Ch33 ISM

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Chapter 33 
Interference and Diffraction 
 
Conceptual Problems 
 
1 • A phase difference due to path-length difference is observed for 
monochromatic visible light. Which phase difference requires the least 
(minimum) path length difference? (a) 90o (b) 180o (c) 270o (d) the answer 
depends on the wavelength of the light. 
 
Determine the Concept The phase difference δ due to a path difference Δr are 
related according to λπδ rΔ=2 . Therefore, the least path length difference 
corresponds to the smallest phase difference. ( )a is correct. 
 
2 • Which of the following pairs of light sources are coherent: (a) two 
candles, (b) one point source and its image in a plane mirror, (c) two pinholes 
uniformly illuminated by the same point source, (d) two headlights of a car, (e) 
two images of a point source due to reflection from the front and back surfaces of 
a soap film. 
 
Determine the Concept Coherent sources have a constant phase difference. The 
pairs of light sources that satisfy this criterion are (b), (c), and (e). 
 
3 • [SSM] The spacing between Newton’s rings decreases rapidly as the 
diameter of the rings increases. Explain qualitatively why this occurs. 
 
Determine the Concept The thickness of the air space between the flat glass and 
the lens is approximately proportional to the square of d, the diameter of the ring. 
Consequently, the separation between adjacent rings is proportional to 1/d. 
 
4 • If the angle of a wedge-shaped air film, such as that in Example 32-2 is 
too large, fringes are not observed. Why? 
 
Determine the Concept There are two possible reasons that fringes might not be 
observed. (1) The distance between adjacent fringes is so small that the fringes 
are not resolved by the eye. (2) Twice the thickness of the air space is greater than 
the coherence length of the light. If this is the case, fringes would be observed in 
the region close to the point where the thickness of the air space approaches zero. 
 
5 • Why must a film that is used to observe interference colors be thin? 
 
Determine the Concept Colors are observed when the light reflected off the front 
and back surfaces of the film interfere destructively for some wavelengths and 
3047 
Chapter 33 
 
 
3048 
constructively for other wavelengths. For this interference to occur, the phase 
difference between the light reflected off the front and back surfaces of the film 
must be constant. This means that twice the thickness of the film must be less 
than the coherence length of the light. The film is called a thin film if twice its 
thickness is less than the coherence length of the light. 
 
6 • A loop of wire is dipped in soapy water and held up so that the soap 
film is vertical. (a) Viewed by reflection with white light, the top of the film 
appears black. Explain why. (b) Below the black region are colored bands. Is the 
first band red or violet? 
 
(a) The phase change due to reflection from the front surface of the film is 180°; 
the phase change due to reflection from the back surface of the film is 0°. As the 
film thins toward the top, the phase change due to the path length difference 
between the two reflected waves (the phase difference associated with the film’s 
thickness) becomes negligible and the two reflected waves interfere destructively. 
 
(b) The first constructive interference will arise when twice the thickness of the 
film is equal to half the wavelength of the color with the shortest wavelength. 
Therefore, the first band will be violet (shortest visible wavelength). 
 
7 • [SSM] A two-slit interference pattern is formed using 
monochromatic laser light with a wavelength of 640 nm. At the second maximum 
from the central maximum, what is the path-length difference between the light 
coming from each of the slits? (a) 640 nm (b) 320 nm (c) 960 nm (d) 1280 nm. 
 
Determine the Concept For constructive interference, the path difference is an 
integer multiple of λ; that is, λmr =Δ . For m = 2, ( )nm 6402=Δr . ( )d is correct.
 
8 • A two-slit interference pattern is formed using monochromatic laser 
light with a wavelength of 640 nm. At the first minimum from the central 
maximum, what is the path-length difference between the light coming from each 
of the slits? (a) 640 nm (b) 320 nm (c) 960 nm (d) 1280 nm. 
 
Determine the Concept For destructive interference, the path difference is an 
odd-integer multiple of λ21 ; that is ( ) ...,5,3,1,21 ==Δ mmr λ . For the first 
minimum, m = 1 and ( ) nm 320nm 64021 ==Δr . ( )b is correct. 
 
9 • A two-slit interference pattern is formed using monochromatic laser 
light with a wavelength of 450 nm. What happens to the distance between the first 
maximum and the central maximum as the two slits are moved closer together? 
Interference and Diffraction 
 
 
3049
(a) The distance increases. (b) The distance decreases. (c) The distance remains 
the same. 
Determine the Concept The relationship between the slit separation d and the 
angular position θm of each maximum is given by ...,2,1,0,sin == mmd m λθ 
(Equation 33-2). Because d and sinθm are inversely proportional for a given 
wavelength and interference maximum (value of m), decreasing d increases sinθm 
and θm. ( )a is correct. 
 
10 • A two-slit interference pattern is formed using two different 
monochromatic lasers, one green and one red. Which color light has its first 
maximum closer to the central maximum? (a) Green, (b) red, (c) both maxima are 
in the same location. 
 
Determine the Concept The relationship between the slit separation d, the 
angular position θm of each maximum, and the wavelength of the light 
illuminating the slits is given by ...,2,1,0 ,sin == mmd m λθ (Equation 33-2). 
Because λ and sinθm are directly proportional for a given interference maximum 
(value of m) and the wavelength of green light is shorter than the wavelength of 
red light, ( )a is correct. 
 
11 • A single slit diffraction pattern is formed using monochromatic laser 
light with a wavelength of 450 nm. What happens to the distance between the first 
maximum and the central maximum as the slit is made narrower? (a) The distance 
increases. (b) The distance decreases. (c) The distance remains the same. 
 
Determine the Concept The relationship between the slit width a, the angular 
position θm of each maximum, and the wavelength of the light illuminating the 
slit is given by ...,3,2,1 ,sin == mma m λθ (Equation 33-11). Because a and 
sinθm are inversely proportional for a given diffraction maximum (value of m), 
narrowing the slit increases mθsin and θm. ( )a is correct. 
 
12 • Equation 33-2 which is d sin θm = mλ, and Equation 33-11, which is 
a sin θm = mλ, are sometimes confused. For each equation, define the symbols 
and explain the equation’s application. 
 
Determine the Concept Equation 33-2 expresses the condition for an intensity 
maximum in two-slit interference. Here d is the slit separation, λ the 
wavelength of the light, m an integer, and θm the angle at which the interference 
maximum appears. Equation 33-11 expresses the condition for an intensity 
minimum in single-slit diffraction. Here a is the width of the slit, λ the 
wavelength of the light, and θ m the angle at which the minimum appears, and m is 
Chapter 33 
 
 
3050 
a nonzero integer. 
 
13 • When a diffraction grating is illuminated by white light, the first-order 
maximum of green light (a) is closer to the central maximum than the first-order 
maximum of red light. (b) is closer to the central maximum than the first-order 
maximum of blue light. (c) overlaps the second-order maximum of red light. 
(d) overlaps the second-order maximum of blue light. 
 
Picture the Problem We can solve λθ md =sin for θ with m = 1 to express the 
location of the first-order maximum as a function of the wavelength of the light. 
 
λθ md =sin 
where d is the separation of the slits 
and m = 0, 1, 2, … 
 
The interference maxima in a 
diffraction pattern are at angles θ 
given by: 
Solvefor the angular location θ1 
of the first-order maximum : ⎟⎠
⎞⎜⎝
⎛= −
d
λθ 11 sin 
 
Because λgreen light < λred light: light redlightgreen θθ < and )(a is correct. 
 
14 • A double-slit interference experiment is set up in a chamber that can be 
evacuated. Using light from a helium-neon laser, an interference pattern is 
observed when the chamber is open to air. As the chamber is evacuated, one will 
note that (a) the interference fringes remain fixed. (b) the interference fringes 
move closer together. (c) the interference fringes move farther apart. (d) the 
interference fringes disappear completely. 
 
Determine the Concept 
 
The distance on the screen to mth 
bright fringe is given by: 
 
d
Lmy nm
λ= 
where L is the distance from the slits to 
the screen, nλ is the wavelength of the 
light in a medium whose index of 
refraction is n, and d is the separation 
of the slits. 
 
The separation of the interference 
fringes is given by: 
 
( )
d
L
d
Lm
d
Lmyy nnnmm
λλλ =−+=−+ 11 
Because the index of refraction of a vacuum is slightly less than the index of 
refraction of air, the removal of air increases nλ and, hence, ym − ym−1. )(c is 
correct. 
Interference and Diffraction 
 
 
3051
 
 
15 • [SSM] True or false: 
 
(a) When waves interfere destructively, the energy is converted into heat 
energy. 
(b) Interference patterns are observed only if the relative phases of the waves 
that superimpose remain constant. 
(c) In the Fraunhofer diffraction pattern for a single slit, the narrower the slit, 
the wider the central maximum of the diffraction pattern. 
(d) A circular aperture can produce both a Fraunhofer diffraction pattern and a 
Fresnel diffraction pattern. 
(e) The ability to resolve two point sources depends on the wavelength of the 
light. 
 
(a) False. When destructive interference of light waves occurs, the energy is no 
longer distributed evenly. For example, light from a two-slit device forms a 
pattern with very bright and very dark parts. There is practically no energy at the 
dark fringes and a great deal of energy at the bright fringe. The total energy over 
the entire pattern equals the energy from one slit plus the energy from the second 
slit. Interference re-distributes the energy. 
 
(b) True. 
 
(c) True. The width of the central maximum in the diffraction pattern is given by 
a
m
m
λθ 1sin −= where a is the width of the slit. Hence, the narrower the slit, the 
wider the central maximum of the diffraction pattern. 
 
(d) True. 
 
(e) True. The critical angle for the resolution of two sources is directly 
proportional to the wavelength of the light emitted by the sources (
D
λα 22.1c = ). 
 
16 • You observe two very closely-spaced sources of white light through a 
circular opening using various filters. Which color filter is most likely to prevent 
your resolving the images on your retinas as coming from two distinct sources? 
(a) red (b) yellow (c) green (d) blue (e) The filter choice is irrelevant. 
 
 
 
 
 
Chapter 33 
 
 
3052 
 
 
Determine the Concept The condition for the resolution of the two sources is 
given by Rayleigh’s criterion: Dc λα 22.1= (Equation 33-25), where αc is the 
critical angular separation and D is the diameter of the aperture. The larger the 
critical angle required for resolution, the less likely it is that you can resolve the 
sources as being two distinct sources. Because αc and λ are directly proportional, 
the filter that passes the shorter wavelength light would be most likely to resolve 
the sources. ( )d is correct. 
 
17 •• Explain why the ability to distinguish the two headlights of an 
oncoming car, at a given distance, is easier for the human eye at night than during 
daylight hours. Assume the headlights of the oncoming car are on during both 
daytime and nighttime hours. 
 
Determine the Concept The condition for the resolution of the two sources is 
given by Rayleigh’s criterion: Dc λα 22.1= (Equation 33-25), where αc is the 
critical angular separation, D is the diameter of the aperture, and λ is the 
wavelength of the light coming from the objects, in this case headlights, to be 
resolved. Because the diameter of the pupils of your eyes are larger at night, the 
critical angle is smaller at night, which means that at night you can resolve the 
light as coming from two distinct sources when they are at a greater distance. 
 
Estimation and Approximation 
 
18 • It is claimed that the Great Wall of China is the only man made object 
that can be seen from space with the naked eye. Check to see if this claim is true, 
based on the resolving power of the human eye. Assume the observers are in low-
Earth orbit that has an altitude of about 250 km. 
 
Picture the Problem We’ll assume that the diameter of the pupil of the eye is 
5.0 mm and use the best-case scenario (the minimum resolvable width varies 
directly with the wavelength of the light reflecting from the object) that the 
wavelength of light is 400 nm (the lower limit for the human eye). Then we can 
use the expression for the minimum angular separation of two objects than can be 
resolved by the eye and the relationship between this angle and the width of an 
object and the distance from which it is viewed to support the claim. 
 
Relate the width w of an object that 
can be seen at an altitude h to the 
critical angular separation αc: 
 
h
w=ctanα ⇒ ctanαhw = 
Interference and Diffraction 
 
 
3053
 
The minimum angular separation αc 
of two point objects that can just be 
resolved by an eye depends on the 
diameter D of the eye and the 
wavelength λ of light: 
 
D
λα 22.1c = 
Substitute for αc in the expression 
for w to obtain: ⎟⎠
⎞⎜⎝
⎛=
D
hw λ22.1tan 
 
Substitute numerical values and evaluate wmin for an altitude of 250 km: 
 
( ) m24
mm0.5
nm40022.1tankm250min ≈⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛=w 
 
This claim is probably false. Because the minimum width that is resolvable from 
low-Earth orbit (250 km) is 24 m and the width of the Great Wall is 5 to 8 m high 
and 5 m wide, so this claim is likely false. However, it is easily seen using 
binoculars, and pictures can be taken of it using a camera. This is because both 
binoculars and cameras have apertures that are larger than the pupil of the human 
eye. (The Chinese astronaut Yang Liwei reported that he was not able to see the 
wall with the naked eye during the first Chinese manned space flight in 2003.) 
 
19 •• [SSM] (a) Estimate how close an approaching car at night on a flat, 
straight stretch of highway must be before its headlights can be distinguished 
from the single headlight of a motorcycle. (b) Estimate how far ahead of you a 
car is if its two red taillights merge to look as if they were one. 
 
Picture the Problem Assume a separation of 1.5 m between typical automobile 
headlights and tail lights, a nighttime pupil diameter of 5.0 mm, 550 nm for the 
wavelength of the light (as an average) emitted by the headlights, 640 nm for red 
taillights, and apply the Rayleigh criterion. 
 
(a) The Rayleigh criterion is given 
by Equation 33-25: 
 
Dc
λα 22.1= 
where D is the separation of the 
headlights (or tail lights). 
 
The critical angular separation is 
also given by: 
 
L
d=α 
where d is the separation of head lights 
(or tail lights) and L is the distance to 
approaching or receding automobile. 
 
Chapter 33 
 
 
3054 
Equate these expressions for αc to 
obtain: 
 
DL
d λ22.1= ⇒ λ22.1
DdL = 
Substitute numerical values and 
evaluate L: 
 
( )( )
( ) km 11nm 55022.1
m 5.1mm 0.5 ≈=L 
(b) For red light: ( )( )
( ) km .69nm 64022.1
m 5.1mm 0.5 ≈=L 
 
20 •• A small loudspeaker is located at a large distance to the east from you. 
The loudspeaker is driven by a sinusoidal current whose frequency can be varied. 
Estimate the lowest frequency for which your ears would receive the sound waves 
exactly out of phase when you are facing north. 
 
Picture the Problem If your ears receive the sound exactly out of phase,the 
waves arriving at your ear that is farthest from the speaker must be traveling one-
half wavelength farther than the waves arriving at your ear that is nearest the 
speaker. The lowest frequency corresponds to the longest wavelength. Assume 
that the speaker and your ears are on the same line and let the distance between 
your ears be about 20 cm. Take the speed of sound in air to be 343 m/s. 
 
The frequency received by your ears 
is given by: λ
vf = 
 
Let Δr be the distance between your 
ears (the path difference) to obtain: 
 
λ21=Δr ⇒ rΔ= 2λ 
Substituting for λ yields: 
r
vf Δ= 2 
 
Substitute numerical values and 
evaluate f: ( ) kHz 86.0cm 202
m/s 343 ==f 
or between 0.80 and 0.90 kHz. 
 
21 •• [SSM] Estimate the maximum distance a binary star system could be 
resolvable by the human eye. Assume the two stars are about fifty times further 
apart than the Earth and Sun are. Neglect atmospheric effects. (A test similar to 
this ″eye test″ was used in ancient Rome to test for eyesight acuity before entering 
the army. A normal eye could just barely resolve two well-known close-together 
stars in the sky. Anyone who could not tell there were two stars was rejected. In 
this case, the stars were not a binary system, but the principle is the same.) 
 
 
 
 
Interference and Diffraction 
 
 
3055
Picture the Problem Assume that the diameter of a pupil at night is 5.0 mm and 
that the wavelength of light is in the middle of the visible spectrum at about 550 
nm. We can use the Rayleigh criterion for the separation of two sources and the 
geometry of the Earth-to-binary star system to derive an expression for the 
distance to the binary stars. 
 
If the distance between the binary 
stars is represented by d and the 
Earth-star distance by L, then their 
angular separation is given by: 
 
L
d=α 
The critical angular separation of the 
two sources is given by the Rayleigh 
criterion: 
 
Dc
λα 22.1= 
For α = αc: 
 DL
d λ22.1= ⇒ λ22.1
DdL = 
 
Substitute numerical values and 
evaluate L: 
( )( )( )
( )
y 5.9
m 109.461
y 1km 1059.5
nm 55022.1
m 105.150mm 0.5
15
13
11
⋅
×
⋅××≈
×=
c
c
L
 
 
Phase Difference and Coherence 
 
22 • Light of wavelength 500 nm is incident normally on a film of water 
1.00 μm thick. (a) What is the wavelength of the light in the water? (b) How 
many wavelengths are contained in the distance 2t, where t is the thickness of the 
film? (c) What is the phase difference between the wave reflected from the top of 
the air–water interface and the wave reflected from the bottom of the water–air 
interface in the region where the two reflected waves superpose? 
 
Picture the Problem The wavelength of light in a medium whose index of 
refraction is n is the ratio of the wavelength of the light in air divided by n. The 
number of wavelengths of light contained in a given distance is the ratio of the 
distance to the wavelength of light in the given medium. The difference in phase 
between the two waves is the sum of a π phase shift in the reflected wave and a 
phase shift due to the additional distance traveled by the wave reflected from the 
bottom of the water−air interface. 
 
(a) Express the wavelength of light 
in water in terms of the wavelength 
of light in air: 
nm376
1.33
nm500
water
air
water === n
λλ 
Chapter 33 
 
 
3056 
(b) Relate the number of 
wavelengths N to the thickness t of 
the film and the wavelength of light 
in water: 
 
( ) 32.5
nm376
m 00.122
water
=== μλ
tN 
(c) Express the phase difference as 
the sum of the phase shift due to 
reflection and the phase shift due to 
the additional distance traveled by 
the wave reflected from the bottom 
of the water−air interface: 
 
Nt πππλπ
δδδ
222
water
 traveleddistance additionalreflection
+=+=
+=
 
( )
rad6.11
rad64.11rad32.52rad
π
πππδ
=
=+=
 
Substitute for N and evaluate δ: 
or, subtracting 11.64π rad from 12π rad,
rad 1.1rad4.0 == πδ 
 
23 •• [SSM] Two coherent microwave sources both produce waves of 
wavelength 1.50 cm. The sources are located in the z = 0 plane, one at 
x = 0, y = 15.0 cm and the other at x = 3.00 cm, y = 14.0 cm. If the sources are in 
phase, find the difference in phase between these two waves for a receiver located 
at the origin. 
 
Picture the Problem The difference in phase depends on the path difference 
according to πλδ 2
rΔ= . The path difference is the difference in the distances of 
(0, 15.0 cm) and (3.00 cm, 14.0 cm) from the origin. 
 
Relate a path difference Δr to a 
phase shift δ: 
 
πλδ 2
rΔ= 
The path difference Δr is: ( ) ( )
cm682.0
cm0.14cm3.00cm0.15 22
=
+−=rΔ
 
Substitute numerical values and 
evaluate δ: rad .922cm50.1
cm682.0 ≈⎟⎟⎠
⎞
⎜⎜⎝
⎛= πδ 
 
Interference in Thin Films 
 
24 • A wedge-shaped film of air is made by placing a small slip of paper 
between the edges of two flat plates of glass. Light of wavelength 700 nm is 
Interference and Diffraction 
 
 
3057
incident normally on the glass plates, and interference fringes are observed by 
reflection. (a) Is the first fringe near the point of contact of the plates dark or 
bright? Why? (b) If there are five dark fringes per centimeter, what is the angle of 
the wedge? 
 
Picture the Problem Because the mth fringe occurs when the path difference 2t 
equals m wavelengths, we can express the additional distance traveled by the light 
in air as an mλ. The thickness of the wedge, in turn, is related to the angle of the 
wedge and the distance from its vertex to the mth fringe. 
 
(a) The first fringe is dark because the phase difference due to reflection by the 
bottom surface of the top plate and the top surface of the bottom plate is 180° 
 
(b) The mth fringe occurs when the 
path difference 2t equals m 
wavelengths: 
λmt =2 
 
 
 
x
t=θ ⇒ θxt = 
where we’ve used a small-angle 
approximation to replace an arc length 
by the length of a chord. 
 
Relate the thickness of the air wedge 
to the angle of the wedge: 
Substitute for t to obtain: λθ mx =2 ⇒ λλθ
x
m
x
m
2
1
2
== 
 
Substitute numerical values and 
evaluate θ : ( ) rad1075.1nm700cm
5
2
1 4−×=⎟⎠
⎞⎜⎝
⎛=θ
 
25 •• [SSM] The diameters of fine fibers can be accurately measured using 
interference patterns. Two optically flat pieces of glass of length L are arranged 
with the wire between them, as shown in Figure 33-40. The setup is illuminated 
by monochromatic light, and the resulting interference fringes are observed. 
Suppose that L is 20.0 cm and that yellow sodium light (wavelength of 590 nm) 
is used for illumination. If 19 bright fringes are seen along this 20.0-cm distance, 
what are the limits on the diameter of the wire? Hint: The nineteenth fringe might 
not be right at the end, but you do not see a twentieth fringe at all. 
 
Picture the Problem The condition that one sees m fringes requires that the path 
difference between light reflected from the bottom surface of the top slide and the 
top surface of the bottom slide is an integer multiple of a wavelength of the light. 
 
Chapter 33 
 
 
3058 
The mth fringe occurs when the path 
difference 2d equals m wavelengths: 
 
λmd =2 ⇒ 
2
λmd = 
Because the nineteenth (but not the 
twentieth) bright fringe can be seen, 
the limits on d must be: 
 
( ) ( )
22 2
1
2
1 λλ +<<− mdm 
where m = 19 
( ) ( )
2
nm59019
2
nm59019 2121 +<<− d Substitute numerical values to obtain: 
or 
m8.5m5.5 μμ << d 
 
26 •• Light that has a wavelength equal to 600 nm is used to illuminate two 
glass plates at normal incidence. The plates are 22 cm in length, touch at one end, 
and are separated at the other end by a wire that has a radius of 0.025 mm. How 
many bright fringes appear along the total length of the plates? 
 
Picture the Problem The light reflected from the top surface of the bottom plate 
(wave 2 in the diagram) is phase shifted relative to the light reflected from the 
bottom surface of the top plate (wave 1 in the diagram). This phase difference is 
thesum of a phase shift of π (equivalent to a λ/2 path difference) resulting from 
reflection plus a phase shift due to the additional distance traveled. 
t
1 2
wire
glass plate
glass plate
 
 
Relate the extra distance traveled by 
wave 2 to the distance equivalent to 
the phase change due to reflection 
and to the condition for constructive 
interference: 
...,3,2,2 21 λλλλ =+t 
or 
...,,,2 252321 λλλ=t 
and ( )λ212 += mt where m = 0, 1, 2, …, 
0 ≤ t ≤ 2r and λ is the wavelength of 
light in air. 
 
 
Interference and Diffraction 
 
 
3059
 
Solving for m gives: 
m = 2tλ −
1
2
 
where m = 0, 1, 2, …and 0 ≤ t ≤ 2r 
 
Solve for the highest value of m: 
mmax = int
2 2r( )
λ −
1
2
⎛
⎝⎜
⎞
⎠⎟ = int
4r
λ −
1
2
⎛
⎝⎜
⎞
⎠⎟ 
where r is the radius of the wire. 
 
Substitute numerical values and 
evaluate m: mmax = int
4 0.025mm( )
600nm
− 1
2
⎛
⎝⎜
⎞
⎠⎟
= int 166.2( )= 166
 
 
Because we start counting from 
m = 0, the number of bright fringes is 
mmax + 1. 
N = mmax + 1 = 167 
 
27 •• A thin film having an index of refraction of 1.50 is surrounded by air. 
It is illuminated normally by white light. Analysis of the reflected light shows that 
the wavelengths 360, 450, and 602 nm are the only missing wavelengths in or 
near the visible portion of the spectrum. That is, for these wavelengths, there is 
destructive interference. (a) What is the thickness of the film? (b) What visible 
wavelengths are brightest in the reflected interference pattern? (c) If this film 
were resting on glass with an index of refraction of 1.60, what wavelengths in the 
visible spectrum would be missing from the reflected light? 
 
Picture the Problem (a) We can use the condition for destructive interference in 
a thin film to find the thickness of the film. (b) and (c) Once we’ve found the 
thickness of the film, we can use the condition for constructive interference to 
find the wavelengths in the visible portion of the spectrum that will be brightest in 
the reflected interference pattern and the condition for destructive interference to 
find the wavelengths of light missing from the reflected light when the film is 
placed on glass with an index of refraction greater than that of the film. 
 
 
 
 
 
 
 
 
Chapter 33 
 
 
3060 
...,,,2 25232121 ''''t λλλλ =+ (a) Express the condition for 
destructive interference in the 
thin film: 
or 
...,3,2,2 '''t λλλ= 
 or 
 
n
m'mt λλ ==2 (1) 
where m = 1, 2, 3, … and λ′ is the 
wavelength of the light in the film. 
 
Solving for λ yields: 
m
nt2=λ 
 
Substitute for the missing 
wavelengths to obtain: m
nt2nm450 = and 
1
2nm360 += m
nt 
 
Divide the first of these equations 
by the second and simplify to 
obtain: 
 
m
m
m
nt
m
nt
1
1
2
2
nm360
nm450 +=
+
= 
 
Solving for m gives: 
 
nm450for 4 == λm 
Solve equation (1) for t to obtain: 
n
mt
2
λ= 
 
Substitute numerical values and 
evaluate t: 
( )
( ) nm60050.12
nm4504 ==t 
 
,...3,2,2 21 ''''t λλλλ =+ (b) The condition for constructive 
interference in the thin film is: or ( ) 'm'''t λλλλ 21252321 ,...,,2 +== 
where λ′ is the wavelength of light in 
the oil and m = 0, 1, 2, … 
 
 
Substitute for λ′ to obtain: ( )
n
mt λ212 += ⇒
2
1
2
+= m
ntλ 
where n is the index of refraction of the 
film. 
 
Substitute numerical values and 
simplify to obtain: 
( )( )
2
1
2
1
nm1800nm60050.12
+=+= mmλ 
Interference and Diffraction 
 
 
3061
 
Substitute numerical values for m and evaluate λ to obtain the following table: 
 
m 0 1 2 3 4 5 
λ (nm) 3600 1200 720 514 400 327 
 
From the table, we see that the only wavelengths in the visible spectrum are 720 
nm, 514 nm, and 400 nm. 
 
...,,,2 252321 '''t λλλ= (c) Because the index of refraction 
of the glass is greater than that of the 
film, the light reflected from the 
film-glass interface will be shifted 
by λ21 (as is the wave reflected from 
the top surface) and the condition for 
destructive interference becomes: 
or 
( )
n
mt λ212 += 
where n is the index of refraction of the 
film and m = 0, 1, 2, … 
 
Solving for λ yields: 
2
1
2
+= m
ntλ 
 
Substitute numerical values and 
simplify to obtain: 
 
( )( )
2
1
2
1
nm1800nm6005.12
+=+= mmλ 
Substitute numerical values for m and evaluate λ to obtain the following table: 
 
m 0 1 2 3 4 5 
λ (nm) 3600 1200 720 514 400 327 
 
From the table we see that the missing wavelengths in the visible spectrum are 
720 nm, 514 nm, and 400 nm. 
 
28 •• A drop of oil (refractive index of 1.22) floats on water (refractive index 
of 1.33). When reflected light is observed from above, as shown in Figure 33-41 
what is the thickness of the drop at the point where the second red fringe, counting 
from the edge of the drop, is observed? Assume red light has a wavelength of 
650 nm. 
 
Picture the Problem Because there is a λ21 phase change due to reflection at 
both the air-oil and oil-water interfaces, the condition for constructive interference 
is that twice the thickness of the oil film equal an integer multiple of the 
wavelength of light in the film. 
 
Chapter 33 
 
 
3062 
,...3,2,2 '''t λλλ= The condition for constructive 
interference is: or 
'mt λ=2 (1) 
where λ′ is the wavelength of light in 
the oil and m = 1, 2, 3, … 
 
 
Substitute for λ′ to obtain: 
n
mt λ=2 ⇒
n
mt
2
λ= 
Substitute numerical values and 
evaluate t: 
( )( )
( ) nm53322.12
nm6502 ==t 
 
29 •• [SSM] A film of oil that has an index of refraction of 1.45 rests on 
an optically flat piece of glass with an index of refraction of 1.60. When 
illuminated by white light at normal incidence, light of wavelengths 690 nm and 
460 nm is predominant in the reflected light. Determine the thickness of the oil 
film. 
 
Picture the Problem Because there is a λ21 phase change due to reflection at 
both the air-oil and oil-glass interfaces, the condition for constructive interference 
is that twice the thickness of the oil film equal an integer multiple of the 
wavelength of light in the film. 
 
Express the condition for 
constructive interference: 
 
'm'''t λλλλ == ,...3,2,2 (1) 
where λ′ is the wavelength of light in 
the oil and m = 0, 1, 2, … 
 
n
mt λ=2 ⇒
m
nt2=λ 
where n is the index of refraction of the 
oil. 
 
Substitute for λ′ to obtain: 
Substitute for the predominant 
wavelengths to obtain: m
nt2nm690 = and 
1
2nm460 += m
nt 
 
Divide the first of these equations by 
the second and simplify to obtain: 
 m
m
m
nt
m
nt
1
1
2
2
nm460
nm690 +=
+
= ⇒ 2=m
 
Solve equation (1) for t: 
n
mt
2
λ= 
 
Interference and Diffraction 
 
 
3063
Substitute numerical values and 
evaluate t: 
( )( )
( ) nm47645.12
nm6902 ==t 
 
30 •• A film of oil that has an index of refraction 1.45 floats on water. When 
illuminated with white light at normal incidence, light of wavelengths 700 nm and 
500 nm is predominant in the reflected light. Determine the thickness of the oil 
film. 
 
Picture the Problem Because the index of refraction of air is less than that of the 
oil, there is a phase shift of π rad ( λ21 ) in the light reflected at the air-oil 
interface. Because the index of refraction of the oil is greater than that of the 
glass, there is no phase shift in the light reflected from the oil-glass interface. We 
can use the condition for constructive interference to determine m for λ = 700 nm 
and then use this value in our equation describing constructive interference to find 
the thickness t of the oil film. 
 
,...3,2,2 21 ''''t λλλλ =+ Express the condition for 
constructive interference between 
the waves reflected from the air-
oil interface and the oil-glass 
interface: 
or ( ) 'm'''t λλλλ 21252321 ,...,,2 +== (1) 
where λ′ is the wavelength of light in 
the oil and m = 0, 1, 2, … 
 
Substitute for 'λ and solve for λ 
to obtain: 21
2
+= m
ntλ 
 
Substitute the predominant 
wavelengths to obtain: 
 
2
1
2nm700 += m
nt and 
2
3
2nm500 += m
nt 
 
Dividethe first of these equations 
by the second to obtain: 
 
2
1
2
3
2
3
2
1
2
2
nm500
nm700
+
+=
+
+=
m
m
m
nt
m
nt
⇒ 2=m
 
Solve equation (1) for t: ( )
n
mt
22
1 λ+= 
 
Substitute numerical values and 
evaluate t: 
( ) ( ) nm60345.12
nm7002 21 =+=t 
 
 
 
Chapter 33 
 
 
3064 
Newton’s Rings 
 
31 •• [SSM] A Newton’s ring apparatus consists of a plano-convex glass 
lens with radius of curvature R that rests on a flat glass plate, as shown in Figure 
33-42. The thin film is air of variable thickness. The apparatus is illuminated from 
above by light from a sodium lamp that has a wavelength of 590 nm. The pattern 
is viewed by reflected light. (a) Show that for a thickness t the condition for a 
bright (constructive) interference ring is ( )λ212 += mt where m = 0, 1, 2, . . . 
(b) Show that for t << R, the radius r of a fringe is related to t by r = 2tR . 
(c) For a radius of curvature of 10.0 m and a lens diameter of 4.00 cm. How many 
bright fringes would you see in the reflected light? (d) What would be the 
diameter of the sixth bright fringe? (e) If the glass used in the apparatus has an 
index of refraction n = 1.50 and water replaces the air between the two pieces of 
glass, explain qualitatively the changes that will take place in the bright-fringe 
pattern. 
 
Picture the Problem This arrangement is essentially identical to a ″thin film″ 
configuration, except that the ″film″ is air. A phase change of 180° ( λ21 ) occurs 
at the top of the flat glass plate. We can use the condition for constructive 
interference to derive the result given in (a) and use the geometry of the lens on 
the plate to obtain the result given in (b). We can then use these results in the 
remaining parts of the problem. 
 
,...3,2,2 21 λλλλ =+t (a) The condition for 
constructive interference is: or ( )λλλλ 21252321 ,...,,2 +== mt 
where λ is the wavelength of light in air 
and m = 0, 1, 2, … 
 
 
Solving for t yields: 
 ( ) ...,2,1,0,221 =+= mmt
λ (1) 
 
( ) 222 RtRr =−+ 
or 
2222 2 tRtRrR +−+= 
 
(b) From Figure 33-42 we have: 
 
For t << R we can neglect the last 
term to obtain: 
 
RtRrR 2222 −+≈ ⇒ Rtr 2= (2) 
(c) Square equation (2) and 
substitute for t from equation (1) to 
obtain: 
( ) λRmr 212 += ⇒ 2
12 −= λR
rm 
Interference and Diffraction 
 
 
3065
Substitute numerical values and 
evaluate m: 
( )
( )( )
fringes.bright 68be will thereso and
67
2
1
nm590m0.10
cm00.2 2 =−=m
 
 
(d) The diameter of the mth fringe is: ( ) λRmrD 2122 +== 
 
Noting that m = 5 for the sixth 
fringe, substitute numerical values 
and evaluate D: 
( )( )( )
cm14.1
nm590m0.1052 21
=
+=D
 
 
(e) The wavelength of the light in the film becomes λair/n = 444 nm. The 
separation between fringes is reduced (the fringes would become more closely 
spaced.) and the number of fringes that will be seen is increased by a factor of 
1.33. 
 
32 •• A plano-convex glass lens of radius of curvature 2.00 m rests on an 
optically flat glass plate. The arrangement is illuminated from above with 
monochromatic light of 520-nm wavelength. The indexes of refraction of the lens 
and plate are 1.60. Determine the radii of the first and second bright fringe from 
the center in the reflected light. 
 
Picture the Problem This arrangement is essentially identical to a ″thin film″ 
configuration, except that the ″film″ is air. A phase change of 180° ( λ21 ) occurs 
at the top of the flat glass plate. We can use the condition for constructive 
interference and the results from Problem 31(b) to determine the radii of the first 
and second bright fringes in the reflected light. 
 
,...3,2,2 21 λλλλ =+t The condition for constructive 
interference is: or ( )λλλλ 21252321 ,...,,2 +== mt 
where λ is the wavelength of light in air 
and m = 0, 1, 2, … 
 
 
Solving for t gives: 
 
( ) ...,2,1,0 ,
22
1 =+= mmt λ 
 
In Problem 31(b) it was shown 
that: 
 
tRr 2= 
 
Substitute for t to obtain: 
 
( ) Rmr λ21+= 
 
Chapter 33 
 
 
3066 
The first fringe corresponds to 
m = 0: 
 
( )( ) mm721.0m00.2nm52021 ==r 
The second fringe corresponds to 
m = 1: 
( )( ) mm25.1m00.2nm52023 ==r 
 
33 ••• Suppose that before the lens of Problem 32 is placed on the plate, a 
film of oil of refractive index 1.82 is deposited on the plate. What will then be the 
radii of the first and second bright fringes? 
 
Picture the Problem This arrangement is essentially identical to a ″thin film″ 
configuration, except that the ″film″ is oil. A phase change of 180° ( λ21 ) occurs 
at lens-oil interface. We can use the condition for constructive interference and 
the results from Problem 31(b) to determine the radii of the first and second 
bright fringes in the reflected light. 
 
,...3,2,2 21 ''''t λλλλ =+ The condition for constructive 
interference is: or ( ) 'm'''t λλλλ 21252321 ,...,,2 +== 
where λ′ is the wavelength of light in 
the oil and m = 0, 1, 2, … 
 
 
Substitute for λ′ and solve for t: 
 
( ) ...,2,1,0,
22
1 =+= m
n
mt λ 
where λ is the wavelength of light in air.
 
In Problem 31(b) it was shown that: 
 
tRr 2= 
 
Substitute for t to obtain: 
 ( ) n
Rmr λ21+= 
 
The first fringe corresponds to 
m = 0: 
 
( )( ) mm535.0
82.1
m00.2nm520
2
1 ==r
 
The second fringe corresponds to 
m = 1: 
( )( ) mm926.0
82.1
m00.2nm520
2
3 ==r
 
Two-Slit Interference Patterns 
 
34 • Two narrow slits separated by 1.00 mm are illuminated by light of 
wavelength 600 nm, and the interference pattern is viewed on a screen 2.00 m 
Interference and Diffraction 
 
 
3067
away. Calculate the number of bright fringes per centimeter on the screen in the 
region near the center fringe. 
 
Picture the Problem The number of bright fringes per unit distance is the 
reciprocal of the separation of the fringes. We can use the expression for the 
distance on the screen to the mth fringe to find the separation of the fringes. 
 
Express the number N of bright 
fringes per centimeter in terms of 
the separation of the fringes: 
 
y
N Δ=
1 (1) 
Express the distance on the screen to 
the mth and (m + 1)st bright fringe: 
 
d
Lmym
λ= and ( )
d
Lmym
λ11 +=+ 
Subtract the first of these equations 
from the second to obtain: d
Ly λ=Δ 
 
Substitute in equation (1) to obtain: 
 L
dN λ= 
 
Substitute numerical values and 
evaluate N: ( )( ) 1cm33.8m00.2nm600
mm00.1 −==N 
 
35 • [SSM] Using a conventional two-slit apparatus with light of 
wavelength 589 nm, 28 bright fringes per centimeter are observed near the center 
of a screen 3.00 m away. What is the slit separation? 
 
Picture the Problem We can use the expression for the distance on the screen to 
the mth and (m + 1)st bright fringes to obtain an expression for the separation Δy 
of the fringes as a function of the separation of the slits d. Because the number of 
bright fringes per unit length N is the reciprocal of Δy, we can find d from N, λ, 
and L. 
 
Express the distance on the screen to 
the mth and (m + 1)st bright fringe: 
 
d
Lmym
λ= and ( )
d
Lmym
λ11 +=+ 
Subtract the first of these equations 
from the second to obtain: 
 
d
Ly λ=Δ ⇒
y
Ld Δ=
λ 
Because the number of fringes per 
unit length N is the reciprocal of Δy: 
 
LNd λ= 
Chapter 33 
 
 
3068 
Substitute numerical values and 
evaluate d: 
( )( )( )
mm95.4
m00.3nm589cm28 1
=
= −d
 
 
36 • Light of wavelength 633 nm from a helium–neon laser is shone 
normally on a plane containing two slits. The first interference maximum is 82 cm 
from the central maximum on a screen 12 m away. (a) Find the separation of the 
slits. (b) How many interference maxima is it, in principle, possible to observe? 
 
Picture the Problem We can use the 
geometry of the setup, represented to 
the right, to find the separation of the 
slits. To find the number of interference 
maxima that, in principle, can be 
observed, we can apply the equation 
describing two-slit interferencemaxima 
and require that sinθ ≤ 1. 
d
 cm 821 =y
 m 12=L
θ
θ1
λ
0
 
 
Because d << L, we can approximate 
sinθ1 as: 
 
d
λθ ≈1sin ⇒
1sinθ
λ≈d (1) 
From the right triangle whose sides 
are L and y1 we have: 
 
( ) ( ) 06817.0cm82m12
cm82sin
221
=
+
=θ 
 
Substitute numerical values in 
equation (1) and evaluate d: 
m3.9m29.9
06817.0
nm633 μμ ==≈d 
 
(b) The equation describing two-slit 
interference maxima is: 
 
...,2,1,0sin == m,md λθ 
Because sinθ ≤ 1 determines the 
maximum number of interference 
fringes that can be seen: 
 
λmaxmd = ⇒ λ
dm =max 
Substitute numerical values and 
evaluate mmax: 
14
nm633
m29.9
max == μm because m must 
be an integer. 
 
Because there are 14 fringes on 
either side of the central maximum: 
( ) 29114212 max =+=+= mN 
 
Interference and Diffraction 
 
 
3069
37 •• Two narrow slits are separated by a distance d. Their interference 
pattern is to be observed on a screen a large distance L away. (a) Calculate the 
spacing between successive maxima near the center fringe for light of wavelength 
500 nm, when L is 1.00 m and d is 1.00 cm. (b) Would you expect to be able to 
observe the interference of light on the screen for this situation? (c) How close 
together should the slits be placed for the maxima to be separated by 1.00 mm for 
this wavelength and screen distance? 
 
Picture the Problem We can use the equation for the distance on a screen to the 
mth bright fringe to derive an expression for the spacing of the maxima on the 
screen. In (c) we can use this same relationship to express the slit separation d. 
 
(a) Express the distance on the 
screen to the mth and (m + 1)st bright 
fringe: 
 
d
Lmym
λ= and ( )
d
Lmym
λ11 +=+ 
Subtract the first of these equations 
from the second to obtain: 
 
d
Ly λ=Δ (1) 
 
Substitute numerical values and 
evaluate Δy: 
 
( )( ) m0.50
cm00.1
m00.1nm500 μ==Δy 
(b) According to the Raleigh criterion you could resolve them, but not by much. 
 
(c) Solve equation (1) for d to obtain: 
y
Ld Δ=
λ 
 
Substitute numerical values and 
evaluate d: 
( )( ) mm500.0
mm00.1
m00.1nm500 ==d 
 
38 •• Light is incident at an angle φ with the normal to a vertical plane 
containing two slits of separation d (Figure 33-43). Show that the interference 
maxima are located at angles θm given by sin θm + sin φ = mλ/d. 
 
Picture the Problem Let the separation of the slits be d. We can find the total 
path difference when the light is incident at an angle φ and set this result equal to 
an integer multiple of the wavelength of the light to obtain the given equation. 
 
Express the total path difference: 
 
msinsin θφ dd +=Δl 
The condition for constructive 
interference is: 
λm=Δl 
where m is an integer. 
Chapter 33 
 
 
3070 
Substitute to obtain: λθφ mdd =+ msinsin 
 
Divide both sides of the equation by 
d to obtain: d
mλθφ =+ msinsin 
 
39 •• [SSM] White light falls at an angle of 30º to the normal of a plane 
containing a pair of slits separated by 2.50 μm. What visible wavelengths give a 
bright interference maximum in the transmitted light in the direction normal to the 
plane? (See Problem 38.) 
 
Picture the Problem Let the separation of the slits be d. We can find the total 
path difference when the light is incident at an angle φ and set this result equal to 
an integer multiple of the wavelength of the light to relate the angle of incidence 
on the slits to the direction of the transmitted light and its wavelength. 
 
Express the total path difference: 
 
θφ sinsin dd +=Δl 
The condition for constructive 
interference is: 
 
λm=Δl 
where m is an integer. 
Substitute to obtain: λθφ mdd =+ sinsin 
 
Divide both sides of the equation 
by d to obtain: d
mλθφ =+ sinsin 
 
Set θ = 0 and solve for λ: 
m
d φλ sin= 
 
Substitute numerical values and 
simplify to obtain: 
( )
mm
m25.130sinm50.2 μμλ =°= 
 
Evaluate λ for positive integral values of m: 
 
m λ (nm) 
1 1250 
2 625 
3 417 
4 313 
 
From the table we can see that 625 nm and 417 nm are in the visible portion of the 
electromagnetic spectrum. 
 
Interference and Diffraction 
 
 
3071
40 •• Two small loudspeakers are separated by 5.0 cm, as shown in Figure 
33-44. The speakers are driven in phase with a sine wave signal of frequency 10 
kHz. A small microphone is placed a distance 1.00 m away from the speakers on 
the axis running through the middle of the two speakers, and the microphone is 
then moved perpendicular to the axis. Where does the microphone record the first 
minimum and the first maximum of the interference pattern from the speakers? 
The speed of sound in air is 343 m/s. 
 
Picture the Problem The diagram 
shows the two speakers, S1 and S2, the 
central-bright image and the first-order 
image to the left of the central-bright 
image. The distance y is measured from 
the center of the central-bright image. 
We can apply the conditions for 
constructive and destructive 
interference from two sources and use 
the geometry of the speakers and 
microphone to find the distance to the 
first interference minimum and the 
distance to the first interference 
maximum. S S21
θ
d
L
y
 
 
Relate the distance Δy to the first 
minimum from the center of the 
central maximum to θ and the 
distance L from the speakers to the 
plane of the microphone: 
 
L
y=θtan ⇒ θtanLy = (1) 
Interference minima occur where: 
 
( )λθ 21sin += md 
where m = 0, 1, 2, 3, … 
 
Solve for θ to obtain: 
 
( )
⎥⎦
⎤⎢⎣
⎡ += −
d
m λθ 211sin 
 
Relate the wavelength λ of the sound 
waves to the speed of sound v and 
the frequency f of the sound: 
 
f
v=λ 
Substitute for λ in the expression for 
θ to obtain: 
 
( ) ⎥⎦
⎤⎢⎣
⎡ += −
df
vm 211sinθ 
 
Substituting for θ in equation (1) 
yields: 
 
( )
⎭⎬
⎫
⎩⎨
⎧
⎥⎦
⎤⎢⎣
⎡ += −
df
vmLy 2
1
1sintan (2) 
 
Chapter 33 
 
 
3072 
Noting that the first minimum corresponds to m = 0, substitute numerical values 
and evaluate Δy: 
 
( ) ( )( )( )( ) cm37kHz10cm0.5
m/s343
sintanm00.1 2
1
1
min1st =⎭⎬
⎫
⎩⎨
⎧
⎥⎦
⎤⎢⎣
⎡= −y 
 
The maxima occur where: 
 
λθ md =sin 
where m = 1, 2, 3, … 
 
For diffraction maxima, equation (2) 
becomes: 
 ⎭
⎬⎫⎩⎨
⎧
⎥⎦
⎤⎢⎣
⎡= −
af
mvLy 1sintan 
 
Noting that the first maximum corresponds to m = 1, substitute numerical values 
and evaluate y: 
 
( ) ( )( )( )( ) cm94kHz10cm0.5
m/s3431sintanm00.1 1max1st =⎭⎬
⎫
⎩⎨
⎧
⎥⎦
⎤⎢⎣
⎡= −y 
 
Diffraction Pattern of a Single Slit 
 
41 • Light that has a 600-nm wavelength is incident on a long narrow slit. 
Find the angle of the first diffraction minimum if the width of the slit is 
(a) 1.0 mm, (b) 0.10 mm, and (c) 0.010 mm. 
 
Picture the Problem We can use the expression locating the first zeroes in the 
intensity to find the angles at which these zeroes occur as a function of the slit 
width a. 
 
The first zeroes in the intensity occur 
at angles given by: 
 
a
λθ =sin ⇒ ⎟⎠
⎞⎜⎝
⎛= −
a
λθ 1sin 
(a) For a = 1.0 mm: 
 mrad60.0mm0.1
nm600sin 1 =⎟⎟⎠
⎞
⎜⎜⎝
⎛= −θ 
 
(b) For a = 0.10 mm: 
 mrad0.6mm10.0
nm600sin 1 =⎟⎟⎠
⎞
⎜⎜⎝
⎛= −θ 
 
(c) For a = 0.010 mm: 
 mrad60mm010.0
nm600sin 1 =⎟⎟⎠
⎞
⎜⎜⎝
⎛= −θ 
 
Interference and Diffraction 
 
 
3073
42 • Plane microwaves are incident on the thin metal sheet that has a long, 
narrow slit of width 5.0 cm in it. The microwave radiation strikes the sheet at 
normal incidence. The first diffraction minimum is observed at θ = 37º. What is 
the wavelength of the microwaves? 
 
Picture the Problem We can use the expression locating the first zeroes in the 
intensity to find the wavelength of the radiation as a function of the angle at 
which the first diffraction minimum is observed and the width of the plate. 
 
The first zeroes in the intensity occur 
at angles given by: 
 
a
λθ =sin ⇒ θλ sina= 
Substitute numerical values and 
evaluate λ: 
( ) cm0.337sincm0.5 =°=λ 
 
43 ••• [SSM] Measuring the distance to the moon (lunarranging) is 
routinely done by firing short-pulse lasers and measuring the time it takes for the 
pulses to reflect back from the moon. A pulse is fired from Earth. To send the 
pulse out, the pulse is expanded so that it fills the aperture of a 6.00-in-diameter 
telescope. Assuming the only thing spreading the beam out is diffraction and that 
the light wavelength is 500 nm, how large will the beam be when it reaches the 
Moon, 3.82 × 105 km away? 
 
Picture the Problem The diagram shows the beam expanding as it travels to the 
moon and that portion of it that is reflected from the mirror on the moon 
expanding as it returns to Earth. We can express the diameter of the beam at the 
moon as the product of the beam divergence angle and the distance to the moon 
and use the equation describing diffraction at a circular aperture to find the beam 
divergence angle. 
L Ddtelescope dmirror
 
 
Relate the diameter D of the beam 
when it reaches the moon to the 
distance to the moon L and the beam 
divergence angle θ : 
 
LD θ≈ (1) 
Chapter 33 
 
 
3074 
The angle θ subtended by the first 
diffraction minimum is related to the 
wavelength λ of the light and the 
diameter of the telescope opening 
dtelescope by: 
 
telescope
22.1sin
d
λθ = 
Because θ << 1, sinθ ≈ θ and: 
 
telescope
22.1
d
λθ ≈ 
 
Substitute for θ in equation (1) to 
obtain: 
telescope
22.1
d
LD λ= 
 
Substitute numerical values and evaluate D: 
 
( ) ( ) km53.1
cm10
m1
in
cm2.54in00.6
nm50022.1m1082.3
2
8 =
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
××
×=D 
 
 
Interference-Diffraction Pattern of Two Slits 
 
44 • How many interference maxima will be contained in the central 
diffraction maximum in the interference–diffraction pattern of two slits if the 
separation of the slits is exactly 5 times their width? How many will there be if 
the slit separation is an integral multiple of the slit width (that is d ) for any 
value of n? 
= na
 
Picture the Problem We need to find the value of m for which the mth 
interference maximum coincides with the first diffraction minimum. Then there 
will be fringes in the central maximum. 12 −= mN
 
The number of fringes N in the 
central maximum is: 
 
12 −= mN (1) 
Relate the angle θ1 of the first 
diffraction minimum to the width a 
of the slits of the diffraction grating: 
 
a
λθ =1sin 
Express the angle θm corresponding 
to the mth interference maxima in 
terms of the separation d of the slits: 
d
m
m
λθ =sin 
Interference and Diffraction 
 
 
3075
Because we require that θ1 = θm, 
we can equate these expressions 
to obtain: 
 
ad
m λλ = ⇒ 
a
dm = 
Substituting for d and simplifying 
yields: 
55 ==
a
am
 
 
Substitute for m in equation (1) to 
obtain: 
 
( ) 9152 =−=N
 
 
If d = na: n
a
na
a
dm ===
 
and 12 −= nN 
 
45 •• [SSM] A two-slit Fraunhofer interference–diffraction pattern is 
observed using light that has a wavelength equal to 500 nm. The slits have a 
separation of 0.100 mm and an unknown width. (a) Find the width if the fifth 
interference maximum is at the same angle as the first diffraction minimum. 
(b) For that case, how many bright interference fringes will be seen in the central 
diffraction maximum? 
 
Picture the Problem We can equate the sine of the angle at which the first 
diffraction minimum occurs to the sine of the angle at which the fifth interference 
maximum occurs to find a. We can then find the number of bright interference 
fringes seen in the central diffraction maximum using .12 −= mN 
 
(a) Relate the angle θ1 of the first 
diffraction minimum to the width a 
of the slits of the diffraction grating: 
 
a
λθ =1sin 
Express the angle θ5 corresponding 
to the mth fifth interference maxima 
maximum in terms of the separation 
d of the slits: 
 
d
λθ 5sin 5 = 
Because we require that θ1 = θm5, we 
can equate these expressions to 
obtain: 
 
ad
λλ =5 ⇒ 
5
da = 
Substituting the numerical value of d 
yields: 
m0.20
5
mm100.0 μ==a 
 
(b) Because m = 5: ( ) 915212 =−=−= mN 
Chapter 33 
 
 
3076 
46 •• A two-slit Fraunhofer interference–diffraction pattern is observed 
using light that has a wavelength equal to 700 nm. The slits have widths of 
0.010 mm and are separated by 0.20 mm. How many bright fringes will be seen in 
the central diffraction maximum? 
 
Picture the Problem We can equate the sine of the angle at which the first 
diffraction minimum occurs to the sine of the angle at which the mth interference 
maximum occurs to find m. We can then find the number of bright interference 
fringes seen in the central diffraction maximum using .12 −= mN 
 
The number of fringes N in the 
central maximum is: 
 
12 −= mN (1) 
Relate the angle θ1 of the first 
diffraction minimum to the width a 
of the slits of the diffraction grating: 
 
a
λθ =1sin 
Express the angle θm corresponding 
to the mth interference maxima in 
terms of the separation d of the slits: 
 
d
m
m
λθ =sin 
Because we require that θ1 = θm, we 
can equate these expressions to 
obtain: 
 
ad
m λλ = ⇒
a
dm = 
Substitute for m in equation (1) to 
obtain: 
12 −=
a
dN 
 
Substitute numerical values and 
evaluate N: 
( ) 391
mm010.0
mm20.02 =−=N 
 
47 •• Suppose that the central diffraction maximum for two slits has 17 
interference fringes for some wavelength of light. How many interference fringes 
would you expect in the diffraction maximum adjacent to one side of the central 
diffraction maximum? 
 
Picture the Problem There are 8 interference fringes on each side of the central 
maximum. The secondary diffraction maximum is half as wide as the central one. 
It follows that it will contain 8 interference maxima. 
 
 
 
Interference and Diffraction 
 
 
3077
48 •• Light that has a wavelength equal to 550 nm illuminates two slits that 
both have widths equal to 0.030 mm and separations equal to 0.15 mm. 
(a) How many interference maxima fall within the full width of the central 
diffraction maximum? (b) What is the ratio of the intensity of the third 
interference maximum to one side of the center interference maximum to the 
intensity of the center interference maximum? 
 
Picture the Problem We can equate the sine of the angle at which the first 
diffraction minimum occurs to the sine of the angle at which the mth interference 
maximum occurs to find m. We can then find the number of bright interference 
fringes seen in the central diffraction maximum using .12 −= mN In (b) we can 
use the expression relating the intensity in a single-slit diffraction pattern to phase 
constant θλ
πφ sin2 a= to find the ratio of the intensity of the third interference 
maximum to one side of the center interference maximum. 
 
(a) The number of fringes N in the 
central maximum is: 
 
12 −= mN (1) 
Relate the angle θ1 of the first 
diffraction minimum to the width a 
of the slits of the diffraction grating: 
 
a
λθ =1sin 
Express the angle θm corresponding 
to the mth interference maxima in 
terms of the separation d of the slits: 
 
d
m
m
λθ =sin 
Because we require that θ1 = θm, we 
can equate these expressions to 
obtain: 
 
ad
m λλ = ⇒
a
dm = 
Substitute in equation (1) to obtain: 12 −=
a
dN 
 
Substitute numerical values and 
evaluate N: 
( ) 91
mm030.0
mm15.02 =−=N 
 
(b) Express the intensity for a single-
slit diffraction pattern as a function 
of the phase difference φ: 
2
2
1
2
1
0
sin
⎟⎟⎠
⎞
⎜⎜⎝
⎛= φ
φII (2) 
where θλ
πφ sin2 a=
 
 
Chapter 33 
 
 
3078 
d
λθ 3sin 3 = For m = 3: 
and 
⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞⎜⎝
⎛==
d
a
d
aa πλλ
πθλ
πφ 632sin2 3 
 
Substitute numerical values and 
evaluate φ: 5
6
mm15.0
mm030.06 ππφ =⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
 
Solve equation (2) for the ratio of I3 
to I0: 
 
2
2
1
2
1
0
3 sin ⎟⎟⎠
⎞
⎜⎜⎝
⎛= φ
φ
I
I
 
 
Substitute numerical values and 
evaluate I3/I0: 
25.05
6
2
1
5
6
2
1sin
2
0
3 =
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
⎟⎠
⎞⎜⎝
⎛
⎟⎠
⎞⎜⎝
⎛
= π
π
I
I 
 
Using Phasors to Add Harmonic Waves 
 
49 • [SSM] Find the resultant of the two waves whose electric fields at a 
given location vary with time as follows: iE ˆsin2 01 tA ω=
r
and 
( )iE ˆsin3 2302 πω += tAr . 
 
Picture the Problem Chose the 
coordinate system shown in the 
phasor diagram. We can use the 
standard methods of vector addition 
to find the resultant of the two waves. 
y
x
δ
1E
r
2E
r
 E
r
 
 
The resultant of the two waves is of 
the form: 
 
( )iEE δω += tsinrr (1) 
The magnitude of E
r
is: ( ) ( ) 02020 6.332 AAA =+=Er 
 
The phase angle δ is: 
rad 98.0
2
3tan
0
01 −=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= −
A
Aδ 
 
Interference and Diffraction 
 
 
3079
Substitute for E
r
and δ in equation 
(1) to obtain: 
( )iE rad 98.0sin6.3 0 −= tA ωr 
 
50 • Find the resultant of the two waves whose electric fields at a given 
location vary with time as follows: iE ˆsin4 01 tA ω=
r
and ( )iE ˆsin3 6102 πω += tAr . 
 
Picture the Problem Chose the coordinate system shown in the phasor diagram. 
We can use the standard methods of vector addition to find the resultant of the 
two waves. 
y
x
δ °60
 
1E
r
2E
r
 E
r
 
 
The resultant of the two waves is of 
the form: 
 
( )iEE δω += tsinrr (1) 
°−+= 120cos2 1
22
1
2
22 EEEEE
rrrrr
 
or 
°−+= 120cos2 1
22
1 22 EEEEE
rrrrr
 
Apply the law of cosines to the 
magnitudes of the scalars to obtain: 
 
 
Substitute for 1E
r
and 2E
r
 and evaluate E
r
to obtain: 
 
( ) ( ) ( )( ) 0002020 08.6120cos34234 AAAAA =°−+=Er 
 
Applying the law of sines yields: 
EE
rr °= 120sinsin
2
δ 
 
Solve for δ to obtain: 
⎥⎥⎦
⎤
⎢⎢⎣
⎡ °= −
E
E
r
r
120sin
sin
21δ 
 
Substitute numerical values and 
evaluate δ: rad 43.008.6
120sin3sin
0
01 =⎥⎦
⎤⎢⎣
⎡ °= −
A
Aδ 
 
Chapter 33 
 
 
3080 
Substitute for E
r
and δ in equation 
(1) to obtain: 
 
( )iE rad 43.0sin1.6 0 += tA ωr 
Remarks: We could have used the law of cosines to find R and the law of 
sines to find δ. 
 
51 •• Monochromatic light is incident on a sheet with a long narrow slit 
(Figure 33-45). Let I0 be the intensity at the central maximum of the diffraction 
pattern on a distant screen, and let I be the intensity at the second intensity 
maximum from the central intensity maximum. The distance from this second 
intensity maximum to the far edge of the slit is longer than the distance from the 
second intensity maximum to the near edge of the slit by approximately 2.5 
wavelengths. What is the ratio if I to I0? 
 
Picture the Problem We can evaluate the expression for the intensity for a 
single-slit diffraction pattern at the second secondary maximum to express I2 in 
terms of I0. 
 
The intensity at the second secondary 
maximum is given by: 
 
⇒⎥⎦
⎤⎢⎣
⎡=
2
2
1
2
1
0
sin
φ
φ
II
2
2
1
2
1
0
sin ⎥⎦
⎤⎢⎣
⎡= φ
φ
I
I 
where 
θλ
πφ sin2 a=
 
 
At this second secondary maximum: λθ
2
5sin =a and πλλ
πφ 5
2
52 =⎟⎠
⎞⎜⎝
⎛= 
 
Substitute for φ and evaluate 
0I
I : 
0162.0
2
5
2
5sin
2
0
=
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡ ⎟⎠
⎞⎜⎝
⎛
= π
π
I
I 
 
52 •• Monochromatic light is incident on a sheet that has three long narrow 
parallel equally spaced slits a distance d apart. (a) Show that the positions of the 
interference minima on a screen a large distance L away from the sheet that has 
the three equally spaced slits (spacing d, with d >> λ) are given approximately by 
dLmym 3λ= where m = 1, 2, 4, 5, 7, 8, 10, . . . that is, m is not a multiple of 3. 
(b) For a screen distance of 1.00 m, a light wavelength of 500 nm, and a source 
spacing of 0.100 mm, calculate the width of the principal interference maxima 
(the distance between successive minima) for three sources. 
Interference and Diffraction 
 
 
3081
Picture the Problem We can use phasor concepts to find the phase angle δ in 
terms of the number of phasors N (three in this problem) forming a closed 
polygon of N sides at the minima and then use this information to express the path 
difference Δr for each of these locations. Applying a small angle approximation, 
we can obtain an expression for y that we can evaluate for enough of the path 
differences to establish the pattern given in the problem statement. 
 
(a) Express the phase angle δ in 
terms of the number of phasors N 
forming a closed polygon of N sides: 
 
⎟⎠
⎞⎜⎝
⎛=
N
m πδ 2 
where m = 1, 2, 3, 4, 5, 6, ,7, … 
For three equally spaced sources, 
the phase angle is: 
 
⎟⎠
⎞⎜⎝
⎛=
3
2πδ m 
 
Express the path difference 
corresponding to this phase angle to 
obtain: 
 
32
λδπ
λ mr =⎟⎠
⎞⎜⎝
⎛=Δ (1) 
Interference maxima occur for: 
 
m = 3, 6, 9, 12, … 
Interference minima occur for: 
 
m = 1, 2, 4, 5, 7, 8, … 
(Note that m is not a multiple of 3.) 
 
θsindr =Δ 
or, provided the small angle 
approximation is valid, 
L
ydr =Δ ⇒ r
d
Ly Δ= 
 
Express the path difference Δr in 
terms of sinθ and the separation 
d of the slits: 
 
Substituting for Δr from equation (1) 
yields: 
 
...,8,7,5,4,2,1 ,
3min
== m
d
Lmy λ 
 
(b) For L = 1.00 m, λ = 500 nm, and 
d = 0.100 mm: 
( )( )
( ) mm33.3mm0.1003
m00.1nm50022 min ==y
 
53 •• [SSM] Monochromatic light is incident on a sheet that has four long 
narrow parallel equally spaced slits a distance d apart. (a) Show that the positions 
of the interference minima on a screen a large distance L away from four equally 
spaced sources (spacing d, with d >> λ) are given approximately by 
dLmym 4λ= where m = 1, 2, 3, 5, 6, 7, 9, 10, . . . that is, m is not a multiple of 
4. (b) For a screen distance of 2.00 m, light wavelength of 600 nm, and a source 
Chapter 33 
 
 
3082 
spacing of 0.100 mm, calculate the width of the principal interference maxima 
(the distance between successive minima) for four sources. Compare this width 
with that for two sources with the same spacing. 
 
Picture the Problem We can use phasor concepts to find the phase angle δ in 
terms of the number of phasors N (four in this problem) forming a closed polygon 
of N sides at the minima and then use this information to express the path 
difference Δr for each of these locations. Applying a small angle approximation, 
we can obtain an expression for y that we can evaluate for enough of the path 
differences to establish the pattern given in the problem statement. 
 
(a) Express the phase angle δ in 
terms of the number of phasors N 
forming a closed polygon of N sides: 
 
⎟⎠
⎞⎜⎝
⎛=
N
m πδ 2 
where m = 1, 2, 3, 4, 5, 6, ,7, … 
For four equally spaced sources, the 
phase angle is: 
 
⎟⎠
⎞⎜⎝
⎛=
2
πδ m 
 
Express the path difference 
corresponding to this phase angle to 
obtain: 
 
42
λδπ
λ mr =⎟⎠
⎞⎜⎝
⎛=Δ (1) 
Interference maximum occur for: m = 3, 6, 9, 12, … 
 
Interference minima occur for: 
 
m = 1, 2, 4, 5, 7, 8, … 
(Note that m is not a multiple of 3.) 
 
θsindr =Δ 
or, provided the small angle 
approximation is valid, 
L
ydr =Δ ⇒ r
d
Ly Δ= 
 
Express the path difference Δr in 
terms of sinθ and the separation 
d of the slits: 
 
Substituting for Δr from equation (1) 
yields: 
 
,...9,7,6,5,3,2,1 ,
4min
== m
d
Lmy λ 
 
(b) For L = 2.00 m, λ = 600 nm, 
d = 0.100 mm, and n = 1: 
 
( )( )
( ) mm00.6mm0.1004
m00.2nm60022 min ==y
For two slits: ( )
d
Lmy λ21min 22 += 
 
Interference and Diffraction 
 
 
3083
For L = 2.00 m, λ = 600 nm, 
d = 0.100 mm, and m = 0: 
 
( )( ) mm0.12
mm0.100
m00.2nm6002 min ==y 
The width for four sources is half the width for two sources. 
 
54 •• Light of wavelength 480 nm falls normally on four slits. Each slit is 
2.00 μm wide and the center-to-center separation between it and the next slit is 
6.00 μm. (a) Find the angular width of the of the central intensity maximum of 
the single-slit diffraction pattern on a distant screen. (b) Find the angular position 
of all interference intensity maxima that lie inside the central diffraction 
maximum.(c) Find the angular width of the central interference. That is, find the 
angle between the first intensity minima on either side of the central intensity 
maximum. (d) Sketch the relative intensity as a function of the sine of the angle. 
 
Picture the Problem We can use aλθ =sin to find the first zeros in the intensity 
pattern. The four-slit interference maxima occur at angles given by 
,md λθ =sin where m = 0,1,2, … . In (c) we can use the result of Problem 53 to 
find the angular spread between the central interference maximum and the first 
interference minimum on either side of it. In (d) we’ll use a phasor diagram for a 
four-slit grating to find the resultant amplitude at a given point in the intensity 
pattern as a function of the phase constant δ, that, in turn, is a function of the angle 
θ that determines the location of a point in the interference pattern. 
 
(a) The first zeros in the intensity 
occur at angles given by: a
λθ =sin ⇒ ⎟⎠
⎞⎜⎝
⎛= −
a
λθ 1sin 
 
Substitute numerical values and 
evaluate θ : mrad242m00.2
nm480sin 1 =⎟⎟⎠
⎞
⎜⎜⎝
⎛= − μθ 
 
(b) The four-slit interference 
maxima occur at angles given by: 
 
λθ md =sin ⇒ ⎥⎦
⎤⎢⎣
⎡= −
d
m
m
λθ 1sin 
where m = 0, 1, 2, 3, … 
 
Substitute numerical values to 
obtain: 
( )
( )m
m
m
0800.0sin
m00.6
nm480sin
1
1
−
−
=
⎥⎦
⎤⎢⎣
⎡= μθ 
 
Chapter 33 
 
 
3084 
 
( )[ ] 00800.00sin 10 == −θ Evaluate θm for m = 0, 1, 2, and 3: 
( )[ ] mrad1.800800.01sin 11 ±== −θ 
( )[ ] mrad1610800.02sin 12 ±== −θ 
( )[ ] rad.24200800.03sin 13 ±== −θ 
where θ3 will not be seen as it 
coincides with the first minimum in the 
diffraction pattern. 
 
(c) From Problem 53: 
d
n
4min
λθ = 
 
For n = 1: ( ) mrad20m00.64
nm480
min == μθ 
 
(d) Use the phasor method to show the superposition of four waves of the same 
amplitude A0 and constant phase difference .sin
2 θλ
πδ d= 
 
A0
A0
A
A0
A0
f
α
δ δ
δ
δ
δ
"
"
δ '
φ
φ
α
 
 
Express A in terms of δ ′ and δ ′′: 
 
( )'A''AA δδ coscos2 00 += (1) 
Because the sum of the external 
angles of a polygon equals 2π: 
 
πδα 232 =+ 
 
Interference and Diffraction 
 
 
3085
Examining the phasor diagram we 
see that: 
 
πδα =+ '' 
Eliminate α and solve for ''δ to 
obtain: 
 
δδ 23='' 
Because the sum of the internal 
angles of a polygon of n sides is 
(n − 2)π : 
 
πδφ 323 =+ '' 
From the definition of a straight 
angle we have: 
 
πδδφ =+− ' 
Eliminate φ between these equations 
to obtain: 
δδ 21=' 
 
Substitute for ''δ and 'δ in equation 
(1) to obtain: 
 
( )δδ 21230 coscos2 += AA 
Because the intensity is 
proportional to the square of the 
amplitude of the resultant wave: 
 
( )221230 coscos4 δδ += II 
The following graph of I/I0 as a function of sinθ was plotted using a spreadsheet 
program. The diffraction envelope was plotted using ,sin4
2
2
1
2
1
2
0
⎟⎟⎠
⎞
⎜⎜⎝
⎛= φ
φ
I
I where 
.sin2 θλ
πφ a= Note the excellent agreement with the results calculated in (a), (b) 
and (c). 
 
Chapter 33 
 
 
3086 
-2
0
2
4
6
8
10
12
14
16
18
-0.3 -0.2 -0.1 0 0.1 0.2 0.3
sin(theta)
I /I 0
intensity
diffraction envelope
 
 
55 ••• [SSM] Three slits, each separated from its neighbor by 60.0 μm, are 
illuminated at the central intensity maximum by a coherent light source of 
wavelength 550 nm. The slits are extremely narrow. A screen is located 2.50 m 
from the slits. The intensity is 50.0 mW/m2. Consider a location 1.72 cm from the 
central maximum. (a) Draw a phasor diagram suitable for the addition of the three 
harmonic waves at that location. (b) From the phasor diagram, calculate the 
intensity of light at that location. 
 
Picture the Problem We can find the 
phase constant δ from the geometry of 
the diagram to the right. Using the 
value of δ found in this fashion we can 
express the intensity at the point 1.72 
cm from the centerline in terms of the 
intensity on the centerline. On the 
centerline, the amplitude of the 
resultant wave is 3 times that of each 
individual wave and the intensity is 9 
times that of each source acting 
separately. 
 m 50.2=L
 cm .721y
θ
 
 
(a) Express δ for the adjacent slits: 
 
θλ
πδ sin2 d= 
 
For θ << 1, θθθ ≈≈ tansin : 
L
y=≈ θθ tansin 
 
Interference and Diffraction 
 
 
3087
Substitute to obtain: 
 L
dy
λ
πδ 2= 
 
Substitute numerical values and 
evaluate δ : 
( )( )
( )( )
°==
=
270rad
2
3
m50.2nm550
cm72.1m0.602
π
μπδ
 
 
The three phasors, 270° apart, are 
shown in the diagram to the right. 
Note that they form three sides of a 
square. Consequently, their sum, 
shown as the resultant R, equals the 
magnitude of one of the phasors. 
δ
δ
A
A
A
0
0
0
0R = A
 
(b) Express the intensity at the point 
1.72 cm from the centerline: 
 
2RI ∝ 
Because I0 ∝ 9R2: 
2
2
0 9R
R
I
I = ⇒ 
9
0II = 
 
Substitute for I0 and evaluate I: 22 mW/m56.5
9
mW/m0.50 ==I 
 
56 ••• In single-slit Fraunhofer diffraction, the intensity pattern (Figure 33-
11) consists of a broad central maximum with a sequence of secondary maxima to 
either side of the central maximum. This intensity is given by 
2
2
1
2
1
0
sin
⎟⎟⎠
⎞
⎜⎜⎝
⎛= φ
φ
II , 
where φ is the phase difference between the wavelets arriving from the opposite 
edges of the slits. Calculate the values of φ for the first three secondary maxima to 
one side of the central maximum by finding the values of φ for which dI/dφ is 
equal to zero. Check your results by comparing your answers with approximate 
values for φ of 3π, 5π and 7π. (That these values for φ are approximately correct 
at the secondary intensity maxima is discussed in the discussion surrounding 
Figure 33-27.) 
 
 
Chapter 33 
 
 
3088 
Picture the Problem We can use the phasor diagram shown in Figure 33-26 to 
determine the first three values of φ that produce secondary intensity maxima. 
Setting the derivative of Equation 33-19 equal to zero will yield a transcendental 
equation whose roots are the values of φ corresponding to the intensity maxima in 
the diffraction pattern. 
 
Referring to Figure 33-26 we see 
that the first subsidiary maximum 
occurs where: 
 
πφ 3= 
An intensity minimum occurs where: 
 
πφ 4= 
 
Another intensity maximum occurs 
where: 
 
πφ 5= 
 
Thus, secondary intensity maxima 
occur where: 
 
( ) ...,3,2,112 =+= n,n πφ 
and the first three secondary intensity 
maxima are at πππφ 7 and ,5,3= 
 
The intensity in the single-slit 
diffraction pattern is given by: 
 
2
2
1
2
1
0
sin
⎟⎟⎠
⎞
⎜⎜⎝
⎛= φ
φ
II 
 
Set the derivative of this expression equal to zero for extreme values (relative 
minima and maxima): 
 
( ) extremafor 0
sincossin
2 2
2
1
2
1
2
1
2
1
4
1
2
1
2
1
0 =⎥⎥⎦
⎤
⎢⎢⎣
⎡ −
⎟⎟⎠
⎞
⎜⎜⎝
⎛= φ
φφφ
φ
φ
φ Id
dI 
 
Simplify to obtain the transcendental 
equation: 
 
φφ 2121tan = 
Solve this equation numerically 
(use the ″Solver″ function of your 
calculator or trial-and-error 
methods) to obtain: 
 
πππφ 6.94 and 92.4 86.2 ,,= 
At the three intensity minima φ = 2π, 4π, and 6π , and at the three intensity 
maxima φ = 2.86π, 4.92π, and 6.94π. At the intensity maxima φ ≈ 3π, 5π, and 7π. 
 
 
Interference and Diffraction 
 
 
3089
Remarks: Note that our results in (b) are smaller than the approximate 
values found in (a) by 4.9%, 1.6%, and 0.86% and that the agreement 
improves as n increases. 
 
Diffraction and Resolution 
 
57 • [SSM] Light that has a wavelength equal to 700 nm is incident on a 
pinhole of diameter 0.100 mm. (a) What is the angle between the central 
maximum and the first diffraction minimum for a Fraunhofer diffraction pattern? 
(b) What is the distance between the central maximum and the first diffraction 
minimum on a screen 8.00 m away? 
 
Picture the Problem We can use 
D
λθ 22.1= to find the angle between 
the central maximum and the first 
diffraction minimum for a Fraunhofer 
diffraction pattern and the diagram to 
the right to find the distance between 
the central maximum and the firstdiffraction minimum on a screen 8 m 
away from the pinhole. 
L
miny
θ
D
 
 
(a) The angle between the central 
maximum and the first diffraction 
minimum for a Fraunhofer 
diffraction pattern is given by: 
 
D
λθ 22.1= 
Substitute numerical values and 
evaluate θ : mrad54.8mm100.0
nm70022.1 =⎟⎟⎠
⎞
⎜⎜⎝
⎛=θ 
 
(b) Referring to the diagram, we see 
that: 
 
θtanmin Ly = 
Substitute numerical values and 
evaluate ymin: 
( ) ( )
cm83.6
mrad54.8tanm00.8min
=
=y
 
 
58 • Two sources of light that both have wavelengths equal to 700 nm are 
10.0 m away from the pinhole of Problem 57. How far apart must the sources be 
for their diffraction patterns to be resolved by Rayleigh’s criterion? 
 
Chapter 33 
 
 
3090 
 
Picture the Problem We can apply Rayleigh’s criterion to the overlapping 
diffraction patterns and to the diameter D of the pinhole to obtain an expression 
that we can solve for Δy. 
Pinhole
α
α
Δy
L
c
c
 
 
Rayleigh’s criterion is satisfied 
provided: 
 
D
λα 22.1c = 
Relate αc to the separation Δy of the 
light sources: 
 
L
yΔ≈cα provided αc << 1. 
Equate these expressions to obtain: 
 DL
y λ22.1=Δ ⇒
D
Ly λ22.1=Δ 
 
Substitute numerical values and 
evaluate Δy: 
( )( ) cm54.8
mm100.0
m0.10nm70022.1 ==Δy
 
59 • Two sources of light that both have wavelengths of 700 nm are 
separated by a horizontal distance x. They are 5.00 m from a vertical slit of width 
0.500 mm. What is the smallest value of x for which the diffraction pattern of the 
sources can be resolved by Rayleigh’s criterion? 
 
Picture the Problem We can use 
Rayleigh’s criterion for slits and the 
geometry of the diagram to the right 
showing the overlapping diffraction 
patterns to express x in terms of λ, L, 
and the width a of the slit. 
c
c
Lα
α
x 
Interference and Diffraction 
 
 
3091
 
Referring to the diagram, relate αc, 
L, and x: L
x≈cα 
 
For slits, Rayleigh’s criterion is: 
 a
λα =c 
 
Equate these two expressions to 
obtain: 
 
aL
x λ= ⇒
a
Lx λ= 
Substitute numerical values and 
evaluate x: 
( )( ) mm00.7
mm500.0
m00.5nm700 ==x 
 
60 •• The ceiling of your lecture hall is probably covered with acoustic tile, 
which has small holes separated by about 6.0 mm. (a) Using light that has a 
wavelength of 500 nm, how far could you be from this tile and still resolve these 
holes? Assume the diameter of the pupil of your eye is about 5.0 mm. (b) Could 
you resolve these holes better using red light or using violet light? Explain your 
answer. 
 
Picture the Problem We can use Rayleigh’s criterion for circular apertures and 
the geometry of the diagram to the right showing the overlapping diffraction 
patterns to express L in terms of λ, x, and the diameter D of your pupil. 
Your pupil
α
α
x
L
c
c
 
 
(a) Referring to the diagram, relate 
αc, L, and x: L
x≈cα provided α << 1 
 
For circular apertures, Rayleigh’s 
criterion is: 
 
D
λα 22.1c = 
 
Equate these two expressions to 
obtain: 
 
DL
x λ22.1= ⇒ λ22.1
xDL = 
Chapter 33 
 
 
3092 
Substitute numerical values and 
evaluate L: 
( )( )
( ) m49nm50022.1
mm0.5mm0.6 ==L 
 
(b) Because L is inversely proportional to λ, the holes can be resolved better with 
violet light which has a shorter wavelength. The critical angle for resolution is 
proportional to the wavelength. Thus, the shorter the wavelength the farther away 
you can be and still resolve the two images. 
 
61 •• [SSM] The telescope on Mount Palomar has a diameter of 200 in. 
Suppose a double star were 4.00 light-years away. Under ideal conditions, what 
must be the minimum separation of the two stars for their images to be resolved 
using light that has a wavelength equal to 550 nm? 
 
Picture the Problem We can use Rayleigh’s criterion for circular apertures and 
the geometry of the diagram to obtain an expression we can solve for the 
minimum separation Δx of the stars. 
Your pupil
Δ
α
α
x
L
c
c
 
 
Rayleigh’s criterion is satisfied 
provided: 
 
D
λα 22.1c = 
Relate αc to the separation Δx of 
the light sources: 
 
L
xΔ≈cα because αc << 1 
Equate these expressions to obtain: 
DL
x λ22.1=Δ ⇒
D
Lx λ22.1=Δ 
 
Substitute numerical values and evaluate Δx: 
 
( )
m1000.5
in1
cm2.54in200
y1
m10461.9y4nm550
22.1 9
15
×=
⎟⎟
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜⎜
⎜
⎝
⎛
×
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⋅
××⋅
=Δ c
c
x 
Interference and Diffraction 
 
 
3093
62 •• The star Mizar in Ursa Major is a binary system of stars of nearly 
equal magnitudes. The angular separation between the two stars is 14 seconds of 
arc. What is the minimum diameter of the pupil that allows resolution of the two 
stars using light that has a wavelength equal to 550 nm? 
 
Picture the Problem We can use Rayleigh’s criterion for circular apertures and 
the geometry of the diagram to obtain an expression we can solve for the 
minimum diameter D of the pupil that allows resolution of the binary stars. 
Your pupil
Δ
α
α
x
L
c
cD
 
 
Rayleigh’s criterion is satisfied 
provided: 
 
D
λα 22.1c = ⇒
c
22.1 α
λ=D 
Substitute numerical values and 
evaluate D: 
 
cm1mm9.9
180
rad
3600
114
nm55022.1
≈=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
°×
°×
= π
''
''
D
 
 
Diffraction Gratings 
 
63 • [SSM] A diffraction grating that has 2000 slits per centimeter is used 
to measure the wavelengths emitted by hydrogen gas. (a) At what angles in the 
first-order spectrum would you expect to find the two violet lines that have 
wavelengths of 434 nm and 410 nm? (b) What are the angles if the grating has 15 
000 slits per centimeter? 
 
Picture the Problem We can solve λθ md =sin for θ with m = 1 to express the 
location of the first-order maximum as a function of the wavelength of the light. 
 
λθ md =sin 
where d is the separation of the slits 
and m = 0, 1, 2, … 
 
(a) The interference maxima in a 
diffraction pattern are at angles θ 
given by: 
Chapter 33 
 
 
3094 
Solve for the angular location θm of 
the maxima : ⎟⎠
⎞⎜⎝
⎛= −
d
m
m
λθ 1sin 
 
Relate the number of slits N per 
centimeter to the separation d of the 
slits: 
 
d
N 1= 
Substitute for d to obtain: 
 
( )λθ mNm 1sin−= (1) 
Evaluateθ for λ = 434 nm and m =1: ( )( )[ ]
mrad9.86
nm434cm2000sin 111
=
= −−θ
 
 
Evaluateθ for λ = 410 nm and m = 1: ( )( )[ ]
mrad1.82
nm410cm2000sin 111
=
= −−θ
 
 
(b) Use equation (1) to evaluateθ for 
λ = 434 nm and m = 1: 
( )( )[ ]
mrad 709
nm434cm15000sin 111
=
= −−θ
 
 
Evaluate θ for λ = 410 nm and m = 1: ( )( )[ ]
mrad 662
nm410cm15000sin 111
=
= −−θ
 
 
64 • Using a diffraction grating that has 2000 lines per centimeter, two 
other lines in the first-order hydrogen spectrum are found at angles of 
9.72 × 10–2 rad and 1.32 × 10–1 rad. What are the wavelengths of these lines? 
 
Picture the Problem We can solve λθ md =sin for λ with m = 1 to express the 
location of the first-order maximum as a function of the angles at which the first-
order images are found. 
 
The interference maxima in a 
diffraction pattern are at angles θ 
given by: 
λθ md =sin ⇒
m
d θλ sin= 
where d is the separation of the slits 
and m = 0, 1, 2, … 
 
Relate the number of slits N per 
centimeter to the separation d of 
the slits: 
 
d
N 1= 
Interference and Diffraction 
 
 
3095
Let m = 1 and substitute for d to 
obtain: N
θλ sin= 
 
Substitute numerical values and 
evaluate λ1 for θ1 = 9.72 ×10–2 rad: 
 
( ) nm485
cm2000
rad1072.9sin
1
2
1 =×= −
−
λ 
Substitute numerical values and 
evaluate λ1 for θ 2 = 1.32 ×10–1 rad: 
( ) nm658
cm2000
rad1032.1sin
1
1
1 =×= −
−
λ 
 
65 • The colors of many butterfly wings and beetle carapaces are due to 
effects of diffraction. The Morpho butterfly has structural elements on its wings 
that effectively act as a diffraction grating with spacing 880 nm. At what angle 
will the first diffraction maximum occur for normally incident light diffracted by 
the butterfly’s wings? Assume the light is blue with a wavelength of 440 nm. 
 
Picture the Problem We can use the gratingequation to find the angle at which 
normally incident blue light will be diffracted by the butterfly’s wings. 
 
The grating equation is: 
 
λθ md =sin , where m = 1, 2, 3, … 
 
Solve for θ to obtain: 
 ⎥⎦
⎤⎢⎣
⎡= −
d
mλθ 1sin 
 
Substitute numerical values and 
evaluate θ1: 
( )( ) °=⎥⎦
⎤⎢⎣
⎡= − 0.30
nm880
nm4401sin 1θ 
 
66 •• A diffraction grating that has 2000 slits per centimeter is used to 
analyze the spectrum of mercury. (a) Find the angular separation in the first-order 
spectrum of the two lines of wavelength 579 nm and 577 nm. (b) How wide must 
the beam on the grating be for these lines to be resolved? 
 
Picture the Problem We can use the grating equation to find the angular 
separation of the first-order spectrum of the two lines. In Part (b) we can apply the 
definition of the resolving power of the grating to find the width of the grating 
that must be illuminated for the lines to be resolved. 
 
(a) Express the angular separation in 
the first-order spectrum of the two 
lines: 
 
577579 θθθ −=Δ 
Solve the grating equation for θ : ⎟⎠
⎞⎜⎝
⎛= −
d
mλθ 1sin 
 
Chapter 33 
 
 
3096 
Substitute for θ579 and θ577 to obtain: 
 
( ) ( ) [ ] [ mmmm 1154.0sin1158.0sin
cm2000
1
nm577sin
cm2000
1
nm579sin 11
1
1
1
1 −−
−
−
−
− −=
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
=Δθ ] 
 
For m = 1: 
 
[ ] [ ]
°=°−°=
−=Δ −−
02.063.665.6
1154.0sin1158.0sin 11 mmθ
 
 
(b) The width of the beam necessary 
for these lines to be resolved is given 
by: 
 
Ndw = (1) 
Relate the resolving power of the 
diffraction grating to the number of 
slits N that must be illuminated in 
order to resolve these wavelengths in 
the mth order: 
 
mN=Δλ
λ 
For m = 1: 
λ
λ
Δ=N 
 
Substitute for N in equation (1) to 
obtain: λ
λ
Δ=
dw 
 
Letting λ be the average of the 
two wavelengths, substitute 
numerical values and evaluate w: 
( )
mm1
nm2
cm2000
1nm578 1
=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
=
−
w 
 
67 •• [SSM] A diffraction grating that has 4800 lines per centimeter is 
illuminated at normal incidence with white light (wavelength range of 400 nm to 
700 nm). How many orders of spectra can one observe in the transmitted light? 
Do any of these orders overlap? If so, describe the overlapping regions. 
 
Picture the Problem We can use the grating equation ... 3, 2, 1, sin == m,md λθ 
to express the order number in terms of the slit separation d, the wavelength of 
the light λ, and the angle θ. 
 
 
Interference and Diffraction 
 
 
3097
The interference maxima in the 
diffraction pattern are at angles θ 
given by: 
 
λθ md =sin ⇒ λ
θsindm = 
where m = 1, 2, 3, … 
If one is to see the complete 
spectrum, it must be true that: 
 
1sin ≤θ ⇒ λ
dm ≤ 
 
Evaluate mmax: 
98.2
nm700
cm4800
1
cm4800
1
1
max
1
max ===
−−
λm 
 
Because mmax = 2.98, one can see the complete spectrum only for m = 1 and 2. 
 
Express the condition for overlap: 
 
2211 λλ mm ≥ 
One can see the complete spectrum for only the first and second order spectra. 
That is, only for m = 1 and 2. Because 700 nm < 2 × 400 nm, there is no overlap 
of the second-order spectrum into the first-order spectrum; however, there is 
overlap of long wavelengths in the second order with short wavelengths in the 
third-order spectrum. 
 
68 •• A square diffraction grating that has an area of 25.0 cm2 has a 
resolution of 22 000 in the fourth order. At what angle should you look to see a 
wavelength of 510 nm in the fourth order? 
 
Picture the Problem We can use the grating equation and the resolving power of 
the grating to derive an expression for the angle at which you should look to see a 
wavelength of 510 nm in the fourth order. 
 
The interference maxima in the 
diffraction pattern are at angles θ 
given by: 
 
... 3, 2, 1, where
 sin
=
=
m
,md λθ
 (1) 
 
The resolving power R is given by: 
 
mNR = 
where N is the number of slits and m is 
the order number. 
 
Relate d to the width w of the 
grating: 
 
N
wd = 
Substitute for N and simplify to 
obtain: R
mwd = 
Chapter 33 
 
 
3098 
Substitute for d in equation (1) to 
obtain: 
λθ m
R
mw =sin ⇒ ⎟⎠
⎞⎜⎝
⎛= −
w
Rλθ 1sin 
 
Substitute numerical values and 
evaluate θ : 
( )( ) °=⎥⎦
⎤⎢⎣
⎡= − 0.13
cm00.5
nm510000,22sin 1θ 
 
69 •• Sodium light that has a wavelength equal to 589 nm falls normally on 
a 2.00-cm-square diffraction grating ruled with 4000 lines per centimeter. The 
Fraunhofer diffraction pattern is projected onto a screen a distance of 1.50 m from 
the grating by a 1.50-m-focal-length lens that is placed immediately in front of the 
grating. Find (a) the distance of the first and second order intensity maxima from 
the central intensity maximum, (b) the width of the central maximum, and (c) the 
resolution in the first order. (Assume the entire grating is illuminated.) 
 
Picture the Problem The distance on the screen to the mth bright fringe can be 
found using λθ md =sin (where d is the slit separation and m = 0, 1, 2, …) and 
the geometry of the grating and projection screen. We can use 
LyNd 2min Δ== λθ to find the width of the central maximum and R = mN, 
where N is the number of slits in the grating, to find the resolving power in the 
first order. 
 
(a) The angle θ at which maxima 
occur is related to the slit separation 
d, the wavelength of the incident 
light λ, and the order number m 
according to: 
 
λθ md =sin (1) 
where m = 0, 1, 2, … . 
 
θ is also related to the distance to 
the screen L and the positions of 
the intensity maxima ym: 
 
22
sin
m
m
yL
y
+=θ 
Substituting for sin θ in equation 
(1) yields: 
 
λm
yL
dy
m
m =+ 22 ⇒ 222 λ
λ
md
Lmym −= 
Substituting numerical values 
yields: 
 
( )( )
( )222 nm589
4000
cm00.1
m50.1nm589
m
mym
−⎟⎠
⎞⎜⎝
⎛
= 
 
Interference and Diffraction 
 
 
3099
 
Evaluate this expression for m = 1 
and m = 2 to obtain: 
 
cm4.361 =y and cm1.802 =y 
 
(b) The angle θmin that locates the 
first minima in the diffraction 
pattern is given by: 
 
L
y
Nd 2min
Δ== λθ ⇒
Nd
Ly λ2=Δ 
where Δy is the width of the central 
maximum. 
 
Substitute numerical values and 
evaluate Δy: 
 
( )( )
( )
m4.88
cm4000
1lines8000
nm589m50.12
1
μ=
⎟⎟⎠
⎞
⎜⎜⎝
⎛=Δ
−
y
 
 
(c) The resolving power R in the 
mth order is given by: 
 
mNR = 
Substitute numerical values and 
evaluate R: 
( )( ) 800080001 ==R 
 
70 •• The spectrum of neon is exceptionally rich in the visible region. 
Among the many lines are two lines at wavelengths of 519.313 nm and 
519.322 nm. If light from a neon discharge tube is normally incident on a 
transmission grating with 8400 lines per centimeter and the spectrum is observed 
in second order, what must be the width of the grating that is illuminated, so that 
these two lines can be resolved? 
 
Picture the Problem The width of the grating w is the product of its number of 
lines N and the separation of its slits d. Because the resolution of the grating is a 
function of the average wavelength, the difference in the wavelengths, and the 
order number, we can express w in terms of these quantities. 
 
Express the width w of the grating as 
a function of the number of lines N 
and the slit separation d: 
 
Ndw = 
The resolving power R of the grating 
is given by: 
 
mNR =Δ= λ
λ ⇒ λ
λ
Δ= mN 
Substitute for N in the expression for 
w to obtain: 
 
λ
λ
Δ= m
dw 
Chapter 33 
 
 
3100 
Letting λ be the average of the given wavelengths, substitute numerical values 
and evaluate w: 
 
( )
( ) cm3nm519.313nm322.5192
cm8400
1nm519.322nm313.519 12
1
=−
⎟⎟⎠
⎞
⎜⎜⎝
⎛+
=
−
w 
 
71 •• [SSM] Mercury has several stable isotopes, among them 198Hg and 
202Hg. The strong spectral line of mercury, at about 546.07 nm, is a composite of 
spectral lines from the various mercury isotopes. The wavelengths of this line for 
198Hg and 202Hg are 546.07532 nm and 546.07355 nm, respectively. What must be 
the resolving power ofa grating capable of resolving these two isotopic lines in 
the third-order spectrum? If the grating is illuminated over a 2.00-cm-wide region, 
what must be the number of lines per centimeter of the grating? 
 
Picture the Problem We can use the expression for the resolving power of a 
grating to find the resolving power of the grating capable of resolving these two 
isotopic lines in the third-order spectrum. Because the total number of the slits of 
the grating N is related to width w of the illuminated region and the number of 
lines per centimeter of the grating and the resolving power R of the grating, we 
can use this relationship to find the number of lines per centimeter of the grating. 
 
The resolving power of a 
diffraction grating is given by: 
 
mNR =Δ= λ
λ (1) 
 
Substitute numerical values and 
evaluate R: 
 55 1009.3100852.3
07355.54607532.546
07532.546
×=×=
−=R 
 
Express n, be the number of lines 
per centimeter of the grating, in 
terms of the total number of slits 
N of the grating and the width w 
of the grating: 
 
w
Nn = 
From equation (1) we have: 
 m
RN = 
 
Substitute for N to obtain: 
 mw
Rn = 
 
Interference and Diffraction 
 
 
3101
Substitute numerical values and 
evaluate n: ( )( ) 14
5
cm1014.5
cm00.23
100852.3 −×=×=n 
 
72 ••• A diffraction grating has n lines per unit length. Show that the angular 
separation (Δθ ) of two lines of wavelengths λ and λ + Δλ is approximately 
2
22
1 λλθ −=
mn
ΔΔ where m is the order number. 
 
Picture the Problem We can differentiate the grating equation implicitly and 
approximate dθ /dλ by Δθ /Δλ to obtain an expression Δθ as a function of m, n, 
Δλ, and cosθ. We can use the Pythagorean identity sin2θ + cos2θ = 1 and the 
grating equation to write cosθ in terms of n, m, and λ. Making these substitutions 
will yield the given equation. 
 
The grating equation is: 
 
... 2, 1, 0, sin == , mmd λθ (1) 
( ) ( )λλθλ md
dd
d
d =sin Differentiate both sides of this 
equation with respect to λ: 
or 
m
d
dd =λ
θθcos 
 
Because n = 1/d: 
 
nm
d
d =λ
θθcos ⇒ λ
θθ
d
d
m
n cos1= 
 
Approximate dθ /dλ by Δθ /Δλ: θλ
θ cos1 Δ
Δ=
m
n 
 
Solving for Δθ yields: 
 θ
λθ
cos
Δ=Δ nm 
 
Substitute for cosθ to obtain: 
θ
λθ
2sin1−
Δ=Δ nm 
 
From equation (1): λλθ nm
d
m ==sin 
 
Substituting for sinθ yields: 
2221 λ
λθ
mn
nm
−
Δ=Δ 
Chapter 33 
 
 
3102 
 
Simplify by dividing the numerator and denominator by nm to obtain: 
 
2
2222
222222 1111 λ
λ
λ
λ
λ
λθ
−
Δ=−
Δ=
−
Δ=Δ
mnmn
mnmn
nm
 
 
73 ••• [SSM] For a diffraction grating in which all the surfaces are normal 
to the incident radiation, most of the energy goes into the zeroth order, which is 
useless from a spectroscopic point of view, since in zeroth order all the 
wavelengths are at 0º. Therefore, modern reflection gratings have shaped, or 
blazed, grooves, as shown in Figure 33-45. This shifts the specular reflection, 
which contains most of the energy, from the zeroth order to some higher order. 
(a) Calculate the blaze angle φm in terms of the groove separation d, the 
wavelength λ, and the order number m in which specular reflection is to occur for 
m = 1, 2, . . . . (b) Calculate the proper blaze angle for the specular reflection to 
occur in the second order for light of wavelength 450 nm incident on a grating 
with 10 000 lines per centimeter. 
 
Picture the Problem We can use the grating equation and the geometry of the 
grating to derive an expression for φm in terms of the order number m, the 
wavelength of the light λ, and the groove separation d. 
 
(a) Because θi = θr, application of 
the grating equation yields: 
 
( )
... 2, 1, 0, where
2sin i
=
=
m
, md λθ
 (1) 
Because φ and θi have their left and 
right sides mutually perpendicular: 
 
mi φθ = 
Substitute for θi to obtain: 
 
( ) λφ md =m2sin 
Solving for φm yields: ⎟⎠
⎞⎜⎝
⎛= −
d
m λφ 121m sin 
 
(b) For m = 2: 
°=
⎟⎟
⎟⎟
⎠
⎞
⎜⎜
⎜⎜
⎝
⎛
=
−
− 1.32
cm000,10
1
nm4502sin
1
1
2
1
2φ 
 
74 ••• In this problem, you will derive the relation mNR == λλ Δ 
(Equation 33-27) for the resolving power of a diffraction grating containing N 
slits separated by a distance d. To do this, you will calculate the angular 
Interference and Diffraction 
 
 
3103
separation between the intensity maximum and intensity minimum for some 
wavelength λ and set it equal to the angular separation of the mth-order maximum 
for two nearby wavelengths. (a) First show that the phase difference φ between 
the waves from two adjacent slits is given by 
 
φ = 2πdλ sin θ . (b) Next differentiate 
this expression to show that a small change in angle dθ results in a change in 
phase of dφ given by
 
dφ = 2πdλ cosθ dθ . (c) Then for N slits, the angular 
separation between an interference maximum and an interference minimum 
corresponds to a phase change of dφ = 2 π/N. Use this to show that the angular 
separation dθ between the intensity maximum and intensity minimum for some 
wavelength λ is given by 
 
dθ = λ
Nd cosθ . (d) Next use the fact that the angle of 
the mth-order interference maximum for wavelength λ is specified by 
λθ md =sin (Equation 33-26). Compute the differential of each side of this 
equation to show that angular separation of the mth-order maximum for two 
nearly equal wavelengths differing by dλ is given by 
 
dθ = mdλ
d cosθ . (e) According 
to Rayleigh’s criterion, two wavelengths will be resolved in the mth order if the 
angular separation of the wavelengths, given by the Part (d) result, equals the 
angular separation of the interference maximum and the interference minimum 
given by the Part (c) result. Use this to arrive at mNR == λλ Δ (Equation 33-
27) for the resolving power of a grating. 
 
Picture the Problem We can follow the procedure outlined in the problem 
statement to obtain mNR == λλ Δ . 
 
(a) Express the relationship between 
the phase difference φ and the path 
difference Δr: 
 
λπ
φ rΔ=
2
 ⇒ λ
πφ rΔ= 2 
Because Δr = dsinθ : θλ
πφ sin2 d= 
 
(b) Differentiate this expression with 
respect to θ to obtain: 
 
θλ
πθλ
π
θθ
φ cos2sin2 dd
d
d
d
d =⎥⎦
⎤⎢⎣
⎡= 
Solve for dφ: θθλ
πφ ddd cos2= 
 
(c) From Part (b): 
θπ
φλθ
cos2 d
dd = 
Chapter 33 
 
 
3104 
Substitute 2π/N for dφ to obtain: 
θ
λθ
cosNd
d = 
 
(d) Equation 33-26 is: ... 2, 1, 0, sin == m,md λθ 
 
[ ] [ ]λλθλ md
dd
d
d =sin 
or 
m
d
dd =λ
θθcos 
 
Differentiate this expression 
implicitly with respect to λ to obtain: 
 
Solve for dθ to obtain: 
 θ
λθ
cosd
mdd = 
 
(e) Equate the two expressions for 
dθ obtained in (c) and (d): θ
λ
θ
λ
coscos d
md
Nd
= 
 
Approximating dλ by Δλ and 
allowing for the possibility that 
Δλ < 0 yields: 
 
θ
λ
θ
λ
coscos d
m
Nd
Δ= 
Solving for R = λ/Δλ yields: 
mNR == λ
λ
Δ 
 
General Problems 
 
75 • [SSM] Naturally occurring coronas (brightly colored rings) are 
sometimes seen around the Moon or the Sun when viewed through a thin cloud. 
(Warning: When viewing a sun corona, be sure that the entire sun is blocked by 
the edge of a building, a tree, or a traffic pole to safeguard your eyes.) These 
coronas are due to diffraction of light by small water droplets in the cloud. A 
typical angular diameter for a coronal ring is about 10º. From this, estimate the 
size of the water droplets in the cloud. Assume that the water droplets can be 
modeled as opaque disks with the same radius as the droplet, and that the 
Fraunhofer diffraction pattern from an opaque disk is the same as the pattern from 
an aperture of the same diameter. (This last statement is known as Babinet’s 
principle.) 
 
Picture the Problem We can use Dλθ 22.1sin = to relate the diameter D of the 
opaque-disk water droplets to the angular diameter θ of a coronal ring and to the 
wavelength of light. We’ll assume a wavelength of 500 nm. 
 
Interference and Diffraction 
 
 
3105
The angle θsubtended by the first 
diffraction minimum is related to the 
wavelength λ of light and the 
diameter D of the opaque-disk water 
droplet: 
 
D
λθ 22.1sin = 
 
Because of the great distance to the 
cloud of water droplets, θ << 1 and: 
 
D
λθ 22.1≈ ⇒ θ
λ22.1=D 
 
Substitute numerical values and 
evaluate D: 
( ) m5.3
180
rad10
nm50022.1 μπ =
°×°
=D 
 
76 • An artificial corona (see Problem 75) can be made by placing a 
suspension of polystyrene microspheres in water. Polystyrene microspheres are 
small, uniform spheres made of plastic with an index of refraction equal to 1.59. 
Assuming that the water has an index of refractive equal to 1.33, what is the 
angular diameter of such an artificial corona if 5.00-μm-diameter particles are 
illuminated by light from a helium–neon laser with wavelength in air of 
632.8 nm? 
 
Picture the Problem We can use Dnλθ 22.1sin = to relate the diameter D of a 
microsphere to the angular diameter θ of a coronal ring and to the wavelength of 
light in water. 
 
The angle θ subtended by the first 
diffraction minimum is related to the 
wavelength λn of light in water and 
the diameter D of the microspheres: 
 
nDD
n λλθ 22.122.1sin == 
 
Because θ << 1: 
 nD
λθ 22.1≈ 
 
Substitute numerical values and 
evaluate θ : 
( )( )
( )( )
°=
=≈
65.6
rad116.0
m00.533.1
nm8.63222.1
μθ 
 
77 • Coronas (see Problem 74) can be caused by pollen grains, typically of 
birch or pine. Such grains are irregular in shape, but they can be treated as if they 
had an average diameter of about 25 μm. What is the angular diameter (in 
degrees) of the corona for blue light? What is the diameter (in degrees) of the 
corona for red light? 
 
 
Chapter 33 
 
 
3106 
Picture the Problem We can use Dλθ 22.1sin = to relate the diameter D of a 
pollen grain to the angular diameter θ of a coronal ring and to the wavelength of 
light. We’ll assume a wavelength of 450 nm for blue light and 650 nm for red 
light. 
 
D
λα 22.1sin = 
or, because θα 21= where θ is the 
angular diameter of the coronal ring, 
D
λθ 22.1sin 21 = 
 
The angleα subtended by the first 
diffraction intensity minima is 
related to the wavelength λ of light 
and to the diameter D of the 
microspheres: 
 
Because θ << 1, sinθ ≈ θ and: 
 D
λθ 22.121 ≈ ⇒ D
λθ 44.2≈ 
 
Substitute numerical values and 
evaluate θ for red light: 
( )
°=
×=≈ −
6.3
rad10344.6
m25
nm65044.2 2
red μθ 
 
Substitute numerical values and 
evaluate θ for blue light: 
( )
°=
×=≈ −
5.2
rad10392.4
m25
nm45044.2 2
blue μθ 
 
78 • Light from a He-Ne laser (632.8 nm wavelength) is directed upon a 
human hair, in an attempt to measure its diameter by examining the diffraction 
pattern. The hair is mounted in a frame 7.5 m from a wall, and the central 
diffraction maximum is measured to be 14.6 cm wide. What is the diameter of 
the hair? (The diffraction pattern of a hair with diameter d is the same as the 
diffraction pattern of a single slit with width a = d. See Babinet’s principle, 
Problem 75.) 
 
Picture the Problem The diagram shows the hair whose diameter d = a, the 
screen a distance L from the hair, and the separation Δy of the first diffraction 
peak from the center. We can use the geometry of the experiment to relate Δy to L 
and a and the condition for diffraction maxima to express θ1 in terms of the 
diameter of the hair and the wavelength of the light illuminating the hair. 
ya
θ Δ
L
1
 
Interference and Diffraction 
 
 
3107
 
Relate θ to Δy: 
 L
yΔ= 2
1
tanθ ⇒ ⎟⎠
⎞⎜⎝
⎛ Δ= −
L
y
2
tan 1θ 
 
Diffraction maxima occur where: 
 ( )λθ 21sin += ma ⇒ ( )θ
λ
sin
2
1+= ma 
where m = 1, 2, 3, … 
 
Substituting for θ yields: 
 
( )
⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛ Δ
+=
−
L
y
m
a
2
tansin 1
2
1 λ
 
 
Substitute numerical values and 
evaluate a for m = 1: 
( )( )
( )
m 98
m 5.72
cm 6.14tansin
nm 8.6321
1
2
1 μ=
⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛
+=
−
a 
 
79 • [SSM] A long, narrow horizontal slit lies 1.00 μm above a plane 
mirror, which is in the horizontal plane. The interference pattern produced by the 
slit and its image is viewed on a screen 1.00 m from the slit. The wavelength of 
the light is 600 nm. (a) Find the distance from the mirror to the first maximum. 
(b) How many dark bands per centimeter are seen on the screen? 
 
Picture the Problem We can apply the condition for constructive interference to 
find the angular position of the first maximum on the screen. Note that, due to 
reflection, the wave from the image is 180o out of phase with that from the source.
 
(a) Because y0 << L, the distance 
from the mirror to the first maximum 
is given by: 
 
00 θLy = (1) 
Express the condition for 
constructive interference: 
 
( )λθ 21sin += md 
where m = 0, 1, 2, … 
Solving for θ yields: 
 
( ) ⎥⎦
⎤⎢⎣
⎡ += −
d
m λθ 211sin 
 
For the first maximum, m = 0 and: ⎥⎦
⎤⎢⎣
⎡= −
d2
sin 10
λθ 
 
Substitute in equation (1) to obtain: 
 ⎥⎦
⎤⎢⎣
⎡= −
d
Ly
2
sin 10
λ 
Chapter 33 
 
 
3108 
Because the image of the slit is as far 
behind the mirror’s surface as the slit 
is in front of it, d = 2.00 μm. 
Substitute numerical values and 
evaluate y0: 
 
( ) ( )
cm1.15
m00.22
nm600sinm00.1 10
=
⎥⎦
⎤⎢⎣
⎡= − μy 
 
(b) The separation of the fringes on 
the screen is given by: 
 
d
Ly λ=Δ 
The number of dark bands per unit 
length is the reciprocal of the fringe 
separation: 
 
L
d
y
n λ=Δ=
1 
Substitute numerical values and 
evaluate n: ( )( ) 1m33.3m00.1nm600
m00.2 −== μn 
 
80 • A radio telescope is situated at the edge of a lake. The telescope is 
looking at light from a radio galaxy that is just rising over the horizon. If the 
height of the antenna is 20 m above the surface of the lake, at what angle above 
the horizon will the radio galaxy be when the telescope is centered in the first 
intensity interference maximum of the radio waves? Assume the wavelength of 
the radio waves is 20 cm. Hint: The interference is caused by the light reflecting 
off the lake and remember that this reflection will result in a 180º phase shift. 
 
Picture the Problem The radio waves from the galaxy reach the telescope by two 
paths; one coming directly from the galaxy and the other reflected from the 
surface of the lake. The radio waves reflected from the surface of the lake are 
phase shifted 180°, relative to the radio waves reaching the telescope directly, by 
reflection from the surface of the lake. We can use the condition for constructive 
interference of two waves to find the angle above the horizon at which the radio 
waves from the galaxy will interfere constructively. 
θ
d
θ
r
θ
P
Telescope
Δ
Radio waves directly from galaxy
Ra
dio
 w
av
es 
ref
lec
ted
 fro
m 
the
 la
ke
 
 
 
 
 
Interference and Diffraction 
 
 
3109
Because the reflected radio waves are 
phase shifted by 180°, the condition 
for constructive interference at point 
P is: 
 
( )λ21+=Δ mr 
where m = 0, 1, 2, … 
Referring to the figure, note that: 
d
rΔ≈θsin ⇒ ⎥⎦
⎤⎢⎣
⎡Δ= −
d
r1sinθ 
 
Substitute for Δr to obtain: ( )
⎥⎦
⎤⎢⎣
⎡ += −
d
m λθ 211sin 
 
Noting that m = 0 for the first 
interference maximum, substitute 
numerical values and evaluate θ0: 
( )
°=
×=⎥⎦
⎤⎢⎣
⎡= −−
29.0
rad1000.5
m20
cm20
sin 32
1
1
0θ
 
 
81 • The diameter of the radio telescope at Arecibo, Puerto Rico, is 300 m. 
What is the smallest angular separation of two objects that this telescope can 
detect when it is tuned to detect microwaves of 3.2-cm wavelength? 
 
Picture the Problem The resolving power of a telescope is the ability of the 
instrument to resolve two objects that are close together. Hence we can use 
Rayleigh’s criterion to find the resolving power of the Arecibo telescope. 
 
Rayleigh’s criterion for resolution is: 
 D
λα 22.1c = 
 
Substitute numerical values and 
evaluate αc: mrad13.0m300
cm2.322.1c =⎟⎟⎠
⎞
⎜⎜⎝
⎛=α 
 
82 •• A thin layer of a transparent material that has an index of refraction of 
1.30 is used as a nonreflectivecoating on the surface of glass that has an index of 
refraction of 1.50. What should the minimum thickness of the material be for the 
material to be nonreflecting for light that has a wavelength 600 nm? 
 
Picture the Problem Note that 
reflection at both surfaces involves a 
phase shift of π rad. We can apply the 
condition for destructive interference to 
find the thickness t of the nonreflective 
coating. 
= 600 nm
t
Air
Coating
π
π
λ
Glass 
Chapter 33 
 
 
3110 
 
The condition for destructive 
interference is: 
 
( ) ( )
coating
air
2
1
coating2
12
n
mmt λλ +=+= 
Solve for t to obtain: ( )
coating
air
2
1
2n
mt λ+= 
 
Evaluate t for m = 0: ( ) ( ) nm11530.12
nm600
2
1 ==t 
 
83 •• [SSM] A Fabry–Perot interferometer (Figure 33-47) consists of two 
parallel, half-silvered mirrors that face each other and are separated by a small 
distance a. A half-silvered mirror is one that transmits 50% of the incident 
intensity and reflects 50% of the incident intensity. Show that when light is 
incident on the interferometer at an angle of incidence θ, the transmitted light will 
have maximum intensity when 2a = mλ/cos θ. 
 
Picture the Problem The Fabry-Perot 
interferometer is shown in the figure. 
For constructive interference in the 
transmitted light the path difference 
must be an integral multiple of the 
wavelength of the light. This path 
difference can be found using the 
geometry of the interferometer. 
θ
a
 
 
Express the path difference between 
the two rays that emerge from the 
interferometer: 
 
θcos
2ar =Δ 
For constructive interference we 
require that: 
 
...,2,1,0==Δ m,mr λ 
Equate these expressions to 
obtain: 
 
θλ cos
2am = 
 
Solve for 2a to obtain: θλ cos2 ma = 
 
84 •• A mica sheet 1.20 μm thick is suspended in air. In reflected light, there 
are gaps in the visible spectrum at 421, 474, 542, and 633 nm. Find the index of 
refraction of the mica sheet. 
 
Interference and Diffraction 
 
 
3111
Picture the Problem The gaps in the 
spectrum of the visible light are the 
result of destructive interference 
between the incident light and the 
reflected light. Noting that there is a π 
rad phase shift at the first air-mica 
interface, we can use the condition for 
destructive interference to find the 
index of refraction n of the mica sheet. 
Air Mica Air
n
t
π
 
 
Because there is a π rad phase shift 
at the first air-mica interface, the 
condition for destructive interference 
is: 
 
...,32,1,,2 airmica === mnmmt
λλ 
Solving for n yields: 
 t
mn
2
airλ= (1) 
 
For λ = 474 nm: ( )mt nm4742 = 
 
For λ = 421 nm: ( )( )1nm4212 += mt 
 
Equate these two expressions and 
solve for m to obtain: 
 
m = 8 for λ = 474 nm 
Substitute numerical values in 
equation (1) and evaluate n: ( ) 58.1m20.12
nm4748 =⎟⎟⎠
⎞
⎜⎜⎝
⎛= μn 
 
85 •• [SSM] A camera lens is made of glass that has an index of refraction 
of 1.60. This lens is coated with a magnesium fluoride film (index of refraction 
1.38) to enhance its light transmission. The purpose of this film is to produce zero 
reflection for light of wavelength 540 nm. Treat the lens surface as a flat plane 
and the film as a uniformly thick flat film. (a) What minimum thickness of this 
film will accomplish its objective? (b) Would there be destructive interference for 
any other visible wavelengths? (c) By what factor would the reflection for light of 
400 nm wavelength be reduced by the presence of this film? Neglect the variation 
in the reflected light amplitudes from the two surfaces. 
 
 
 
 
 
Chapter 33 
 
 
3112 
Picture the Problem Note that the 
light reflected at both the air-film and 
film-lens interfaces undergoes a π rad 
phase shift. We can use the condition 
for destructive interference between the 
light reflected from the air-film 
interface and the film-lens interface to 
find the thickness of the film. In (c) we 
can find the factor by which light of the 
given wavelengths is reduced by this 
film from .cos 21
2 δ∝I 
 
Air
n
t
π
Film Lens
π
 
 
(a) Express the condition for 
destructive interference between the 
light reflected from the air-film 
interface and the film-lens interface: 
 
( ) ( )
n
mmt air21film212
λλ +=+= (1) 
where m = 0, 1, 2, … 
Solving for t gives: 
 
( )
n
mt
2
air
2
1 λ+= 
 
Evaluate t for m = 0: 
( ) nm8.97nm83.9738.12
nm540
2
1 ==⎟⎠
⎞⎜⎝
⎛=t
 
(b) Solve equation (1) for λair to 
obtain: 21
air
2
+= m
tnλ 
 
Evaluate λair for m = 1: ( )( ) nm180
1
38.1nm8.972
2
1air
=+=λ 
 
No; because 180 nm is not in the visible portion of the spectrum. 
 
(c) Express the reduction factor f as 
a function of the phase difference δ 
between the two reflected waves: 
 
δ212cos=f (2) 
Relate the phase difference to the 
path difference Δr: 
 
film2 λπ
δ rΔ= ⇒ ⎟⎟⎠
⎞
⎜⎜⎝
⎛ Δ=
film
2 λπδ
r 
 
Because Δr = 2t: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
film
22 λπδ
t 
Interference and Diffraction 
 
 
3113
Substitute in equation (2) to obtain: 
⎥⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=⎥⎦
⎤⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛=
air
2
film
2
film
2
12
2cos
2cos22cos
λ
π
λ
π
λπ
nt
ttf
 
 
Evaluate f for λ = 400 nm: ( )( )
273.0
nm400
nm83.9738.12cos2400
=
⎥⎦
⎤⎢⎣
⎡= πf
 
 
86 •• In a pinhole camera, the image is fuzzy because of geometry (rays arrive 
at the film after passing through different parts of the pinhole) and because of 
diffraction. As the pinhole is made smaller, the fuzziness due to geometry is 
reduced, but the fuzziness due to diffraction is increased. The optimum size of the 
pinhole for the sharpest possible image occurs when the spread due to diffraction 
equals the spread due to the geometric effects of the pinhole. Estimate the 
optimum size of the pinhole if the distance from the pinhole to the film is 10.0 cm 
and the wavelength of the light is 550 nm. 
 
Picture the Problem As indicated in 
the problem statement, we can find the 
optimal size of the pinhole by equating 
the angular width of the object at the 
film and the angular width of the 
diffraction pattern. 
2θD
L
Object Film
 
 
Express the angular width of the a 
distant object at the film in terms of 
the diameter D of the pinhole and the 
distance L from the pinhole to the 
object: 
 
L
D=θ2 ⇒ 
L
D
2
=θ 
Using Rayleigh’s criterion, express 
the angular width of the diffraction 
pattern: 
 
D
λθ 22.1ndiffractio = 
Equate these two expressions to 
obtain: 
 
DL
D λ22.1
2
= ⇒ LD λ44.2= 
Chapter 33 
 
 
3114 
Substitute numerical values and 
evaluate D: 
( )( )
mm366.0
cm0.10nm55044.2
=
=D
 
 
87 •• [SSM] The Impressionist painter Georges Seurat used a technique 
called pointillism, in which his paintings are composed of small, closely spaced 
dots of pure color, each about 2.0 mm in diameter. The illusion of the colors 
blending together smoothly is produced in the eye of the viewer by diffraction 
effects. Calculate the minimum viewing distance for this effect to work properly. 
Use the wavelength of visible light that requires the greatest distance between 
dots, so that you are sure the effect will work for all visible wavelengths. Assume 
the pupil of the eye has a diameter of 3.0 mm. 
 
Picture the Problem We can use the geometry of the dots and the pupil of the 
eye and Rayleigh’s criterion to find the greatest viewing distance that ensures that 
the effect will work for all visible wavelengths. 
Dots of paint
Pupil
θ
cα
d
L
 
 
Referring to the diagram, express 
the angle subtended by the adjacent 
dots: 
 
L
d≈θ 
Letting the diameter of the pupil of 
the eye be D, apply Rayleigh’s 
criterion to obtain: 
 
D
λα 22.1c = 
Set θ = αc to obtain: 
DL
d λ22.1= ⇒ λ22.1
DdL = 
 
Evaluate L for the shortest 
wavelength light in the visible 
portion of the spectrum: 
( )( )
( )( ) m12nm40022.1
mm0.2mm0.3 ==L