) to obtain: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛= −− red ,2
air
red1
 violet,2
air
violet1 sinsinsinsinΔ θθθ
n
n
n
n
 
 
For °== 30red 2, violet2, θθ : 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛ °−⎟⎟⎠
⎞
⎜⎜⎝
⎛ °= −−−−
air
red1
air
violet1
air
red1
air
violet1
2
sin
2
sin30sinsin30sinsinΔ
n
n
n
n
n
n
n
nθ 
 
Substitute numerical values and evaluate Δθ: 
 
( ) ( ) °=°−°=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛= −− 49.261.5310.56
00.12
61.1sin
00.12
66.1sinΔ 11θ 
 
Chapter 31 
 
2906 
 
Remarks: Note that Δθ is positive. This means that the angle for violet light is 
greater than that for red light and confirms our answer in Part (a). 
 
31 •• [SSM] A slab of glass that has an index of refraction of 1.50 is 
submerged in water that has an index of refraction of 1.33. Light in the water is 
incident on the glass. Find the angle of refraction if the angle of incidence is 
(a) 60º, (b) 45º, and (c) 30º. 
 
Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to 
the glass and apply Snell’s law to the water-glass interface. 
 
Apply Snell’s law to the water-
glass interface to obtain: 
 
2211 sinsin θθ nn = 
Solving for θ2 yields: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 1
2
11
2 sinsin θθ n
n 
 
(a) Evaluate θ2 for θ1 = 60°: 
 
°=⎟⎠
⎞⎜⎝
⎛ °= − 5060sin
50.1
33.1sin 12θ 
 
(b) Evaluate θ2 for θ1 = 45°: 
 
°=⎟⎠
⎞⎜⎝
⎛ °= − 3945sin
50.1
33.1sin 12θ 
 
(c) Evaluate θ2 for θ1 = 30°: 
 
°=⎟⎠
⎞⎜⎝
⎛ °= − 2630sin
50.1
33.1sin 12θ 
 
32 •• Repeat Problem 31 for a beam of light initially in the glass that is 
incident on the glass–water interface at the same angles. 
 
Picture the Problem Let the subscript 1 refer to the glass and the subscript 2 to 
the water and apply Snell’s law to the glass-water interface. 
 
Apply Snell’s law to the water-
glass interface to obtain: 
 
2211 sinsin θθ nn = 
Solving for θ2 yields: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 1
2
11
2 sinsin θθ n
n 
 
(a) Evaluate θ2 for θ1 = 60°: 
 
°=⎟⎠
⎞⎜⎝
⎛ °= − 7860sin
33.1
50.1sin 12θ 
 
Properties of Light 
 
 
2907
(b) Evaluate θ2 for θ1 = 45°: 
 
°=⎟⎠
⎞⎜⎝
⎛ °= − 5345sin
33.1
50.1sin 12θ 
 
(c) Evaluate θ2 for θ1 = 30°: 
 
°=⎟⎠
⎞⎜⎝
⎛ °= − 3430sin
33.1
50.1sin 12θ 
 
33 •• A beam of light in air strikes a glass slab at normal incidence. The 
glass slab has an index of refraction of 1.50. (a) Approximately what percentage 
of the incident light intensity is transmitted through the slab (in one side and out 
the other)? (b) Repeat Part (a) if the glass slab is immersed in water. 
 
Picturethe Problem Let the subscript 
1 refer to the medium to the left (air) of 
the first interface, the subscript 2 to 
glass, and the subscript 3 to the 
medium (air) to the right of the second 
interface. Apply the equation relating 
the intensity of reflected light at normal 
incidence to the intensity of the 
incident light and the indices of 
refraction of the media on either side of 
the interface to both interfaces. We’ll 
neglect multiple reflections at glass-air 
interfaces. 
 
 
 
1I
1r,I 2r,I
2I 3I
00.11 =n 50.12 =n 00.13 =n
 
 
(a) Express the intensity of the 
transmitted light in the second 
medium: 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−=−=
2
21
21
1
1
2
21
21
1r,112
1
nn
nnI
I
nn
nnIIII
 
 
Express the intensity of the 
transmitted light in the third 
medium: 
 
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−=−=
2
32
32
2
2
2
32
32
2r,223
1
nn
nnI
I
nn
nnIIII
 
 
Substitute for I2 to obtain: 
 ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−=
2
32
32
2
21
21
13 11 nn
nn
nn
nnII
 
Chapter 31 
 
2908 
 
Solve for the ratio I3/I1 to obtain: 
 ⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+
−−=
2
32
32
2
21
21
1
3 11
nn
nn
nn
nn
I
I 
 
Substitute numerical values and evaluate I3/I1: 
 
%92
00.150.1
00.150.11
50.100.1
50.100.11
22
1
3 =⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
+
−−⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
+
−−=
I
I 
 
(b) With n1 = n3 = 1.33: 
 
%99
33.150.1
33.150.11
50.133.1
50.133.11
22
1
3 =⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
+
−−⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
+
−−=
I
I 
 
34 •• This problem is a refraction analogy. A band is marching down a 
football field with a constant speed v1. About midfield, the band comes to a 
section of muddy ground that has a sharp boundary making an angle of 30º with 
the 50-yd line, as shown in Figure 31-55. In the mud, each marcher moves at a 
speed equal to 
1
2 v1 in a direction perpendicular to the row of markers they are in. 
(a) Diagram how each line of marchers is bent as it encounters the muddy section 
of the field so that the band is eventually marching in a different direction. 
Indicate the original direction by a ray and the final direction by a second ray. 
(b) Find the angles between these rays and the line normal to the boundary. Is 
their direction of motion ″bent″ toward the normal or away from it? Explain your 
answer in terms of refraction. 
 
Picture the Problem We can apply Snell’s law to find the angle of refraction of 
the line of marchers as they enter the muddy section of the field 
 
(a) 
°30
θ
θ
1
2
 
(b) Apply Snell’s law at the interface 
to obtain: 
 
2211 sinsin θθ nn = 
 
Properties of Light 
 
 
2909
Solving for θ2 yields: ⎥⎦
⎤⎢⎣
⎡= − 1
2
11
2 sinsin θθ n
n 
 
The ratio of the indices of refraction 
is the reciprocal of the ratio of the 
speeds of the marchers in the two 
media: 
 
2
1
1
12
1
1
2
2
1
2
1 ====
v
v
v
v
v
v
v
v
n
n 
Because the left and right sides of 
the 30° angle and θ1 are mutually 
perpendicular, θ1 = 30°. Substitute 
numerical values and evaluate θ2: 
 
[ ] °=°= − 1430sinsin 2112θ 
As the line enters the muddy field, its speed is reduced by half and the direction of 
the forward motion of the line is changed. In this case, the forward motion in the 
muddy field makes an angle θ2 with respect to the normal to the boundary line. 
Note that the separation between successive lines in the muddy field is half that in 
the dry field. 
 
35 •• [SSM] In Figure 31-56, light is initially in a medium that has an 
index of refraction n1. It is incident at angle θ1 on the surface of a liquid that has 
an index of refraction n2. The light passes through the layer of liquid and enters 
glass that has an index of refraction n3. If θ3 is the angle of refraction in the glass, 
show that n1 sin θ1 = n3 sin θ3. That is, show that the second medium can be 
neglected when finding the angle of refraction in the third medium. 
 
Picture the Problem We can apply Snell’s law consecutively, first to the n1-n2 
interface and then to the n2-n3 interface. 
 
Apply Snell’s law to the n1-n2 
interface: 
 
2211 sinsin θθ nn = 
Apply Snell’s law to the n2-n3 
interface: 
 
3322 sinsin θθ nn = 
Equate the two expressions for 
22 sinθn to obtain: 3311
sinsin θθ nn = 
 
36 •• On a safari, you are spear fishing while wading in a river. You observe 
a fish gliding by you. If your line of sight to the fish is 64.0o degrees below the 
horizontal in air, and assuming the spear follows a straight-line path through the 
air and water after it is released, determine the angle below the horizontal that you 
Chapter 31 
 
2910 
 
should aim your spear gun in order to catch dinner. Assume the spear gun barrel 
is 1.50 m above the water surface, the fish is 1.20 m below the surface, and the 
spear travels in a straight line all the way to the fish. 
 
Picture the Problem The following pictorial representation summarizes the 
information given in the problem statement. We can use the geometry of the 
diagram and apply Snell’s law at the air-water interface to find the aiming angle 
α. 
θ
θ
1
2
α°0.64m 
50.1=
h
m 
20.1=
d
lL
Air
Water
1n
2n
 
 
Use the pictorial representation to 
express the aiming angle α: 
 
⎥⎦
⎤⎢⎣
⎡
+
+= − lL
dh1tanα 
Referring to the diagram, note that: 
 
1tanθhL = and 2tanθd=l 
Substituting for L and l yields: 
⎥⎦
⎤⎢⎣
⎡
+
+= −
21
1
tantan
tan θθα dh
dh 
 
Apply Snell’s law at the air-water 
interface to obtain: 
 
2211 sinsin θθ nn = 
Solving for θ2 yields: 
 ⎥⎦
⎤⎢⎣
⎡= − 1
2
11
2 sinsin θθ n
n
 
 
Properties of Light 
 
 
2911
 
Substitute for θ2 to obtain: 
 
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢
⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡+
+=
−
−
1
2
11
1
1
sinsintantan
tan
θθ
α
n
ndh
dh 
 
Noting that θ1 is the complement of 64.0°, substitute numerical values and 
evaluate α: 
 
( ) ( )
°=
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎦
⎤⎢⎣
⎡ °+°
+=
−
− 9.66
0.26sin
33.1
00.1sintanm 20.10.26tanm 50.1
m 20.1m 50.1tan
1
1α 
That is, you should aim 66.9° below the horizontal. 
 
37 ••• You are standing on the edge of a swimming pool and looking directly 
across at the opposite side. You notice that the bottom edge of the opposite side 
of the pool appears to be at an angle of 28° degrees below the horizontal. 
However, when you sit on the pool edge, the bottom edge of the opposite side of 
the pool appears to be at an angle of only 14o below the horizontal. Use these 
observations to determine the width and depth of the pool. Hint: You will need to 
estimate the height of your eyes above the surface of the water when standing and 
sitting. 
 
Picture the Problem The following diagrams represent the situations when you 
are standing on the edge of the pool (the diagram to the left) and when you are 
sitting on the edge of the pool (the diagram to the right). We can use Snell’s law 
and the geometry of the pool to determine the width and depth of the pool. 
θ
θ
1
2
α
h
L l
d
Standing
Water
Air
2n
1n
standingh
 
θ
θ
1
2
α
h
L l
d
Sitting
'
'
' '
'
Air
Water
1n
2n
sittingh
 
 
 
 
Chapter 31 
 
2912 
 
°=°−°=−°= 622890901 αθ 
and 
°=°−°=−°= 761490'90'1 αθ 
Use the fact that angles α and θ1 are 
complementary, as are α′ and θ1′ to 
determine θ1 and θ1′: 
 
1standing tanθhL = 
and 
'tan' 1sitting θhL = 
Express the distances L and L′ in 
terms of θ1 and θ1′: 
 
 ( ) m 197.362tanm 7.1 =°=L 
and ( ) m 808.276tanm 7.0' =°=L 
Assuming that your eyes are 1.7 m 
above the level of the water when 
you are standing and 0.7 m above the 
water when you are sitting, evaluate 
L and L′: 
 
Referring to the pictorial 
representations, note that: 
 
h
Ld
h
−== l2tanθ (1) 
and 
h
Ld
h
'''tan 2
−== lθ 
 
Divide the first of these equations by 
the second to obtain: 
 
''tan
tan
2
2
Ld
Ld
−
−=θ
θ 
Solving for d yields: 
 'tantan
'tantan'
22
22
θθ
θθ
−
−= LLd (2) 
 
ApplySnell’s law to the air-water 
interface when you are standing: 
 
⎥⎦
⎤⎢⎣
⎡= − 1
2
11
2 sinsin θθ n
n 
Substitute numerical values and 
evaluate 2θ : 
 
°=⎥⎦
⎤⎢⎣
⎡ °= − 60.4162sin
33.1
00.1sin 12θ 
Apply Snell’s law to the air-water 
interface when you are sitting: 
 
⎥⎦
⎤⎢⎣
⎡= − 'sinsin' 1
2
11
2 θθ n
n 
°=⎥⎦
⎤⎢⎣
⎡ °= − 85.4676sin
33.1
00.1sin' 12θ 
 
Substitute numerical values and 
evaluate '2θ : 
Properties of Light 
 
 
2913
 
Substitute numerical values in equation (2) and evaluate d: 
 ( ) ( ) widem 5.1m 130.5
85.46tan60.41tan
85.46tanm 197.360.41tanm 808.2 ==°−°
°−°=d 
 
Solving equation (1) for h yields: 
2tanθ
Ldh −= 
 
Substitute numerical values and 
evaluate h: 
deep m 2.2
60.41tan
m 197.3m 130.5 =°
−=h 
 
38 ••• Figure 31-57 shows a beam of light incident on a glass plate of 
thickness d and index of refraction n. (a) Find the angle of incidence so that the 
separation b between the ray reflected from the top surface and the ray reflected 
from the bottom surface and exiting the top surface is a maximum. (b) What is 
this angle of incidence if the index of refraction of the glass is 1.60? (c) What is 
the separation of the two beams if the thickness of the glass plate is 4.0 cm? 
 
Picture the Problem Let x be the 
perpendicular separation between the 
two rays and let l be the separation 
between the points of emergence of the 
two rays on the glass surface. We can 
use the geometry of the refracted and 
reflected rays to express x as a function 
of l, d, θr, and θi. Setting the 
derivative of the resulting equation 
equal to zero will yield the value of θi 
that maximizes x. Air
Glass
Air
θ θ
θ
θ
θ
r
r
i i
i
l
d
x
 
 
(a) Express l in terms of d and the 
angle of refraction θr: 
 
rtan2 θd=l 
 
Express x as a function of l, d, 
θr, and θi: 
 
ir costan2 θθdx = 
Differentiate x with respect toθi: 
 
( ) ⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−==
i
r
ir
2
irir
ii
cossecsintan2costan2 θ
θθθθθθθθθ d
dd
d
dd
d
dx (1) 
 
Chapter 31 
 
2914 
 
Apply Snell’s law to the air-glass 
interface: 
 
r2i1 sinsin θθ nn = (2) 
or, because n1 = 1 and n2 = n, 
ri sinsin θθ n= 
 
rrii coscos θθθθ dnd = 
or 
r
i
i
r
cos
cos1
θ
θ
θ
θ
nd
d = 
Differentiate implicitly with 
respect toθI to obtain: 
 
 
Substitute in equation (1) to obtain: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−=
r
ir
r
3
i
2
r
i
r
2
i
i
r
r
i cos
sinsin
cos
cos12
cos
cos
cos
cos1sin
cos
sin2 θ
θθ
θ
θ
θ
θ
θ
θθθ
θ
θ ndndd
dx 
 
Substitute for 
and 
i
2sin1 θ− i2cos θ
isin
1 θ
n
for rsinθ to obtain: 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
r
i
2
r
3
i
2
i cos
sin
cos
sin12 θ
θ
θ
θ
θ nndd
dx 
Multiply the second term in parentheses by r
2
r
2 coscos θθ and simplify to 
obtain: 
 
( )r2i2i2
r
3
r
3
r
2
i
2
r
3
i
2
i
cossinsin1
cos
2
cos
cossin
cos
sin1
2 θθθθθ
θθ
θ
θ
θ −−=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−=
n
d
nn
d
d
dx 
 
Substitute for : r
2sin1 θ− r2cos θ
 
( )[ ]r2i2i2
r
3
i
sin1sinsin1
cos
2 θθθθθ −−−= n
d
d
dx 
 
Substitute isin
1 θ
n
for rsinθ to obtain: 
 
⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛ −−−= i22i2i2
r
3
i
sin11sinsin1
cos
2 θθθθθ nn
d
d
dx 
 
Factor out 1/n2, simplify, and set equal to zero to obtain: 
 
[ ] extremafor 0sin2sin
cos
2 2
i
22
i
4
r
33
i
=+−= nn
n
d
d
dx θθθθ 
Properties of Light 
 
 
2915
If dx/dθ1 = 0, then it must be true 
that: 
 
0sin2sin 2i
22
i
4 =+− nn θθ 
 
Solve this quartic equation for θi 
to obtain: ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−−= −
2
1
i
1
11sin
n
nθ 
 
(b) Evaluate θI for n = 1.60: 
( )
°=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −−= −
5.48
60.1
1116.1sin 2
1
iθ
 
 
(c) In (a) we showed that: 
 
ir costan2 θθdx = 
 
Solve equation (2) for θr: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= − i
2
11
r sinsin θθ n
n 
 
Substitute numerical values and 
evaluate θr: °=⎟⎠
⎞⎜⎝
⎛ °= − 9.275.48sin
60.1
1sin 1rθ 
 
Substitute numerical values and 
evaluate x: 
( )
cm8.2
5.48cos9.27tancm0.42
=
°°=x
 
 
Total Internal Reflection 
 
39 • [SSM] What is the critical angle for light traveling in water that is 
incident on a water–air interface? 
 
Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to 
the air and use Snell’s law under total internal reflection conditions. 
 
Use Snell’s law to obtain: 
 
2211 sinsin θθ nn = 
When there is total internal 
reflection: 
 
c1 θθ = and °= 902θ 
 
Substitute to obtain: 22c1 90sinsin nnn =°=θ 
 
Solving for θc yields: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
1
21
c sin n
nθ 
Chapter 31 
 
2916 
 
Substitute numerical values and 
evaluate θc: °=⎟⎠
⎞⎜⎝
⎛= − 8.48
33.1
00.1sin 1cθ 
 
40 • A glass surface (n = 1.50) has a layer of water (n = 1.33) on it. Light in 
the glass is incident on the glass–water interface. Find the critical angle for total 
internal reflection. 
 
Picture the Problem Let the index of 
refraction of glass be represented by n1 
and the index of refraction of water by 
n2 and apply Snell’s law to the glass-
water interface under total internal 
reflection conditions. 
Glass
Water
50.11 =n
33.12 =n
θ
θ
1
2
 
 
Apply Snell’s law to the glass-
water interface: 
 
2211 sinsin θθ nn = 
At the critical angle, θ1 = θc and 
θ2 = 90°: 
 
°= 90sinsin 2c1 nn θ 
Solve for θc: 
 ⎥⎦
⎤⎢⎣
⎡ °= − 90sinsin
1
21
c n
nθ 
 
Substitute numerical values and 
evaluate θc: 
 
°=⎥⎦
⎤⎢⎣
⎡ °= − 5.6290sin
50.1
33.1sin 1cθ 
 
41 • A point source of light is located 5.0 m below the surface of a large 
pool of water. Find the area of the largest circle on the pool’s surface through 
which light coming directly from the source can emerge. 
 
Picture the Problem We can apply 
Snell’s law to the water-air interface 
to express the critical angle θc in 
terms of the indices of refraction of 
water (n1) and air (n2) and then relate 
the radius of the circle to the depth d 
of the point source and θc. 
90º
Air
Water
33.11 =n
00.12 =n
m 0.5=d
r
θ
θ
c
c
 
 
Express the area of the circle whose 
radius is r: 
2rA π= 
Properties of Light 
 
 
2917
Relate the radius of the circle to the 
depth d of the point source and the 
critical angle θc: 
 
ctanθdr = 
Apply Snell’s law to the water-air 
interface to obtain: 
 
22c1 90sinsin nnn =°=θ 
Solving for θc yields: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
1
21
c sin n
nθ 
 
Substitute for r and θc to obtain: 
 
[ ]
2
1
21
2
c
sintan
tan
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎭⎬
⎫
⎩⎨
⎧=
=
−
n
nd
dA
π
θπ
 
 
Substitute numerical values and 
evaluate A: ( )
22
2
1
m100.1
33.1
1sintanm0.5
×=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎭⎬
⎫
⎩⎨
⎧= −πA
 
 
42 •• Light traveling in air strikes the largest face of an isosceles-right-
triangle prism at normal incidence. What is the speed of light in this prism if the 
prism is just barely able to produce total internal reflection? 
 
Picture the Problem We can use the 
definition of the index of refraction to 
express the speed of light in the prism 
in terms of the index of refraction n1 of 
the prism. The application of Snell’s 
law at the prism-air interface will allow 
us to relate the index of refraction of 
the prism to the critical angle for total 
internal reflection. Finally, we can use 
the geometry of the isosceles-right-
triangle prism to conclude that θc = 45°. 
45º
45º
θ
00.12 =n
1n
c
 
 
Express the speed of light v in the 
prism in terms of its index of 
refraction n1: 
 
1n
cv = 
Chapter 31 
 
2918 
 
 
Apply Snell’s law to the prism-air 
interface to obtain: 
 
190sinsin 2c1 =°= nn θ 
Solving for n1 yields: 
 c
1 sin
1
θ=n 
 
Substitute for n1 and simplify to 
obtain: 
 
csinθcv = 
 
Substitute numerical values and 
evaluate v: 
( )
m/s101.2
45sinm/s10998.2
8
8
×=
°×=v
 
 
43 •• A point source of light is located at the bottom of a steel tank, and an 
opaque circular card of radius 6.00 cm is placed horizontally over it. A 
transparent fluid is gently added to the tank so that the card floats on the fluid 
surface with its center directly above the light source. No light is seen by an 
observer above the surface untilthe fluid is 5.00 cm deep. What is the index of 
refraction of the fluid? 
 
Picture the Problem The observer above the surface of the fluid will not see any 
light until the angle of incidence of the light at the fluid-air interface is less than 
or equal to the critical angle for the two media. We can use Snell’s law to express 
the index of refraction of the fluid in terms of the critical angle and use the 
geometry of card and light source to express the critical angle. 
θ
θ
c
c
θ
2
r
d
n
n
1
2
 
 
Apply Snell’s law to the fluid-air 
interface to obtain: 
 
2211 sinsin θθ nn = 
Light is seen by the observer 
when θ1 = θc and θ2 = 90°: 
 
22c1 90sinsin nnn =°=θ 
Properties of Light 
 
 
2919
Because the medium above the 
interface is air, n2 = 1. Solve for 
n1 to obtain: 
 
c
1 sin
1
θ=n 
From the geometry of the 
diagram: 
 
d
r=ctanθ ⇒ ⎟⎠
⎞⎜⎝
⎛= −
d
r1
c tanθ 
 
Substitute for cθ to obtain: 
⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛= −
d
r
n
1
1
tansin
1 
 
Substitute numerical values and 
evaluate n1: 
30.1
cm00.5
cm00.6
tansin
1
1
1 =
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
n 
 
44 •• An optical fiber allows rays of light to propagate long distances by 
using total internal reflection. Optical fibers are used extensively in medicine and 
in digital communications. As shown in Figure 31-58 the fiber consists of a core 
material that has an index of refraction n2 and radius b surrounded by a cladding 
material that has an index of refraction n3 < n2. The numerical aperture of the 
fiber is defined as sinθ1, where θ1 is the angle of incidence of a ray of light that 
impinges on the center of the end of the fiber and then reflects off the core-
cladding interface just at the critical angle. Using the figure as a guide, show that 
the numerical aperture is given by sinθ1 = n22 − n32 assuming the ray is initially 
in air. Hint: Use of the Pythagorean theorem may be required. 
 
Picture the Problem We can use the geometry of the figure, the law of refraction 
at the air-n1 interface, and the condition for total internal reflection at the n1-n2 
interface to show that the numerical aperture is given by 23
2
21sin nn −=θ . 
 
Referring to the figure, note that: 
 c
a
n
n ==
2
3
csinθ and c
b=2sinθ 
 
Apply the Pythagorean theorem 
to the right triangle to obtain: 
 
222 cba =+ or 12
2
2
2
=+
c
b
c
a 
 
Solving for 
c
b yields: 
2
2
1
c
a
c
b −= 
 
Chapter 31 
 
2920 
 
Substitute for 
c
a and 
c
b to obtain: 
 
2
2
2
3
2 1sin n
n−=θ 
Use the law of refraction to relate 
θ1 and θ2: 
 
2211 sinsin θθ nn = 
Substitute for sinθ2, let n1 = 1 
(air), and simplify to obtain: 
2
3
2
22
2
2
3
21 1sin nnn
nn −=−=θ 
 
45 •• [SSM] Find the maximum angle of incidence θ1 of a ray that would 
propagate through an optical fiber that has a core index of refraction of 1.492, a 
core radius of 50.00 μm, and a cladding index of 1.489. See Problem 44. 
 
Picture the Problem We can use the result of Problem 44 to find the maximum 
angle of incidence under the given conditions. 
 
From Problem 44: 2
3
2
21sin nn −=θ 
 
Solve for θ1 to obtain: ( )232211 sin nn −= −θ 
 
Substitute numerical values and 
evaluate θ1: ( ) ( )
°=
⎟⎠
⎞⎜⎝
⎛ −= −
5
489.1492.1sin 2211θ
 
 
46 •• Calculate the difference in time needed for two pulses of light to travel 
down 15.0 km of the fiber that is described in Problem 44. Assume that one pulse 
enters the fiber at normal incidence, and the second pulse enters the fiber at the 
maximum angle of incidence calculated in Problem 45. In fiber optics, this effect 
is known as modal dispersion. 
 
Picture the Problem We can derive an expression for the difference in the travel 
times by expressing the travel time for a pulse that enters the fiber at the 
maximum angle of incidence and a pulse that enters the fiber at normal incidence. 
Examination of Figure 31-58 reveals that, if the length of the tube is L, the 
distance traveled by the pulse that enters at an angle θ1 is the ratio of c to a 
multiplied by L. 
 
The difference in the travel times 
Δt is given by: 
 
incidence
 normal1
ttt −=Δ θ (1) 
Properties of Light 
 
 
2921
The travel time for the pulse that 
enters the fiber at the maximum 
angle of incidence is: 
2
medium in the speed
 traveleddistance
1
n
c
a
cL
t ==θ 
 
The travel time for the pulse that 
enters the fiber normally is: 
2
incidence 
normal
n
c
Lt = 
 
Substitute for and in 
equation (1) to obtain: 
1θt
incidence 
normalt
22 n
c
L
n
c
a
cL
t −=Δ 
 
Simplify to obtain: 
 ⎟⎠
⎞⎜⎝
⎛ −=Δ 12
a
c
c
Lnt (2) 
 
Referring to the figure, note that: 
 c
a=csinθ 
 
From Snell’s law, the sine of the 
critical angle is also given by: 
 
2
3
csin n
n=θ ⇒ 
2
3
n
n
c
a = 
Substitute for c/a in equation (2) 
to obtain: ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=Δ 1
3
22
n
n
c
Lnt 
 
Substitute numerical values and 
evaluate Δt: 
( )( )
ns150
1
489.1
492.1
m/s10998.2
km15492.1
Δ 8
=
⎟⎠
⎞⎜⎝
⎛ −×=t 
 
47 ••• Investigate how a thin film of water on a glass surface affects the 
critical angle for total reflection. Use n = 1.50 for glass and n = 1.33 for water. 
(a) What is the critical angle for total internal reflection at the glass–water 
interface? (b) Does a range of incident angles exist such that the angles are greater 
than θc for glass-to-air refraction and for which the light rays will leave the glass, 
travel through the water and then pass into the air? 
 
Picture the Problem Let the index of refraction of glass be represented by n1, the 
index of refraction of water by n2, and the index of refraction of air by n3. We can 
apply Snell’s law to the glass-water interface under total internal reflection 
conditions to find the critical angle for total internal reflection. The application of 
Chapter 31 
 
2922 
 
Snell’s law to glass-air and glass-water interfaces will allow us to decide whether 
there are angles of incidence greater than θc for glass-to-air refraction for which 
light rays will leave the glass and the water and pass into the air. 
00.13 =n
33.12 =n
50.11 =n
Air
Water
Glass
θ
θ
θ
θ1
2
2
3
 
 
(a) Apply Snell’s law to the glass-
water interface: 
 
2211 sinsin θθ nn = 
At the critical angle, θ1 = θc and 
θ2 = 90°: 
 
°= 90sinsin 2c1 nn θ 
Solving for θc yields: 
 ⎥⎦
⎤⎢⎣
⎡ °= − 90sinsin
1
21
c n
nθ 
 
Substitute numerical values and 
evaluate θc: °=⎥⎦
⎤⎢⎣
⎡ °= − 5.6290sin
50.1
33.1sin 1cθ 
 
(b) Apply Snell’s law to a water-air 
interface at the critical angle for a 
water-air interface: 
 
°= 90sinsin 3c2 nn θ 
 
 
 
Solving for cθ yields: ⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
2
31
c sin n
nθ 
 
Substitute numerical values and 
evaluate cθ : 
 
°=⎟⎠
⎞⎜⎝
⎛= − 8.48
33.1
00.1sin 1cθ 
2211 sinsin θθ nn = 
and 
⎟⎟⎠
⎞
⎜⎜⎝
⎛= − 2
1
21
1 sinsin θθ n
n 
Apply Snell’s law to a ray incident at 
the glass-water interface: 
 
Properties of Light 
 
 
2923
 
For c2 θθ = : 
⎟⎟⎠
⎞
⎜⎜⎝
⎛⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛= −−
1
31
2
3
1
21
1 sinsin n
n
n
n
n
nθ 
 
Substitute numerical values and 
evaluate θ1: °=⎟⎠
⎞⎜⎝
⎛= − 8.41
53.1
00.1sin 11θ 
 
Yes, if θ ≥ 41.8°, where is the angle of incidence for the rays in glass that are 
incident on the glass-water boundary, the rays will leave the glass through the 
water and pass into the air. 
 
48 ••• A laser beam is incident on a plate of glass that is 3.0-cm thick (Figure 
31-57). The glass has an index of refraction of 1.5 and the angle of incidence is 
40º. The top and bottom surfaces of the glass are parallel. What is the distance b 
between the beam formed by reflection off the top surface of the glass and the 
beam reflected off the bottom surface of the glass. 
 
Picture the Problem The situation 
is shown in the adjacent figure. We 
can use the geometry of the diagram 
and trigonometric relationships to 
derive an expression for d in terms of 
the angles of incidence and 
refraction. Applying Snell’s law will 
yield θr. 
GlassAir
θ
θ
θ
θ
i
i
r
r
θi
x
5.1=n
t
b
 
Express the distance x in terms of t 
and θr: 
 
rtan2 θtx = 
The separation of the reflected rays 
is: 
 
icosθxb = 
 
Substitute for x to obtain: ir costan2 θθtb = (1) 
 
Apply Snell’s law at the air-glass 
interface to obtain: ri
sinsin θθ n= ⇒ ⎟⎠
⎞⎜⎝
⎛= −
n
i1
r
sinsin θθ 
 
Substitute for rθ in equation (1) to 
obtain: 
 
i
i1 cossinsintan2 θθ ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛= −
n
tb 
 
Chapter 31 
 
2924 
 
Substitute numerical values and 
evaluate b: 
 
( )
cm2.2
40cos
5.1
40sinsintancm0.32 1
=
°⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛ °= −b
 
Dispersion 
 
49 • A beam of light strikes the plane surface of silicate flint glass at an 
angle of incidence of 45º. The index of refraction of the glass varies with 
wavelength (see Figure 31-59). How much smaller is the angle of refraction for 
violet light of wavelength 400 nm than the angle of refraction for red light of 
wavelength 700 nm? 
 
Picture the Problem We can apply Snell’s law of refraction to express the angles 
of refraction for red and violet light in silicate flint glass. 
 
Express the difference between the 
angle of refraction for violet light 
and for red light: 
 
violetr,redr, θθθ −=Δ (1) 
Apply Snell’s law of refraction to 
the interface to obtain: 
 
rsin45sin θn=° ⇒ ⎟⎠
⎞⎜⎝
⎛= −
n2
1sin 1rθ 
Substituting for rθ in equation (1) 
yields: ⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=Δ −−
violet
1
red
1
2
1sin
2
1sin
nn
θ 
 
Substitute numerical values and 
evaluate Δθ : ( ) ( )
°=°−°=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛= −−
0.12.252.26
66.12
1sin
60.12
1sinΔ 11θ
 
50 •• In many transparent materials, dispersion causes different colors 
(wavelengths) of light to travel at different speeds. This can cause problems in 
fiber-optic communications systems where pulses of light must travel very long 
distances in glass. Assuming a fiber is made of silicate crown glass (see Figure 
31-19), calculate the difference in travel times that two short pulses of light take 
to travel 15.0 km in this fiber if the first pulse has a wavelength of 700 nm and the 
second pulse has a wavelength of 500 nm. 
 
Picture the Problem The transit times will be different because the speed with 
which light of various wavelengths propagates in silicate crown glass is 
dependent on the index of refraction. We can use Figure 31-19 to estimate the 
indices of refraction for pulses of wavelengths 500 and 700 nm. 
Properties of Light 
 
 
2925
 
Express the difference in time 
needed for two short pulses of 
light to travel a distance L in the 
fiber: 
 
700500 v
L
v
Lt −=Δ 
Substitute for L, v500, and v700 and 
simplify to obtain: 
( )700500700500 nnc
L
c
Ln
c
Lnt −=−=Δ 
 
Use Figure 31-19 to find the 
indices of refraction of silicate 
crown glass for the two 
wavelengths: 
 
55.1500 ≈n 
and 
50.1700 ≈n 
Substitute numerical values and 
evaluate Δt: ( )
s3
50.155.1
m/s10998.2
km0.15
Δ 8
μ≈
−×=t 
 
Polarization 
 
51 • [SSM] What is the polarizing angle for light in air that is incident on 
(a) water (n = 1.33), and (b) glass (n = 1.50)? 
 
Picture the Problem The polarizing angle is given by Brewster’s law: 
12ptan nn=θ where n1 and n2 are the indices of refraction on the near and far 
sides of the interface, respectively. 
 
Use Brewster’s law to obtain: 
 ⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
1
21
p tan n
nθ 
 
(a) For n1 = 1 and n2 = 1.33: °=⎟⎠
⎞⎜⎝
⎛= − 1.53
00.1
33.1tan 1pθ 
 
(b) For n1 = 1 and n2 = 1.50: °=⎟⎠
⎞⎜⎝
⎛= − 3.56
00.1
50.1tan 1pθ 
 
52 • Light that is horizontally polarized is incident on a polarizing sheet. It 
is observed that only 15 percent of the intensity of the incident light is transmitted 
through the sheet. What angle does the transmission axis of the sheet make with 
the horizontal? 
 
Chapter 31 
 
2926 
 
Picture the Problem The intensity of the transmitted light I is related to the 
intensity of the incident light I0 and the angle the transmission axis makes with the 
horizontal θ according to .cos20 θII =
 
Express the intensity of the 
transmitted light in terms of the 
intensity of the incident light and the 
angle the transmission axis makes 
with the horizontal: 
 
θ20 cosII = ⇒ ⎟⎟⎠
⎞
⎜⎜⎝
⎛= −
0
1cos
I
Iθ 
Substitute numerical values and 
evaluate θ : 
( ) °== − 6715.0cos 1θ 
 
 
53 • Two polarizing sheets have their transmission axes crossed so that no 
light gets through. A third sheet is inserted between the first two so that its 
transmission axis makes an angle θ with the transmission axis of the first sheet. 
Unpolarized light of intensity I0 is incident on the first sheet. Find the intensity of 
the light transmitted through all three sheets if (a) θ = 45º and (b) θ = 30º. 
 
Picture the Problem Let In be the intensity after the nth polarizing sheet and use 
 to find the intensity of the light transmitted through all three sheets 
for θ = 45° and θ = 30°. 
θ20 cosII =
 
(a) The intensity of the light between 
the first and second sheets is: 
 
02
1
1 II = 
The intensity of the light between the 
second and third sheets is: 
 
04
12
02
1
2,1
2
12 45coscos IIII =°== θ 
The intensity of the light that has 
passed through the third sheet is: 
 
08
12
04
1
3,2
2
23 45coscos IIII =°== θ 
(b) The intensity of the light between 
the first and second sheets is: 
 
02
1
1 II = 
The intensity of the light between the 
second and third sheets is: 
 
08
32
02
1
2,1
2
12 30coscos IIII =°== θ 
The intensity of the light that has 
passed through the third sheet is: 
032
32
08
3
3,2
2
23 60coscos IIII =°== θ 
 
Properties of Light 
 
 
2927
54 • A horizontal 5.0 mW laser beam that is vertically polarized is incident 
on a sheet that is oriented with its transmission axis vertical. Behind the first sheet 
is a second sheet that is oriented so that its transmission axis makes an angle of 
27º with respect to the vertical. What is the power of the beam transmitted 
through the second sheet? 
 
Picture the Problem Because the light is polarized in the vertical direction and 
the first polarizer is also vertically polarized, no loss of intensity results from the 
first transmission. We can use Malus’s law to find the intensity of the light after it 
has passed through the second polarizer. 
 
The intensity of the beam is the ratio 
of its power to cross-sectional area: 
 
A
PI = 
Express the intensity of the light 
between the first and second 
polarizers: 
 
01 II = and 01 PP = 
Express the law of Malus in terms of 
the power of the beam: 
 
θ20 cos
A
P
A
P = ⇒ θ20 cosPP =
Express the power of the beam after 
the second transmission: 
 
12
2
02,1
2
12 coscos θθ PPP == 
Substitute numerical values and 
evaluate P2: 
( ) mW0.427cosmW0.5 22 =°=P 
 
55 •• [SSM] The polarizing angle for light in air that is incident on a 
certain substance is 60º. (a) What is the angle of refraction of light incident at this 
angle? (b) What is the index of refraction of this substance? 
 
Picture the Problem Assume that light is incident in air (n1 = 1.00). We can use 
the relationship between the polarizing angle and the angle of refraction to 
determine the latter and Brewster’s law to find the index of refraction of the 
substance. 
 
(a) At the polarizing angle, the sum 
of the angles of polarization and 
refraction is 90°: 
 
°=+ 90rp θθ ⇒ pr 90 θθ −°= 
Substitute for θp to obtain: 
 
°=°−°= 306090rθ 
 
Chapter 31 
 
2928 
 
(b) From Brewster’s law we have: 
 1
2
ptan n
n=θ 
or, because n1 = 1.00, 
p2 tanθ=n 
 
Substitute for θp and evaluate n2: 7.160tan2 =°=n 
 
56 •• Two polarizing sheets have their transmission axes crossed so that no 
light is transmitted. A third sheet is inserted so that its transmission axis makes an 
angle θ with the transmission axis of the first sheet. (a) Derive an expression for 
the intensity of the transmitted light as a function of θ. (b) Show that the intensity 
transmitted through all three sheets is maximum whenθ = 45º. 
 
Picture the Problem Let In be the intensity after the nth polarizing sheet and use 
 to find the intensity of the light transmitted through the three sheets. θ20 cosII =
 
(a) The intensity of the light between 
the first and second sheets is: 
 
02
1
1 II = 
The intensity of the light between 
the second and third sheets is: 
 
θθ 20212,1212 coscos III == 
Express the intensity of the light that 
has passed through the third sheet 
and simplify to obtain: 
 
( )
( )
θ
θθ
θθ
θθ
θ
2sin
sincos2
sincos
90coscos
cos
2
08
1
2
08
1
22
02
1
22
02
1
3,2
2
23
I
I
I
I
II
=
=
=
−°=
=
 
 
(b) Because the sine function is a maximum when its argument is 90°, the 
maximum value of I3 occurs when θ = 45°. 
 
57 •• If the middle polarizing sheet in Problem 56 is rotating at an angular 
speed ω about an axis parallel with the light beam, find an expression for the 
intensity transmitted through all three sheets as a function of time. Assume that 
θ = 0 at time t = 0. 
 
Picture the Problem Let In be the intensity after the nth polarizing sheet, use 
 to find the intensity of the light transmitted through each sheet, and 
replace θ with ωt. 
θ20 cosII =
 
Properties of Light 
 
 
2929
The intensity of the light between the 
first and second sheets is: 
 
02
1
1 II = 
The intensity of the light between the 
second and third sheets is: 
 
tIII ωθ 20212,1212 coscos == 
Express the intensity of the light that 
has passed through the third sheet 
and simplify to obtain: 
 
( )
( )
tI
ttI
ttI
ttI
II
ω
ωω
ωω
ωω
θ
2sin
sincos2
sincos
90coscos
cos
2
08
1
2
08
1
22
02
1
22
02
1
3,2
2
23
=
=
=
−°=
=
 
 
58 •• A stack of N + 1 ideal polarizing sheets is arranged so that each sheet 
is rotated by an angle of π/(2N) rad with respect to the preceding sheet. A linearly 
polarized light wave of intensity I0 is incident normally on the stack. The incident 
light is polarized along the transmission axis of the first sheet and is therefore 
perpendicular to the transmission axis of the last sheet in the stack. 
(a) Show that the intensity of the light transmitted through the entire stack is 
given by 
I0 cos
2N π 2N( )⎡⎣ ⎤⎦ . (b) Using a spreadsheet or graphing program, plot 
the transmitted intensity as a function of N for values of N from 2 to 100. (c) What 
is the direction of polarization of the transmitted beam in each case? 
 
Picture the Problem Let In be the intensity after the nth polarizing sheet and use 
 to find the ratio of Iθ20 cosII = n+1 to In. 
 
(a) Find the ratio of In+1 to In: 
NI
I
n
n
2
cos21 π=+ 
 
Because there are N such 
reductions of intensity: 
 
⎟⎠
⎞⎜⎝
⎛== ++
NI
I
I
I NNN
2
cos2
0
1
1
1 π 
and 
⎟⎠
⎞⎜⎝
⎛=+ NII
N
N 2
cos201
π 
 
(b) A spreadsheet program to graph IN+1/I0 as a function of N is shown below. The 
formulas used to calculate the quantities in the columns are as follows: 
 
Cell Content/Formula Algebraic Form 
N A2 2 
A3 A2 + 1 N + 1 
Chapter 31 
 
2930 
 
B2 (cos(PI()/(2*A2))^(2*A2) ⎟⎠
⎞⎜⎝
⎛
N
N
2
cos2 π 
 
 
 A B 
1 N I/I0
2 2 0.250 
3 3 0.422 
4 4 0.531 
5 5 0.605 
 
95 95 0.974 
96 96 0.975 
97 97 0.975 
98 98 0.975 
99 99 0.975 
100 100 0.976 
 
A graph of I/I0 as a function of N follows. 
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0 10 20 30 40 50 60 70 80 90 100
N
I/
I 0
 
(c) The transmitted light, if any, is polarized parallel to the transmission axis of 
the last sheet. (For N = 2 there is no transmitted light.) 
 
59 •• [SSM] The device described in Problem 58 could serve as a 
polarization rotator, which changes the linear plane of polarization from one 
direction to another. The efficiency of such a device is measured by taking the 
ratio of the output intensity at the desired polarization to the input intensity. The 
result of Problem 58 suggests that the highest efficiency is achieved by using a 
large value for the number N. A small amount of intensity is lost regardless of the 
input polarization when using a real polarizer. For each polarizer, assume the 
transmitted intensity is 98 percent of the amount predicted by the law of Malus 
and use a spreadsheet or graphing program to determine the optimum number of 
sheets you should use to rotate the polarization 90º. 
Properties of Light 
 
 
2931
Picture the Problem Let In be the intensity after the nth polarizing sheet and use 
 to find the ratio of Iθ20 cosII = n+1 to In. Because each sheet introduces a 2% 
loss of intensity, the net transmission after N sheets (0.98)N. 
 
Find the ratio of In+1 to In: ( )
NI
I
n
n
2
cos98.0 21 π=+ 
 
Because there are N such 
reductions of intensity: 
( ) ⎟⎠
⎞⎜⎝
⎛=+
NI
I NNN
2
cos98.0 2
0
1 π 
 
A spreadsheet program to graph IN+1/I0 for an ideal polarizer as a function of N, 
the percent transmission, and IN+1/I0 for a real polarizer as a function of N is 
shown below. The formulas used to calculate the quantities in the columns are as 
follows: 
 
Cell Content/Formula Algebraic Form 
N A3 1 
B2 (cos(PI()/(2*A2))^(2*A2) ⎟⎠
⎞⎜⎝
⎛
N
N
2
cos2 π 
( )N98.0C3 (0.98)^A3 
D4 B3*C3 ( ) ⎟⎠
⎞⎜⎝
⎛
N
NN
2
cos98.0 2 π 
 
 
 A B C D 
1 Ideal Percent Real 
2 N Polarizer Transmission Polarizer 
3 1 0.000 0.980 0.000 
4 2 0.250 0.960 0.240 
5 3 0.422 0.941 0.397 
6 4 0.531 0.922 0.490 
7 5 0.605 0.904 0.547 
8 6 0.660 0.886 0.584 
9 7 0.701 0.868 0.608 
10 8 0.733 0.851 0.624 
11 9 0.759 0.834 0.633 
12 10 0.781 0.817 0.638 
13 11 0.798 0.801 0.639 
14 12 0.814 0.785 0.638 
15 13 0.827 0.769 0.636 
16 14 0.838 0.754 0.632 
17 15 0.848 0.739 0.626 
18 16 0.857 0.724 0.620 
19 17 0.865 0.709 0.613 
Chapter 31 
 
2932 
 
20 18 0.872 0.695 0.606 
21 19 0.878 0.681 0.598 
22 20 0.884 0.668 0.590 
 
A graph of I/I0 as a function of N for the quantities described above follows: 
0.0
0.2
0.4
0.6
0.8
1.0
0 2 4 6 8 10 12 14 16 18 20
Number of sheets (N )
I /I 0
Ideal Polarizer
Percent Transmission
Real Polarizer
 
Inspection of the table, as well as of the graph, tells us that the optimum number 
of sheets is .11 
 
60 •• Show mathematically that a linearly polarized wave can be thought of 
as a superposition of a right and a left circularly polarized wave. 
 
Picture the Problem A circularly polarized wave is said to be right circularly 
polarized if the electric and magnetic fields rotate clockwise when viewed along 
the direction of propagation and left circularly polarized if the fields rotate 
counterclockwise. 
 
For a circularly polarized wave, the x 
and y components of the electric 
field are given by: 
 
tEEx ωcos0= 
and 
tEEy ωsin0= or tEEy ωsin0−= 
for left and right circular polarization, 
respectively. 
 
For a wave polarized along the x 
axis: 
 i
iiEE
ˆcos2
ˆcosˆcos
0
00leftright
tE
tEtE
ω
ωω
=
+=+ rr
 
 
61 •• Suppose that the middle sheet in Problem 53 is replaced by two 
polarizing sheets. If the angle between the transmission axes in the second, third, 
and fourth sheets in the stack make angles of 30º, 60º and 90º, respectively, with 
the transmission axis of the first sheet, (a) what is the intensity of the transmitted 
Properties of Light 
 
 
2933
light? (b) How does this intensity compare with the intensity obtained in Part (a) 
of Problem 53? 
 
Picture the Problem Let In be the intensity after the nth polarizing sheet and use 
 to find the intensity of the light transmitted by the four sheets. θ20 cosII =
 
(a) The intensity of the light between 
the first and second sheets is: 
 
02
1
1 II = 
The intensity of the light between the 
second and third sheets is: 
 
08
32
02
1
2,1
2
12 30coscos IIII =°== θ 
The intensity of the light between the 
third and fourth sheets is: 
 
032
92
08
3
3,2
2
23 30coscos IIII =°== θ 
The intensity of the light to the right 
of the fourth sheet is: 
0
0128
272
032
9
4,3
2
34
211.0
30coscos
I
IIII
=
=°== θ
 
 
(b) The intensity with four sheets at angles of 0°, 30°, 30° and 90° is greater theintensity of three sheets at angles of 0°, 45° and 90° by a factor of 1.69. 
 
Remarks: We could also apply the result obtained in Problem 58(a) to solve 
this problem. 
 
62 •• Show that the electric field of a circularly polarized wave propagating 
parallel with the x axis can be expressed by 
( ) ( )kiE ˆcosˆsin 00 tkxEtkxE ωω −+−=r . 
 
Picture the Problem We can use the components of E
r
 to show that E
r
is 
constant in time and rotates with angular frequency ω. 
 
Express the magnitude of E
r
in terms 
of its components: 
 
22
yx EEE += 
Substitute for Ex and Ey to obtain: 
 
( )[ ] ( )[ ] ( ) ( )[ ]
0
222
0
2
0
2
0 cossincossin
E
tkxtkxEtkxEtkxEE
=
−+−=−+−= ωωωω 
and the E
r
vector rotates in the yz plane with angular frequency ω. 
 
Chapter 31 
 
2934 
 
63 •• [SSM] A circularly polarized wave is said to be right circularly 
polarized if the electric and magnetic fields rotate clockwise when viewed along 
the direction of propagation and left circularly polarized if the fields rotate 
counterclockwise. (a) What is the sense of the circular polarization for the wave 
described by the expression in Problem 62? (b) What would be the expression for 
the electric field of a circularly polarized wave traveling in the same direction as 
the wave in Problem 60, but with the fields rotating in the opposite direction? 
 
Picture the Problem We can apply the given definitions of right and left circular 
polarization to the electric field and magnetic fields of the wave. 
 
(a) The electric field of the wave in Problem 62 is: 
 
( ) ( )kjE ˆcosˆsin 00 tkxEtkxE ωω −+−=r 
 
The corresponding magnetic field is: 
 
( ) ( ) jkB ˆcosˆsin 00 tkxBtkxB ωω −−−=r 
 
Because these fields rotate clockwise when viewed along the direction of 
propagation, the wave is right circularly polarized. 
 
(b) For a left circularly polarized wave traveling in the opposite direction: 
 
( ) ( )kjE ˆcosˆsin 00 tkxEtkxE ωω +−+=r 
 
Sources of Light 
 
64 • A helium–neon laser emits light that has a wavelength equal to 632.8 
nm and has a power output of 4.00 mW. How many photons are emitted per 
second by this laser? 
 
Picture the Problem We can express the number of photons emitted per second 
as the ratio of the power output of the laser and energy of a single photon. 
 
Relate the number of photons per 
second n to the power output of the 
pulse and the energy of a single 
photon : photonE
 
photonE
Pn = 
The energy of a photon is given by: 
 λ
hcE =photon 
Properties of Light 
 
 
2935
Substitute for to obtain: photonE
hc
Pn λ= 
 
Substitute numerical values and evaluate n: 
 ( )( ) photons/s1027.1
J10602.1
eV1
nmeV1240
mW00.4nm8.632 16
19 ×=×⋅= −n 
 
65 • The first excited state of an atom of a gas is 2.85 eV above the ground 
state. (a) What is the maximum wavelength of radiation for resonance absorption 
by atoms of the gas that are in the ground state? (b) If the gas is irradiated with 
monochromatic light that has a wavelength of 320 nm, what is the wavelength of 
the Raman scattered light? 
 
Picture the Problem We can use the Einstein equation for photon energy to find 
the wavelength of the radiation for resonance absorption. We can use the same 
relationship, with ERaman = Einc − ΔE where ΔE is the energy for resonance 
absorption, to find the wavelength of the Raman scattered light. 
 
(a) Use the Einstein equation for 
photon energy to relate the 
wavelength of the radiation to 
energy of the first excited state: 
 
E
hc=λ 
Substitute numerical values and 
evaluate λ: 
 
nm435
eV85.2
nmeV1240 =⋅=λ 
(b) The wavelength of the Raman 
scattered light is given by: 
 
Raman
Raman
nmeV1240
E
⋅=λ 
Relate the energy of the Raman 
scattered light ERaman to the 
energy of the incident light Einc: 
 eV1.025
eV85.2
nm 320
nmeV1240
incRaman
=
−⋅=
Δ−= EEE
 
 
Substitute numerical values and 
evaluate λRaman: nm1210eV025.1
nmeV1240
Raman =⋅=λ 
 
66 •• A gas is irradiated with monochromatic ultraviolet light that has a 
wavelength of 368 nm. Scattered light that has a wavelength equal to 368 nm is 
observed, and scattered light that has a wavelength of 658-nm is also observed. 
Assuming that the gas atoms were in their ground state prior to irradiation, find 
Chapter 31 
 
2936 
 
the energy difference between the ground state and the excited state obtained by 
the irradiation. 
 
Picture the Problem The incident radiation will excite atoms of the gas to higher 
energy states. The scattered light that is observed is a consequence of these atoms 
returning to their ground state. The energy difference between the ground state 
and the atomic state excited by the irradiation is given by λ
hchfE ==Δ . 
 
The energy difference between the 
ground state and the atomic state 
excited by the irradiation is given by: 
 
λλ
nmeV1240 ⋅===Δ hchfE 
Substitute 368 nm for λ and evaluate 
ΔE: eV37.3nm368
nmeV1240 =⋅=ΔE 
 
67 •• [SSM] Sodium has excited states 2.11 eV, 3.20 eV, and 4.35 eV 
above the ground state. Assume that the atoms of the gas are all in the ground 
state prior to irradiation. (a) What is the maximum wavelength of radiation that 
will result in resonance fluorescence? What is the wavelength of the fluorescent 
radiation? (b) What wavelength will result in excitation of the state 4.35 eV above 
the ground state? If that state is excited, what are the possible wavelengths of 
resonance fluorescence that might be observed? 
 
Picture the Problem The ground state 
and the three excited energy levels are 
shown in the diagram to the right. 
Because the wavelength is related to the 
energy of a photon by λ = hc/ΔE, 
longer wavelengths correspond to 
smaller energy differences. 
3
2
1
0 0
3.20 eV
2.11 eV
4.35 eV
 
 
(a) The maximum wavelength of 
radiation that will result in resonance 
fluorescence corresponds to an 
excitation to the 3.20 eV level 
followed by decays to the 2.11 eV 
level and the ground state: 
 
nm388
eV3.20
nmeV1240
max =⋅=λ 
Properties of Light 
 
 
2937
 
nm1140
eV2.11eV3.20
nmeV1240
12 =−
⋅=→λ The fluorescence wavelengths are: 
and 
nm588
0eV2.11
nmeV1240
01 =−
⋅=→λ 
 
nm285
eV35.4
nmeV1240
30 =⋅=→λ 
(not in visible the spectrum) 
 
(b) For excitation: 
nm1080
eV3.20eV35.4
nmeV1240
23 =−
⋅=→λ The fluorescence wavelengths corresponding to the possible 
transitions are: (not in visible spectrum) 
 
nm1140
eV11.2eV20.3
nmeV1240
12 =−
⋅=→λ 
(not in visible spectrum) 
 
nm588
0eV11.2
nmeV1240
01 =−
⋅=→λ 
nm554
eV11.2eV35.4
nmeV1240
13 =−
⋅=→λ 
and 
nm388
0eV20.3
nmeV1240
02 =−
⋅=→λ 
(not in visible spectrum) 
 
68 •• Singly ionized helium is a hydrogen-like atom that has a nuclear 
charge of +2e. Its energy levels are given by En = –4E0/n2, where n = 1, 2, … and 
E0 = 13.6 eV. If a beam of visible white light is sent through a gas of singly 
ionized helium, at what wavelengths will dark lines be found in the spectrum of 
the transmitted radiation? (Assume that the ions of the gas are all in the same state 
with energy E1 prior to irradiation.) 
 
Determine the Concept The energy difference between the ground state and the 
first excited state is 3E0 = 40.8 eV, corresponding to a wavelength of 30.4 nm. 
This is in the far ultraviolet, well outside the visible range of wavelengths. There 
will be no dark lines in the transmitted radiation. 
 
Chapter 31 
 
2938 
 
69 • [SSM] A pulse from a ruby laser has an average power of 10 MW 
and lasts 1.5 ns. (a) What is the total energy of the pulse? (b) How many photons 
are emitted in this pulse? 
 
Picture the Problem We can use the definition of power to find the total energy 
of the pulse. The ratio of the total energy to the energy per photon will yield the 
number of photons emitted in the pulse. 
 
(a) Use the definition of power to 
obtain: 
 
tPE Δ= 
Substitute numerical values and 
evaluate E: 
 
( )( ) mJ15ns5.1MW10 ==E 
(b) Relate the number of photons N 
to the total energy in the pulse and 
the