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Chapter 33 Interference and Diffraction

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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 Оґ due to a path difference О”r are
related according to Оґ 2ПЂ = О”r О» . 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
3048
Chapter 33
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 180В°;
the phase change due to reflection from the back surface of the film is 0В°. 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 О»; that is, О”r = mО» . For m = 2, О”r = 2(640 nm) . (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 12 О» ; that is О”r = m( 12 О» ), m = 1, 3, 5, ... . For the first
minimum, m = 1 and О”r =
1
2
(640 nm) = 320 nm . (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 Оёm of each maximum is given by d sin Оё m = mО» , m = 0,1, 2, ...
(Equation 33-2). Because d and sinОёm are inversely proportional for a given
wavelength and interference maximum (value of m), decreasing d increases sinОёm
and Оёm. (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 Оёm of each maximum, and the wavelength of the light
illuminating the slits is given by d sin Оё m = mО» , m = 0,1, 2, ... (Equation 33-2).
Because О» and sinОёm 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 Оёm of each maximum, and the wavelength of the light illuminating the
slit is given by a sin Оё m = mО» , m =1, 2, 3, ... (Equation 33-11). Because a and
sinОёm are inversely proportional for a given diffraction maximum (value of m),
narrowing the slit increases sin Оё m and Оёm. (a ) is correct.
12 •
Equation 33-2 which is d sin Оёm = mО», and Equation 33-11, which is
a sin Оёm = mО», 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, О» the
wavelength of the light, m an integer, and Оёm 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, О» the
wavelength of the light, and Оё m the angle at which the minimum appears, and m is
3050
Chapter 33
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 d sin Оё = mО» for Оё with m = 1 to express the
location of the first-order maximum as a function of the wavelength of the light.
The interference maxima in a
diffraction pattern are at angles Оё
given by:
d sin Оё = mО»
where d is the separation of the slits
and m = 0, 1, 2, …
Solve for the angular location Оё1
of the first-order maximum :
вЋ›О»вЋћ
Оё1 = sin в€’1 вЋњ вЋџ
Because О»green light < О»red light:
Оё green light < Оё red light and (a) is correct.
⎝d ⎠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:
The separation of the interference
fringes is given by:
ym = m
О»n L
d
where L is the distance from the slits to
the screen, О»n is the wavelength of the
light in a medium whose index of
refraction is n, and d is the separation
of the slits.
y m +1 в€’ y m = (m + 1)
О»n L
d
в€’m
О»n L
d
=
О»n L
d
Because the index of refraction of a vacuum is slightly less than the index of
refraction of air, the removal of air increases О»n and, hence, ym в€’ ymв€’1. (c) is
correct.
Interference and Diffraction 3051
•
(a)
When waves interfere destructively, the energy is converted into heat
energy.
Interference patterns are observed only if the relative phases of the waves
that superimpose remain constant.
In the Fraunhofer diffraction pattern for a single slit, the narrower the slit,
the wider the central maximum of the diffraction pattern.
A circular aperture can produce both a Fraunhofer diffraction pattern and a
Fresnel diffraction pattern.
The ability to resolve two point sources depends on the wavelength of the
light.
(b)
(c)
(d)
(e)
[SSM]
True or false:
15
(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
mО»
where a is the width of the slit. Hence, the narrower the slit, the
Оё m = sin в€’1
a
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 ( О± c = 1.22
О»
D
).
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.
3052
Chapter 33
Determine the Concept The condition for the resolution of the two sources is
given by Rayleigh’s criterion: α c = 1.22λ D (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 О±c and О» 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: α c = 1.22 λ D (Equation 33-25), where αc is the
critical angular separation, D is the diameter of the aperture, and О» 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 lowEarth 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 О±c:
tan О± c =
w
в‡’ w = h tan О± c
h
Interference and Diffraction 3053
The minimum angular separation О±c
of two point objects that can just be
resolved by an eye depends on the
diameter D of the eye and the
wavelength О» of light:
Substitute for О±c in the expression
for w to obtain:
О± c = 1.22
О»
D
О»вЋћ
вЋ›
w = h tanвЋњ1.22 вЋџ
DвЋ вЋќ
Substitute numerical values and evaluate wmin for an altitude of 250 km:
вЋ›
вЋ› 400 nm вЋћ вЋћ
⎟⎟ ⎟⎟ ≈ 24 m
wmin = (250 km ) tan вЋњвЋњ1.22вЋњвЋњ
5
.
0
mm
вЋќ
вЋ вЋ вЋќ
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:
The critical angular separation is
also given by:
О± c = 1.22
О»
D
where D is the separation of the
headlights (or tail lights).
d
L
where d is the separation of head lights
(or tail lights) and L is the distance to
approaching or receding automobile.
О±=
3054
Chapter 33
Equate these expressions for О±c to
obtain:
О»
d
Dd
= 1.22 в‡’ L =
L
D
1.22О»
Substitute numerical values and
evaluate L:
L=
(5.0 mm)(1.5 m ) ≈
1.22(550 nm )
11 km
(b) For red light:
L=
(5.0 mm)(1.5 m ) ≈
1.22(640 nm )
9.6 km
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 onehalf 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:
f =
Let О”r be the distance between your
ears (the path difference) to obtain:
О”r = 12 О» в‡’ О» = 2О”r
Substituting for О» yields:
Substitute numerical values and
evaluate f:
f =
v
О»
v
2 О”r
343 m/s
= 0.86 kHz
2(20 cm )
or between 0.80 and 0.90 kHz.
f =
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:
О±=
The critical angular separation of the
two sources is given by the Rayleigh
criterion:
О± c = 1.22
For О± = О±c:
О»
d
Dd
= 1.22 в‡’ L =
L
D
1.22О»
Substitute numerical values and
evaluate L:
L=
d
L
О»
D
(5.0 mm)(50)(1.5 Г— 1011 m )
1.22(550 nm )
≈ 5.59 × 1013 km ×
1c в‹… y
9.461Г— 1015 m
5.9 c в‹… y
Phase Difference and Coherence
22 •
Light of wavelength 500 nm is incident normally on a film of water
1.00 Ојm 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 ПЂ 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 waterв€’air interface.
(a) Express the wavelength of light
in water in terms of the wavelength
of light in air:
О»water =
О»air
nwater
=
500 nm
= 376 nm
1.33
3056
Chapter 33
2t
2(1.00 Ојm )
= 5.32
376 nm
(b) Relate the number of
wavelengths N to the thickness t of
the film and the wavelength of light
in water:
N=
(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 waterв€’air interface:
Оґ = Оґ reflection + Оґ additional distance traveled
Substitute for N and evaluate Оґ:
Оґ = ПЂ rad + 2ПЂ (5.32 rad ) = 11.64ПЂ rad
=
О» water
=ПЂ +
2t
О»water
2ПЂ = ПЂ + 2ПЂ N
= 11.6ПЂ rad
or, subtracting 11.64ПЂ rad from 12ПЂ rad,
Оґ = 0.4ПЂ rad = 1.1 rad
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
О”r
according to Оґ =
2ПЂ . The path difference is the difference in the distances of
О»
(0, 15.0 cm) and (3.00 cm, 14.0 cm) from the origin.
О”r
Relate a path difference О”r to a
phase shift Оґ:
Оґ=
The path difference О”r is:
О”r = 15.0 cm в€’
О»
2ПЂ
(3.00 cm )2 + (14.0 cm )2
= 0.682 cm
Substitute numerical values and
evaluate Оґ:
вЋ› 0.682 cm вЋћ
⎟⎟2π ≈ 2.9 rad
вЋ Оґ = вЋњвЋњ
вЋќ 1.50 cm
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 mО». 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 180В°
(b) The mth fringe occurs when the
path difference 2t equals m
wavelengths:
2t = mО»
Relate the thickness of the air wedge
to the angle of the wedge:
t
в‡’ t = xОё
x
where we’ve used a small-angle
approximation to replace an arc length
by the length of a chord.
Substitute for t to obtain:
2 xОё = mО» в‡’ Оё =
Substitute numerical values and
evaluate Оё :
Оё= вЋњ
Оё=
mО» 1 m
О»
=
2x 2 x
1вЋ› 5 вЋћ
в€’4
вЋџ (700 nm ) = 1.75 Г— 10 rad
2 ⎝ cm ⎠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.
3058
Chapter 33
mО»
2
The mth fringe occurs when the path
difference 2d equals m wavelengths:
2 d = mО» в‡’ d =
Because the nineteenth (but not the
twentieth) bright fringe can be seen,
the limits on d must be:
(m в€’ 12 ) О» < d < (m + 12 ) О»
2
where m = 19
2
(19 в€’ 12 ) 590 nm < d < (19 + 12 ) 590 nm
Substitute numerical values to
obtain:
2
or
5.5 Ојm < d < 5.8 Ојm
2
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
the sum of a phase shift of ПЂ (equivalent to a О»/2 path difference) resulting from
reflection plus a phase shift due to the additional distance traveled.
1
glass plate
wire
2
t
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:
2t + 12 О» = О» , 2О» , 3О» , ...
or
2t = 12 О» , 32 О» , 52 О» , ...
and
2t = (m + 12 )λ 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:
Solve for the highest value of m:
Substitute numerical values and
evaluate m:
1
О» 2
where m = 0, 1, 2, …and 0 ≤ t ≤ 2r
m=
2t
в€’
вЋ› 2 (2r ) 1 вЋћ
вЋ› 4r 1 вЋћ
m max = int вЋњ
в€’ вЋџ = int вЋњ в€’ вЋџ
вЋќ О» 2вЋ 2вЋ вЋќ О»
where r is the radius of the wire.
вЋ› 4 (0.025m m )
m max = int вЋњ
в€’
вЋќ 600 nm
= int (166.2 ) = 166
Because we start counting from
m = 0, the number of bright fringes is
mmax + 1.
1вЋћ
2 вЋџвЋ 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.
3060
Chapter 33
(a) Express the condition for
destructive interference in the
thin film:
Solving for О» yields:
2t + 12 О»' = 12 О»' , 32 О»' , 52 О»' , ...
or
2t = О»' ,2О»' ,3О»' , ...
or
О»
(1)
2t = mО»' = m
n
where m = 1, 2, 3, … and λ′ is the
wavelength of the light in the film.
О»=
2nt
m
2nt
2nt
and 360 nm =
m
m +1
Substitute for the missing
wavelengths to obtain:
450 nm =
Divide the first of these equations
by the second and simplify to
obtain:
2nt
450 nm
m +1
= m =
2nt
360 nm
m
m +1
Solving for m gives:
m = 4 for О» = 450 nm
Solve equation (1) for t to obtain:
t=
mО»
2n
Substitute numerical values and
evaluate t:
t=
4(450 nm )
= 600 nm
2(1.50)
(b) The condition for constructive
interference in the thin film is:
2t + 12 О»' = О»' , 2О»' , 3О»' ,...
or
2t = 12 О»' , 32 О»' , 52 О»' ,... = (m + 12 )О»'
where λ′ is the wavelength of light in
the oil and m = 0, 1, 2, …
Substitute for λ′ to obtain:
2nt
О»
2t = (m + 12 ) в‡’ О» =
m + 12
n
where n is the index of refraction of the
film.
Substitute numerical values and
simplify to obtain:
О»=
2(1.50 )(600 nm ) 1800 nm
=
m + 12
m + 12
Interference and Diffraction 3061
Substitute numerical values for m and evaluate О» to obtain the following table:
m
О» (nm)
0
3600
1
1200
2
720
3
514
4
400
5
327
From the table, we see that the only wavelengths in the visible spectrum are 720
nm, 514 nm, and 400 nm.
2t = 12 О»' , 32 О»' , 52 О»' , ...
(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 12 О» (as is the wave reflected from
or
2t = (m +
1
2
)О»
n
where n is the index of refraction of the
film and m = 0, 1, 2, …
the top surface) and the condition for
destructive interference becomes:
Solving for О» yields:
Substitute numerical values and
simplify to obtain:
О»=
2nt
m + 12
О»=
2(1.5)(600 nm ) 1800 nm
=
m + 12
m + 12
Substitute numerical values for m and evaluate О» to obtain the following table:
m
О» (nm)
0
3600
1
1200
2
720
3
514
4
400
5
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
1
2
О» 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.
3062
Chapter 33
The condition for constructive
interference is:
2t = О»' , 2О»' , 3О»' ,...
or
2t = mО»' (1)
where λ′ is the wavelength of light in
the oil and m = 1, 2, 3, …
О»
mО»
2n
Substitute for λ′ to obtain:
2t = m
Substitute numerical values and
evaluate t:
(2)(650 nm ) =
t=
2(1.22)
n
в‡’t =
533 nm
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
1
2
О» 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.
2t = О»' , 2О»' , 3О»' ,... = mО»'
Express the condition for
constructive interference:
(1)
where λ′ is the wavelength of light in
the oil and m = 0, 1, 2, …
Substitute for λ′ to obtain:
2nt
n
m
where n is the index of refraction of the
oil.
Substitute for the predominant
wavelengths to obtain:
690 nm =
Divide the first of these equations by
the second and simplify to obtain:
2nt
690 nm
m +1
= m =
в‡’m = 2
2nt
460 nm
m
m +1
Solve equation (1) for t:
t=
2t = m
mО»
2n
О»
в‡’О» =
2nt
2nt
and 460 nm =
m
m +1
Interference and Diffraction 3063
Substitute numerical values and
evaluate t:
t=
(2)(690 nm ) =
2(1.45)
476 nm
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 ПЂ rad ( 12 О» ) 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 О» = 700 nm
and then use this value in our equation describing constructive interference to find
the thickness t of the oil film.
2t + 12 О»' = О»' , 2О»' , 3О»' ,...
Express the condition for
constructive interference between
the waves reflected from the airoil interface and the oil-glass
interface:
or
2t = 12 О»' , 32 О»' , 52 О»' ,... = (m + 12 )О»'
Substitute for О»' and solve for О»
to obtain:
О»=
Substitute the predominant
wavelengths to obtain:
700 nm =
Divide the first of these equations
by the second to obtain:
2nt
700 nm m + 12 m + 32
=
=
в‡’m = 2
2nt
m + 12
500 nm
m + 32
Solve equation (1) for t:
t = (m + 12 )
Substitute numerical values and
evaluate t:
(1)
where λ′ is the wavelength of light in
the oil and m = 0, 1, 2, …
2nt
m + 12
t = (2 + 12 )
2nt
2nt
and 500 nm =
1
m+ 2
m + 32
О»
2n
700 nm
= 603 nm
2(1.45)
3064
Chapter 33
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 2t = (m + 12 )О» 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° ( 12 λ ) 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.
(a) The condition for
constructive interference is:
2t + 12 О» = О» , 2О» , 3О» ,...
or
2t = 12 О» , 32 О» , 52 О» ,... = (m + 12 )О»
where О» is the wavelength of light in air
and m = 0, 1, 2, …
Solving for t yields:
(b) From Figure 33-42 we have:
t=
(m + 12 ) О» , m = 0,1, 2, ...
(1)
2
r 2 + (R в€’ t ) = R 2
or
R 2 = r 2 + R 2 в€’ 2 Rt + t 2
2
For t << R we can neglect the last
term to obtain:
R 2 ≈ r 2 + R 2 − 2 Rt ⇒ r =
(c) Square equation (2) and
substitute for t from equation (1) to
obtain:
r 2 = (m + 12 )RО» в‡’ m =
2 Rt
r2 1
в€’
RО» 2
(2)
Interference and Diffraction 3065
Substitute numerical values and
evaluate m:
m=
(2.00 cm)2 в€’ 1 = 67
(10.0 m )(590 nm) 2
and so there will be 68 bright fringes.
(d) The diameter of the mth fringe is:
Noting that m = 5 for the sixth
fringe, substitute numerical values
and evaluate D:
D = 2r = 2
(m + 12 )RО»
D = 2 (5 + 12 )(10.0 m )(590 nm )
= 1.14 cm
(e) The wavelength of the light in the film becomes О»air/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° ( 12 λ ) 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.
The condition for constructive
interference is:
2t + 12 О» = О» , 2О» , 3О» ,...
or
2t = 12 О» , 32 О» , 52 О» ,... = (m + 12 )О»
where О» is the wavelength of light in air
and m = 0, 1, 2, …
О»
Solving for t gives:
t = (m + 12 ) , m = 0,1, 2, ...
2
In Problem 31(b) it was shown
that:
r = 2tR
Substitute for t to obtain:
r=
(m + 12 )О»R
3066
Chapter 33
The first fringe corresponds to
m = 0:
r=
1
2
(520 nm)(2.00 m ) =
0.721mm
The second fringe corresponds to
m = 1:
r=
3
2
(520 nm)(2.00 m ) =
1.25 mm
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° ( 12 λ ) 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.
The condition for constructive
interference is:
2t + 12 О»' = О»' , 2О»' , 3О»' ,...
or
2t = 12 О»' , 32 О»' , 52 О»' ,... = (m + 12 )О»'
where λ′ is the wavelength of light in
the oil and m = 0, 1, 2, …
Substitute for λ′ and solve for t:
t = (m + 12 )
О»
, m = 0,1, 2, ...
2n
where О» is the wavelength of light in air.
In Problem 31(b) it was shown that:
r = 2tR
Substitute for t to obtain:
r=
(m + 12 ) О»R
The first fringe corresponds to
m = 0:
r=
1 (520 nm )(2.00 m )
= 0.535 mm
2
1.82
The second fringe corresponds to
m = 1:
r=
3 (520 nm )(2.00 m )
= 0.926 mm
2
1.82
n
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.
1
О”y
Express the number N of bright
fringes per centimeter in terms of
the separation of the fringes:
N=
Express the distance on the screen to
the mth and (m + 1)st bright fringe:
ym = m
Subtract the first of these equations
from the second to obtain:
О”y =
Substitute in equation (1) to obtain:
N=
d
О»L
Substitute numerical values and
evaluate N:
N=
1.00 mm
= 8.33 cm в€’1
(600 nm )(2.00 m )
(1)
О»L
d
and ym+1 = (m + 1)
О»L
d
О»L
d
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 О”y
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 О”y, we can find d from N, О»,
and L.
О»L
Express the distance on the screen to
the mth and (m + 1)st bright fringe:
ym = m
Subtract the first of these equations
from the second to obtain:
О”y =
Because the number of fringes per
unit length N is the reciprocal of О”y:
d = NО» L
d
О»L
d
and ym+1 = (m + 1)
в‡’d =
О»L
О”y
О»L
d
3068
Chapter 33
Substitute numerical values and
evaluate d:
(
)
d = 28 cm в€’1 (589 nm )(3.00 m )
= 4.95 mm
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 interference maxima
and require that sinθ ≤ 1.
y1 = 82 cm
d
Оё1
Оё
О»
L = 12 m
Because d << L, we can approximate
sinОё1 as:
sin θ1 ≈
From the right triangle whose sides
are L and y1 we have:
sin Оё1 =
Substitute numerical values in
equation (1) and evaluate d:
d≈
(b) The equation describing two-slit
interference maxima is:
0
О»
d
⇒d ≈
О»
sin Оё1
82 cm
(12 m )2 + (82 cm)2
(1)
= 0.06817
633 nm
= 9.29 Ојm = 9.3 Ојm
0.06817
d sin Оё = mО», m = 0,1, 2, ...
Because sinθ ≤ 1 determines the
maximum number of interference
fringes that can be seen:
d = mmax О» в‡’ mmax =
Substitute numerical values and
evaluate mmax:
mmax =
d
О»
9.29 Ојm
= 14 because m must
633 nm
be an integer.
Because there are 14 fringes on
either side of the central maximum:
N = 2mmax + 1 = 2(14) + 1 = 29
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:
ym = m
Subtract the first of these equations
from the second to obtain:
О”y =
Substitute numerical values and
evaluate О”y:
О”y =
О»L
d
and ym+1 = (m + 1)
О»L
О»L
d
(1)
d
(500 nm )(1.00 m ) =
1.00 cm
50.0 Ојm
(b) According to the Raleigh criterion you could resolve them, but not by much.
(c) Solve equation (1) for d to obtain:
Substitute numerical values and
evaluate d:
d=
d=
О»L
О”y
(500 nm )(1.00 m ) =
1.00 mm
0.500 mm
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 Оёm given by sin Оёm + sin П† = mО»/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 П† 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:
О”l = d sin П† + d sin Оё m
The condition for constructive
interference is:
О”l = mО»
where m is an integer.
3070
Chapter 33
d sin П† + d sin Оё m = mО»
Substitute to obtain:
Divide both sides of the equation by
d to obtain:
sin П† + sin Оё m =
mО»
d
39 ••
[SSM] White light falls at an angle of 30Вє to the normal of a plane
containing a pair of slits separated by 2.50 Ојm. 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 П† 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:
О”l = d sin П† + d sin Оё
The condition for constructive
interference is:
О”l = mО»
where m is an integer.
Substitute to obtain:
d sin П† + d sin Оё = mО»
Divide both sides of the equation
by d to obtain:
sin П† + sin Оё =
Set Оё = 0 and solve for О»:
Substitute numerical values and
simplify to obtain:
mО»
d
О»=
d sin П†
m
О»=
(2.50 Ојm )sin 30В° = 1.25 Ојm
m
m
Evaluate О» for positive integral values of m:
m
1
2
3
4
О» (nm)
1250
625
417
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.
Relate the distance О”y to the first
minimum from the center of the
central maximum to Оё and the
distance L from the speakers to the
plane of the microphone:
Interference minima occur where:
Solve for Оё to obtain:
Relate the wavelength О» of the sound
waves to the speed of sound v and
the frequency f of the sound:
y
Оё
S1
tan Оё =
d
y
в‡’ y = L tan Оё
L
L
S2
(1)
d sin Оё = (m + 12 )О»
where m = 0, 1, 2, 3, …
вЋЎ (m + 12 )О» вЋ¤
вЋҐвЋ¦
вЋЈ d
Оё = sin в€’1 вЋў
О»=
v
f
вЋЎ (m + 12 )v вЋ¤
вЋҐ
вЋЈ df вЋ¦
Substitute for О» in the expression for
Оё to obtain:
Оё = sin в€’1 вЋў
Substituting for Оё in equation (1)
yields:
вЋ§
вЋЎ (m + 12 )v вЋ¤ вЋ«
y = L tan вЋЁsin в€’1 вЋў
вЋҐвЋ¬
вЋЈ df вЋ¦ вЋ­
вЋ©
(2)
3072
Chapter 33
Noting that the first minimum corresponds to m = 0, substitute numerical values
and evaluate О”y:
вЋ§
вЋЎ ( 1 )(343 m/s ) вЋ¤ вЋ«
y1st min = (1.00 m ) tan вЋЁsin в€’1 вЋў 2
вЋҐ вЋ¬ = 37 cm
вЋЈ (5.0 cm )(10 kHz ) вЋ¦ вЋ­
вЋ©
The maxima occur where:
d sin Оё = mО»
where m = 1, 2, 3, …
For diffraction maxima, equation (2)
becomes:
вЋ§
вЋЎ mv вЋ¤ вЋ«
y = L tan вЋЁsin в€’1 вЋў вЋҐ вЋ¬
вЋЈ af вЋ¦ вЋ­
вЋ©
Noting that the first maximum corresponds to m = 1, substitute numerical values
and evaluate y:
вЋ§
вЋЎ (1)(343 m/s ) вЋ¤ вЋ«
y1st max = (1.00 m ) tan вЋЁsin в€’1 вЋў
вЋҐ вЋ¬ = 94 cm
вЋЈ (5.0 cm )(10 kHz ) вЋ¦ вЋ­
вЋ©
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.
вЋ›О»вЋћ
в‡’ Оё = sin в€’1 вЋњ вЋџ
a
вЋќaвЋ О»
The first zeroes in the intensity occur
at angles given by:
sin Оё =
(a) For a = 1.0 mm:
Оё = sin в€’1 вЋњвЋњ
(b) For a = 0.10 mm:
Оё = sin в€’1 вЋњвЋњ
(c) For a = 0.010 mm:
Оё = sin в€’1 вЋњвЋњ
вЋ› 600 nm вЋћ
вЋџвЋџ = 0.60 mrad
вЋќ 1.0 mm вЋ вЋ› 600 nm вЋћ
вЋџвЋџ = 6.0 mrad
вЋќ 0.10 mm вЋ вЋ› 600 nm вЋћ
вЋџвЋџ = 60 mrad
вЋќ 0.010 mm вЋ 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 Оё = 37Вє. 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.
О»
The first zeroes in the intensity occur
at angles given by:
sin Оё =
Substitute numerical values and
evaluate О»:
О» = (5.0 cm )sin 37В° = 3.0 cm
a
в‡’ О» = a sin Оё
43 ••• [SSM] Measuring the distance to the moon (lunar ranging) 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 Г— 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.
dtelescope
Relate the diameter D of the beam
when it reaches the moon to the
distance to the moon L and the beam
divergence angle Оё :
L
dmirror D
D ≈θ L
(1)
3074
Chapter 33
The angle Оё subtended by the first
diffraction minimum is related to the
wavelength О» of the light and the
diameter of the telescope opening
dtelescope by:
Because θ << 1, sinθ ≈ θ and:
Substitute for Оё in equation (1) to
obtain:
О»
sin Оё = 1.22
θ ≈ 1.22
D=
d telescope
О»
d telescope
1.22 LО»
d telescope
Substitute numerical values and evaluate D:
вЋЎ
вЋ¤
вЋў
вЋҐ
1.22(500 nm )
вЋҐ = 1.53 km
D = 3.82 Г— 10 8 m вЋў
вЋў 6.00 in Г— 2.54 cm Г— 1 m вЋҐ
вЋўвЋЈ
in
10 2 cm вЋҐвЋ¦
(
)
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 = na ) for any
value of n?
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 N = 2m в€’ 1 fringes in the central maximum.
The number of fringes N in the
central maximum is:
Relate the angle Оё1 of the first
diffraction minimum to the width a
of the slits of the diffraction grating:
Express the angle Оёm corresponding
to the mth interference maxima in
terms of the separation d of the slits:
N = 2m в€’ 1
sin Оё1 =
sin Оё m =
О»
a
mО»
d
(1)
Interference and Diffraction 3075
Because we require that Оё1 = Оёm,
we can equate these expressions
to obtain:
d
mО» О»
= в‡’ m=
d
a
a
Substituting for d and simplifying
yields:
m=
Substitute for m in equation (1) to
obtain:
N = 2(5) в€’ 1 = 9
If d = na:
m=
5a
=5
a
d na
=
= n and N = 2n в€’ 1
a
a
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 N = 2m в€’ 1.
(a) Relate the angle Оё1 of the first
diffraction minimum to the width a
of the slits of the diffraction grating:
Express the angle Оё5 corresponding
to the mth fifth interference maxima
maximum in terms of the separation
d of the slits:
sin Оё1 =
sin Оё 5 =
О»
a
5О»
d
Because we require that Оё1 = Оёm5, we
can equate these expressions to
obtain:
d
5О» О»
= в‡’ a=
d
a
5
Substituting the numerical value of d
yields:
a=
(b) Because m = 5:
N = 2m в€’ 1 = 2(5) в€’ 1 = 9
0.100 mm
= 20.0 Ојm
5
3076
Chapter 33
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 N = 2m в€’ 1.
The number of fringes N in the
central maximum is:
Relate the angle Оё1 of the first
diffraction minimum to the width a
of the slits of the diffraction grating:
Express the angle Оёm corresponding
to the mth interference maxima in
terms of the separation d of the slits:
N = 2m в€’ 1
sin Оё1 =
sin Оё m =
(1)
О»
a
mО»
d
Because we require that Оё1 = Оёm, we
can equate these expressions to
obtain:
mО» О»
d
= в‡’m =
d
a
a
Substitute for m in equation (1) to
obtain:
N=
2d
в€’1
a
Substitute numerical values and
evaluate N:
N=
2(0.20 mm )
в€’ 1 = 39
0.010 mm
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 N = 2m в€’ 1. In (b) we can
use the expression relating the intensity in a single-slit diffraction pattern to phase
2ПЂ
constant П† =
a sin Оё to find the ratio of the intensity of the third interference
О»
maximum to one side of the center interference maximum.
(a) The number of fringes N in the
central maximum is:
Relate the angle Оё1 of the first
diffraction minimum to the width a
of the slits of the diffraction grating:
Express the angle Оёm corresponding
to the mth interference maxima in
terms of the separation d of the slits:
N = 2m в€’ 1
sin Оё1 =
sin Оё m =
(1)
О»
a
mО»
d
Because we require that Оё1 = Оёm, we
can equate these expressions to
obtain:
mО» О»
d
= в‡’m =
d
a
a
Substitute in equation (1) to obtain:
N=
2d
в€’1
a
Substitute numerical values and
evaluate N:
N=
2(0.15 mm )
в€’1 = 9
0.030 mm
(b) Express the intensity for a singleslit diffraction pattern as a function
of the phase difference П†:
вЋ› sin 1 П† вЋћ
I = I 0 вЋњвЋњ 1 2 вЋџвЋџ
вЋќ 2П† вЋ 2ПЂ
where П† =
a sin Оё
О»
2
(2)
3078
Chapter 33
For m = 3:
sin Оё 3 =
and
П†=
2ПЂ
О»
3О»
d
a sin Оё 3 =
2ПЂ вЋ› 3О» вЋћ
вЋ›aвЋћ
aвЋњ вЋџ = 6ПЂ вЋњ вЋџ
О» вЋќ d вЋ вЋќd вЋ вЋ› 0.030 mm вЋћ 6ПЂ
вЋџвЋџ =
вЋќ 0.15 mm вЋ 5
Substitute numerical values and
evaluate П†:
П† = 6ПЂ вЋњвЋњ
Solve equation (2) for the ratio of I3
to I0:
I 3 вЋ› sin 12 П† вЋћ
вЋџ
=вЋњ
I 0 вЋњвЋќ 12 П† вЋџвЋ Substitute numerical values and
evaluate I3/I0:
вЋЎ 1 вЋ› 6ПЂ вЋћ вЋ¤
sin вЋњ
вЋџ
I3 вЋў 2 вЋќ 5 вЋ вЋҐ
вЋҐ = 0.25
=вЋў
I 0 вЋў 1 вЋ› 6ПЂ вЋћ вЋҐ
вЋў 2 вЋњвЋќ 5 вЋџвЋ вЋҐ
вЋ¦
вЋЈ
2
2
Using Phasors to Add Harmonic Waves
49 •
[SSM] Find the resultant of the two waves whose electric fields at a
r
given location vary with time as follows: E1 = 2 A0 sin ωt iˆ and
r
E = 3 A sin (ωt + 3 π ) iˆ .
2
0
2
The resultant of the two waves is of
the form:
r
The magnitude of E is:
The phase angle Оґ is:
r
E1
y
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.
x
Оґ
r
E
r
E2
r
r
E = E sin (П‰t + Оґ ) i
r
E =
(2 A0 )2 + (3 A0 )2
(1)
= 3.6 A0
вЋ› в€’ 3 A0 вЋћ
вЋџвЋџ = в€’0.98 rad
вЋќ 2 A0 вЋ Оґ = tan в€’1 вЋњвЋњ
Interference and Diffraction 3079
r
E = 3.6 A0 sin (П‰t в€’ 0.98 rad ) i
r
Substitute for E and Оґ in equation
(1) to obtain:
50 •
Find the resultant of the two waves whose electric fields at a given
r
r
location vary with time as follows: E1 = 4 A0 sin ωt iˆ and E 2 = 3 A0 sin (ωt + 16 π ) iˆ .
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.
r
E
r
E2
y
Оґ
The resultant of the two waves is of
the form:
Apply the law of cosines to the
magnitudes of the scalars to obtain:
60В°
r
E1
x
r
r
E = E sin (П‰t + Оґ ) i
(1)
r2
r 2 r 2
r r
E = E1 + E 2 в€’ 2 E1 E 2 cos120В°
or
r
E =
r 2 r 2
r r
E1 + E 2 в€’ 2 E1 E 2 cos120В°
r
r
r
Substitute for E1 and E 2 and evaluate E to obtain:
r
E =
(4 A0 )2 + (3 A0 )2 в€’ 2(4 A0 )(3 A0 )cos120В° = 6.08 A0
Applying the law of sines yields:
sin Оґ sin 120В°
r =
r
E2
E
Solve for Оґ to obtain:
r
вЋЎ E 2 sin 120В° вЋ¤
вЋҐ
Оґ = sin в€’1 вЋў
r
вЋҐ
вЋў
E
вЋ¦
вЋЈ
Substitute numerical values and
evaluate Оґ:
Оґ = sin в€’1 вЋў
вЋЎ 3 A0 sin 120В° вЋ¤
вЋҐ = 0.43 rad
вЋЈ 6.08 A0 вЋ¦
3080
Chapter 33
r
Substitute for E and Оґ in equation
r
E = 6.1A0 sin (П‰t + 0.43 rad ) i
(1) to obtain:
Remarks: We could have used the law of cosines to find R and the law of
sines to find Оґ.
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.
2
The intensity at the second secondary
maximum is given by:
вЋЎ sin 1 П† вЋ¤
I вЋЎ sin 12 П† вЋ¤
=вЋў 1 вЋҐ
I = I0 вЋў 1 2 вЋҐ в‡’
П†
I
0
вЋЈ 2П† вЋ¦
вЋЈ 2 вЋ¦
where
2ПЂ
П†=
a sin Оё
О»
At this second secondary maximum:
a sin Оё =
Substitute for П† and evaluate
I
:
I0
2
2ПЂ вЋ› 5О» вЋћ
5
О» and П† =
вЋњ вЋџ = 5ПЂ
О» вЋќ 2 вЋ 2
вЋЎ вЋ› 5ПЂ
sinвЋњ
I вЋў вЋќ 2
=вЋў
I 0 вЋў 5ПЂ
вЋў
2
вЋЈ
2
вЋћвЋ¤
вЋџвЋҐ
вЋ вЋҐ = 0.0162
вЋҐ
вЋҐ
вЋ¦
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 >> О») are given approximately by
y m = mО»L 3d 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 Оґ 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 О”r 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 Оґ in
terms of the number of phasors N
forming a closed polygon of N sides:
вЋ› 2ПЂ вЋћ
вЋџ
⎝ N ⎠where m = 1, 2, 3, 4, 5, 6, ,7, …
For three equally spaced sources,
the phase angle is:
Оґ = mвЋњ
Express the path difference
corresponding to this phase angle to
obtain:
О»
вЋ› О» вЋћ
О”r = вЋњ
вЋџОґ = m
3
вЋќ 2ПЂ вЋ 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.)
Express the path difference О”r in
terms of sinОё and the separation
d of the slits:
О”r = d sin Оё
or, provided the small angle
approximation is valid,
L
yd
в‡’ y = О”r
О”r =
d
L
Оґ = mвЋњ
вЋ› 2ПЂ вЋћ
вЋџ
вЋќ 3 вЋ Substituting for О”r from equation (1)
yields:
y min =
(b) For L = 1.00 m, О» = 500 nm, and
d = 0.100 mm:
2 y min =
(1)
mО» L
, m = 1, 2, 4, 5 ,7, 8, ...
3d
2(500 nm )(1.00 m )
= 3.33 mm
3(0.100 mm )
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 >> О») are given approximately by
y m = mО»L 4d 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
3082
Chapter 33
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 Оґ 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 О”r 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 Оґ in
terms of the number of phasors N
forming a closed polygon of N sides:
вЋ› 2ПЂ вЋћ
вЋџ
⎝ N ⎠where m = 1, 2, 3, 4, 5, 6, ,7, …
For four equally spaced sources, the
phase angle is:
вЋ›ПЂ вЋћ
Оґ = mвЋњ вЋџ
Express the path difference
corresponding to this phase angle to
obtain:
О»
вЋ› О» вЋћ
О”r = вЋњ
вЋџОґ = m
4
вЋќ 2ПЂ вЋ 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.)
Express the path difference О”r in
terms of sinОё and the separation
d of the slits:
О”r = d sin Оё
or, provided the small angle
approximation is valid,
L
yd
в‡’ y = О”r
О”r =
d
L
Substituting for О”r from equation (1)
yields:
Оґ = mвЋњ
вЋќ2вЋ y min =
(1)
mО»L
, m = 1, 2, 3, 5 ,6, 7, 9,...
4d
(b) For L = 2.00 m, О» = 600 nm,
d = 0.100 mm, and n = 1:
2 y min =
2(600 nm )(2.00 m )
= 6.00 mm
4(0.100 mm )
For two slits:
2 ymin =
2(m + 12 )О»L
d
Interference and Diffraction 3083
For L = 2.00 m, О» = 600 nm,
d = 0.100 mm, and m = 0:
2 y min =
(600 nm )(2.00 m ) = 12.0 mm
0.100 mm
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 Ојm wide and the center-to-center separation between it and the next slit is
6.00 Ојm. (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 sin Оё = О» a to find the first zeros in the intensity
pattern. The four-slit interference maxima occur at angles given by
d sin θ = mλ, 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 Оґ, that, in turn, is a function of the angle
Оё that determines the location of a point in the interference pattern.
вЋ›О»вЋћ
в‡’ Оё = sin в€’1 вЋњ вЋџ
a
вЋќaвЋ О»
(a) The first zeros in the intensity
occur at angles given by:
sin Оё =
Substitute numerical values and
evaluate Оё :
Оё = sin в€’1 вЋњвЋњ
(b) The four-slit interference
maxima occur at angles given by:
вЋЎ mО» вЋ¤
d sin Оё = mО» в‡’ Оё m = sin в€’1 вЋў
вЋЈ d вЋҐвЋ¦
where m = 0, 1, 2, 3, …
Substitute numerical values to
obtain:
Оё m = sin в€’1 вЋў
вЋ› 480 nm вЋћ
вЋџвЋџ = 242 mrad
вЋќ 2.00 Ојm вЋ вЋЎ m(480 nm ) вЋ¤
вЋҐ
вЋЈ 6.00 Ојm вЋ¦
= sin в€’1 (0.0800m )
3084
Chapter 33
Оё 0 = sin в€’1 [0(0.0800)] = 0
Evaluate Оёm for m = 0, 1, 2, and 3:
Оё1 = sin в€’1 [1(0.0800)] = В± 80.1mrad
Оё 2 = sin в€’1 [2(0.0800)] = В± 161mrad
Оё 3 = sin в€’1 [3(0.0800)] = В±0.242 rad
where Оё3 will not be seen as it
coincides with the first minimum in the
diffraction pattern.
(c) From Problem 53:
Оё min =
nО»
4d
For n = 1:
Оё min =
480 nm
= 20 mrad
4(6.00 Ојm )
(d) Use the phasor method to show the superposition of four waves of the same
2ПЂ
amplitude A0 and constant phase difference Оґ =
d sin Оё .
О»
О±
Оґ"
A0
A
П†
П†
Оґ
Оґ
A0
Оґ'
Оґ"
О±
f
Оґ
A0
A0
Express A in terms of δ ′ and δ ′′:
A = 2( A0 cos Оґ '' + A0 cos Оґ ' )
Because the sum of the external
angles of a polygon equals 2ПЂ:
2О± + 3Оґ = 2ПЂ
(1)
Interference and Diffraction 3085
Examining the phasor diagram we
see that:
О± + Оґ '' = ПЂ
Eliminate О± and solve for Оґ '' to
obtain:
Оґ '' = 32 Оґ
Because the sum of the internal
angles of a polygon of n sides is
(n в€’ 2)ПЂ :
From the definition of a straight
angle we have:
3П† + 2Оґ '' = 3ПЂ
П† в€’Оґ ' +Оґ = ПЂ
Eliminate П† between these equations
to obtain:
Оґ ' = 12 Оґ
Substitute for Оґ '' and Оґ ' in equation
(1) to obtain:
A = 2 A0 (cos 32 Оґ + cos 12 Оґ )
Because the intensity is
proportional to the square of the
amplitude of the resultant wave:
I = 4 I 0 (cos 32 Оґ + cos 12 Оґ )
2
The following graph of I/I0 as a function of sinОё was plotted using a spreadsheet
2
вЋ› sin 1 П† вЋћ
I
program. The diffraction envelope was plotted using
= 4 2 вЋњвЋњ 1 2 вЋџвЋџ , where
I0
вЋќ 2П† вЋ П†=
2ПЂ
О»
and (c).
a sin Оё . Note the excellent agreement with the results calculated in (a), (b)
3086
Chapter 33
18
16
intensity
14
diffraction envelope
12
10
I /I 0 8
6
4
2
0
-2
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
sin(theta)
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 Оґ from the geometry of
the diagram to the right. Using the
value of Оґ 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.
(a) Express Оґ for the adjacent slits:
For θ << 1, sin θ ≈ tan θ ≈ θ :
y 1.72 cm
Оё
L = 2.50 m
Оґ=
2ПЂ
О»
d sin Оё
sin θ ≈ tan θ =
y
L
Interference and Diffraction 3087
Substitute to obtain:
Оґ=
Substitute numerical values and
evaluate Оґ :
Оґ=
The three phasors, 270В° 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.
2ПЂdy
О»L
2ПЂ (60.0 Ој m )(1.72 cm )
(550 nm )(2.50 m )
3ПЂ
=
rad = 270В°
2
Оґ
A0
R = A0
A0
Оґ
A0
(b) Express the intensity at the point
1.72 cm from the centerline:
I в€ќ R2
Because I0 в€ќ 9R2:
I
I
R2
=
в‡’ I= 0
2
I 0 9R
9
Substitute for I0 and evaluate I:
50.0 mW/m 2
I=
= 5.56 mW/m 2
9
56 ••• In single-slit Fraunhofer diffraction, the intensity pattern (Figure 3311) consists of a broad central maximum with a sequence of secondary maxima to
2
вЋ› sin 12 П† вЋћ
вЋџ ,
either side of the central maximum. This intensity is given by I = I 0 вЋњвЋњ 1
вЋџ
П†
вЋќ 2
вЋ where П† is the phase difference between the wavelets arriving from the opposite
edges of the slits. Calculate the values of П† 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 П† of 3ПЂ, 5ПЂ and 7ПЂ. (That these values for П† are approximately correct
at the secondary intensity maxima is discussed in the discussion surrounding
Figure 33-27.)
3088
Chapter 33
Picture the Problem We can use the phasor diagram shown in Figure 33-26 to
determine the first three values of П† 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 П† corresponding to the intensity maxima in
the diffraction pattern.
Referring to Figure 33-26 we see
that the first subsidiary maximum
occurs where:
An intensity minimum occurs where:
Another intensity maximum occurs
where:
Thus, secondary intensity maxima
occur where:
The intensity in the single-slit
diffraction pattern is given by:
П† = 3ПЂ
П† = 4ПЂ
П† = 5ПЂ
П† = (2n + 1)ПЂ, n = 1, 2, 3, ...
and the first three secondary intensity
maxima are at П† = 3ПЂ , 5ПЂ , and 7ПЂ
вЋ› sin 12 П† вЋћ
вЋџ
I = I 0 вЋњвЋњ 1
вЋџ
П†
вЋќ 2
вЋ 2
Set the derivative of this expression equal to zero for extreme values (relative
minima and maxima):
вЋ› sin 1 П† вЋћ вЋЎ 1 П† cos 12 П† в€’ 12 sin 12 П† вЋ¤
dI
= 2 I 0 вЋњвЋњ 1 2 вЋџвЋџ вЋў 4
вЋҐ = 0 for extrema
2
1
dφ
П†
(
)
П†
вЋў
вЋҐвЋ¦
вЋќ 2
вЋ вЋЈ
2
Simplify to obtain the transcendental
equation:
Solve this equation numerically
(use the ″Solver″ function of your
calculator or trial-and-error
methods) to obtain:
tan 12 П† = 12 П†
П† = 2.86ПЂ, 4.92ПЂ, and 6.94ПЂ
At the three intensity minima П† = 2ПЂ, 4ПЂ, and 6ПЂ , 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
О»
to find the angle between
Оё = 1.22
D
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 first
diffraction minimum on a screen 8 m
away from the pinhole.
D
L
О»
(a) The angle between the central
maximum and the first diffraction
minimum for a Fraunhofer
diffraction pattern is given by:
Оё = 1.22
Substitute numerical values and
evaluate Оё :
Оё = 1.22вЋњвЋњ
(b) Referring to the diagram, we see
that:
ymin = L tanОё
Substitute numerical values and
evaluate ymin:
ymin
Оё
D
вЋ› 700 nm вЋћ
вЋџвЋџ = 8.54 mrad
0
.
100
mm
вЋќ
вЋ y min = (8.00 m ) tan (8.54 mrad)
= 6.83 cm
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?
3090
Chapter 33
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 О”y.
Pinhole
О”y
О±c
О±c
L
О»
Rayleigh’s criterion is satisfied
provided:
О± c = 1.22
Relate О±c to the separation О”y of the
light sources:
αc ≈
Equate these expressions to obtain:
О”y
О»
О»L
= 1.22 в‡’ О”y = 1.22
L
D
D
Substitute numerical values and
evaluate О”y:
О”y = 1.22
D
О”y
provided О±c << 1.
L
(700 nm )(10.0 m ) =
0.100 mm
8.54 cm
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 О», L,
and the width a of the slit.
О±c
О±c
x
L
Interference and Diffraction 3091
Referring to the diagram, relate О±c,
L, and x:
αc ≈
For slits, Rayleigh’s criterion is:
О±c =
x
L
О»
a
Equate these two expressions to
obtain:
О»L
x О»
= в‡’x =
L a
a
Substitute numerical values and
evaluate x:
x=
(700 nm )(5.00 m ) =
0.500 mm
7.00 mm
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 О», x, and the diameter D of your pupil.
Your pupil
О±c
x
О±c
L
x
provided О± << 1
L
(a) Referring to the diagram, relate
О±c, L, and x:
αc ≈
For circular apertures, Rayleigh’s
criterion is:
О± c = 1.22
Equate these two expressions to
obtain:
О»
D
x
О»
xD
= 1.22 в‡’ L =
L
D
1.22О»
3092
Chapter 33
Substitute numerical values and
evaluate L:
L=
(6.0 mm )(5.0 mm ) =
1.22(500 nm )
49 m
(b) Because L is inversely proportional to О», 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 О”x of the stars.
Your pupil
О±c
О±c
О”x
L
О»
Rayleigh’s criterion is satisfied
provided:
О± c = 1.22
Relate О±c to the separation О”x of
the light sources:
αc ≈
Equate these expressions to obtain:
О”x
О»
О»L
= 1.22 в‡’ О”x = 1.22
L
D
D
D
О”x
because О±c << 1
L
Substitute numerical values and evaluate О”x:
15
вЋ›
вЋ›
вЋћвЋћ
вЋњ (550 nm )вЋњ 4 c в‹… y Г— 9.461Г— 10 m вЋџ вЋџ
вЋњ
вЋџвЋџ
1c в‹… y
вЋњ
вЋќ
вЋ = 5.00 Г— 10 9 m
О”x = 1.22вЋњ
вЋџ
2.54 cm
вЋњ
вЋџ
200 in Г—
вЋњ
вЋџ
1in
вЋќ
вЋ 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
О±c
О”x
D
О±c
L
О»
О»
О±c
Rayleigh’s criterion is satisfied
provided:
О± c = 1.22
Substitute numerical values and
evaluate D:
вЋћ
вЋ›
вЋџ
вЋњ
550
nm
вЋџ
D = 1.22вЋњ
вЋњ 14'' Г— 1В° Г— ПЂ rad вЋџ
вЋџ
вЋњ
3600'' 180В° вЋ вЋќ
D
в‡’ D = 1.22
= 9.9 mm ≈ 1cm
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 d sin Оё = mО» for Оё with m = 1 to express the
location of the first-order maximum as a function of the wavelength of the light.
(a) The interference maxima in a
diffraction pattern are at angles Оё
given by:
d sin Оё = mО»
where d is the separation of the slits
and m = 0, 1, 2, …
3094
Chapter 33
Solve for the angular location Оёm of
the maxima :
Relate the number of slits N per
centimeter to the separation d of the
slits:
вЋ› mО» вЋћ
вЋџ
вЋќ d вЋ Оё m = sin в€’1 вЋњ
N=
1
d
Substitute for d to obtain:
Оё m = sin в€’1 (mNО» )
EvaluateОё for О» = 434 nm and m =1:
Оё1 = sin в€’1 [(2000 cm в€’1 )(434 nm )]
(1)
= 86.9 mrad
EvaluateОё for О» = 410 nm and m = 1:
Оё1 = sin в€’1 [(2000 cm в€’1 )(410 nm )]
= 82.1 mrad
(b) Use equation (1) to evaluateОё for
О» = 434 nm and m = 1:
Evaluate Оё for О» = 410 nm and m = 1:
Оё1 = sin в€’1 [(15000 cm в€’1 )(434 nm )]
= 709 mrad
Оё1 = sin в€’1 [(15000 cm в€’1 )(410 nm )]
= 662 mrad
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 d sin Оё = mО» for О» with m = 1 to express the
location of the first-order maximum as a function of the angles at which the firstorder images are found.
The interference maxima in a
diffraction pattern are at angles Оё
given by:
d sin Оё
m
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:
N=
d sin Оё = mО» в‡’ О» =
1
d
Interference and Diffraction 3095
sin Оё
N
Let m = 1 and substitute for d to
obtain:
О»=
Substitute numerical values and
evaluate λ1 for θ1 = 9.72 ×10–2 rad:
sin 9.72 Г—10в€’2 rad
О»1 =
= 485 nm
2000 cmв€’1
Substitute numerical values and
evaluate λ1 for θ 2 = 1.32 ×10–1 rad:
О»1 =
(
)
(
)
sin 1.32 Г—10в€’1 rad
= 658 nm
2000 cmв€’1
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 grating equation to find the angle at which
normally incident blue light will be diffracted by the butterfly’s wings.
The grating equation is:
Solve for Оё to obtain:
Substitute numerical values and
evaluate Оё1:
d sin θ = mλ , where m = 1, 2, 3, …
вЋЎ mО» вЋ¤
вЋЈ d вЋҐвЋ¦
Оё = sin в€’1 вЋў
вЋЎ (1)(440 nm )вЋ¤
вЋҐ = 30.0В°
вЋЈ 880 nm вЋ¦
Оё = sin в€’1 вЋў
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:
Solve the grating equation for Оё :
О”Оё = Оё 579 в€’ Оё 577
вЋ› mО» вЋћ
вЋџ
вЋќ d вЋ Оё = sin в€’1 вЋњ
3096
Chapter 33
Substitute for Оё579 and Оё577 to obtain:
вЋ¤
вЋЎ
вЋ¤
вЋЎ
вЋў m(577 nm ) вЋҐ
вЋў m(579 nm ) вЋҐ
вЋҐ = sin в€’1 [0.1158m] в€’ sin в€’1 [0.1154m]
вЋҐ в€’ sin в€’1 вЋў
О”Оё = sin в€’1 вЋў
1
1
вЋҐ
вЋў
вЋҐ
вЋў
вЋўвЋЈ 2000 cm в€’1 вЋҐвЋ¦
вЋўвЋЈ 2000 cm в€’1 вЋҐвЋ¦
For m = 1:
О”Оё = sin в€’1 [0.1158m] в€’ sin в€’1 [0.1154m]
= 6.65В° в€’ 6.63В° = 0.02В°
(b) The width of the beam necessary
for these lines to be resolved is given
by:
w = Nd
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:
О»
= mN
О”О»
For m = 1:
N=
О»
О”О»
Substitute for N in equation (1) to
obtain:
w=
О»d
О”О»
Letting О» be the average of the
two wavelengths, substitute
numerical values and evaluate w:
(1)
вЋ›
w=
вЋћ
1
вЋџ
в€’1 вЋџ
вЋќ 2000 cm вЋ = 1mm
2 nm
(578 nm )вЋњвЋњ
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 d sin Оё = mО», m = 1, 2, 3, ...
to express the order number in terms of the slit separation d, the wavelength of
the light О», and the angle Оё.
Interference and Diffraction 3097
The interference maxima in the
diffraction pattern are at angles Оё
given by:
If one is to see the complete
spectrum, it must be true that:
d sin Оё = mО» в‡’ m =
d sin Оё
О»
where m = 1, 2, 3, …
sin θ ≤ 1 ⇒ m ≤
Evaluate mmax:
mmax =
d
О»
1
4800 cm в€’1
О»max
1
4800 cm в€’1
=
= 2.98
700 nm
Because mmax = 2.98, one can see the complete spectrum only for m = 1 and 2.
Express the condition for overlap:
m1 λ1 ≥ m2 λ 2
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 Г— 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 Оё
given by:
d sin Оё = mО»,
where m = 1, 2, 3, ...
(1)
The resolving power R is given by:
R = mN
where N is the number of slits and m is
the order number.
Relate d to the width w of the
grating:
d=
w
N
Substitute for N and simplify to
obtain:
d=
mw
R
3098
Chapter 33
Substitute for d in equation (1) to
obtain:
mw
вЋ› RО» вЋћ
sin Оё = mО» в‡’ Оё = sin в€’1 вЋњ
вЋџ
R
вЋќ w вЋ Substitute numerical values and
evaluate Оё :
Оё = sin в€’1 вЋў
вЋЎ (22,000)(510 nm ) вЋ¤
вЋҐ = 13.0В°
5.00 cm
вЋЈ
вЋ¦
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 d sin θ = mλ (where d is the slit separation and m = 0, 1, 2, …) and
the geometry of the grating and projection screen. We can use
Оё min = О» Nd = О”y 2 L 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 Оё at which maxima
occur is related to the slit separation
d, the wavelength of the incident
light О», and the order number m
according to:
Оё is also related to the distance to
the screen L and the positions of
the intensity maxima ym:
Substituting for sin Оё in equation
(1) yields:
Substituting numerical values
yields:
d sin Оё = mО»
where m = 0, 1, 2, … .
ym
sin Оё =
L + ym2
2
dym
L2 + ym2
ym =
(1)
= mО» в‡’ ym =
mО» L
d 2 в€’ m 2О»2
m(589 nm )(1.50 m )
2
вЋ› 1.00 cm вЋћ
2
2
вЋњ
вЋџ в€’ m (589 nm )
4000
вЋќ
вЋ Interference and Diffraction 3099
Evaluate this expression for m = 1
and m = 2 to obtain:
y1 = 36.4 cm and y 2 = 80.1cm
(b) The angle Оёmin that locates the
first minima in the diffraction
pattern is given by:
О”y
2 LО»
в‡’ О”y =
Nd
Nd 2 L
where О”y is the width of the central
maximum.
Substitute numerical values and
evaluate О”y:
О”y =
Оё min =
О»
=
2(1.50 m )(589 nm )
вЋ›
вЋћ
(8000 lines)вЋњвЋњ 1 в€’1 вЋџвЋџ
вЋќ 4000 cm вЋ = 88.4 Ојm
(c) The resolving power R in the
mth order is given by:
R = mN
Substitute numerical values and
evaluate R:
R = (1)(8000) = 8000
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:
w = Nd
The resolving power R of the grating
is given by:
R=
О»
О»
= mN в‡’ N =
О”О»
m О”О»
Substitute for N in the expression for
w to obtain:
w=
О»d
mО”О»
3100
Chapter 33
Letting О» be the average of the given wavelengths, substitute numerical values
and evaluate w:
1
2
w=
вЋ›
вЋћ
1
вЋџ
в€’1 вЋџ
вЋќ 8400 cm вЋ = 3 cm
2(519.322 nm в€’ 519.313 nm )
(519.313 nm + 519.322 nm )вЋњвЋњ
•• [SSM] Mercury has several stable isotopes, among them 198Hg and
202
Hg. 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
198
Hg and 202Hg are 546.07532 nm and 546.07355 nm, respectively. What must be
the resolving power of a 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?
71
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:
R=
О»
= mN
О”О»
Substitute numerical values and
evaluate R:
R=
546.07532
546.07532 в€’ 546.07355
= 3.0852 Г— 10 5 = 3.09 Г— 10 5
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:
n=
N
w
From equation (1) we have:
N=
R
m
Substitute for N to obtain:
n=
R
mw
(1)
Interference and Diffraction 3101
Substitute numerical values and
evaluate n:
n=
3.0852 Г—105
= 5.14 Г—10 4 cm в€’1
(3)(2.00 cm)
72 ••• A diffraction grating has n lines per unit length. Show that the angular
separation (О”Оё ) of two lines of wavelengths О» and О» + О”О» is approximately
1
О”Оё = О”О»
в€’ О»2 where m is the order number.
2
n m2
Picture the Problem We can differentiate the grating equation implicitly and
approximate dОё /dО» by О”Оё /О”О» to obtain an expression О”Оё as a function of m, n,
О”О», and cosОё. We can use the Pythagorean identity sin2Оё + cos2Оё = 1 and the
grating equation to write cosОё in terms of n, m, and О». Making these substitutions
will yield the given equation.
The grating equation is:
d sin Оё = mО», m = 0, 1, 2, ...
Differentiate both sides of this
equation with respect to О»:
d
(d sin Оё ) = d (mО» )
dО»
dО»
or
dОё
d cos Оё
=m
dО»
Because n = 1/d:
cos Оё
n=
Solving for О”Оё yields:
О”Оё =
Substitute for cosОё to obtain:
From equation (1):
Substituting for sinОё yields:
1
dОё
dОё
= nm в‡’ n = cos Оё
m
dО»
dО»
1 О”Оё
cos Оё
m О”О»
Approximate dОё /dО» by О”Оё /О”О»:
nmО”О»
cos Оё
О”Оё =
sin Оё =
О”Оё =
(1)
nmО”О»
1 в€’ sin 2 Оё
mО»
= nmО»
d
nmО”О»
1 в€’ n 2 m 2О»2
3102
Chapter 33
Simplify by dividing the numerator and denominator by nm to obtain:
О”Оё =
О”О»
1
1 в€’ n 2 m 2О»2
nm
=
О”О»
1 в€’ n 2 m 2О»2
n2m2
=
О”О»
1
в€’ О»2
2 2
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 0Вє. 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 П†m in terms of the groove separation d, the
wavelength О», 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 П†m in terms of the order number m, the
wavelength of the light О», and the groove separation d.
(a) Because Оёi = Оёr, application of
the grating equation yields:
d sin (2Оё i ) = mО»,
Because П† and Оёi have their left and
right sides mutually perpendicular:
Оё i = П†m
Substitute for Оёi to obtain:
d sin(2П† m ) = mО»
Solving for П†m yields:
(b) For m = 2:
where m = 0, 1, 2, ...
П†m =
1
2
(1)
вЋ› О»вЋћ
sin в€’1 вЋњ m вЋџ
вЋќ dвЋ вЋ›
вЋњ
450 nm
в€’1 вЋњ
1
П†2 = 2 sin 2
1
вЋњ
вЋњ 10,000 cm в€’1
вЋќ
вЋћ
вЋџ
вЋџ = 32.1В°
вЋџ
вЋџ
вЋ In this problem, you will derive the relation R = О» О”О» = mN
(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
74
•••
Interference and Diffraction 3103
separation between the intensity maximum and intensity minimum for some
wavelength О» and set it equal to the angular separation of the mth-order maximum
for two nearby wavelengths. (a) First show that the phase difference П† between
2ПЂ d
the waves from two adjacent slits is given by П† =
sin Оё . (b) Next differentiate
О»
this expression to show that a small change in angle dОё results in a change in
2ПЂ d
phase of dφ given by dφ =
cos Оё dОё . (c) Then for N slits, the angular
О»
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 dОё between the intensity maximum and intensity minimum for some
wavelength О» is given by dОё =
О»
. (d) Next use the fact that the angle of
Nd cos Оё
the mth-order interference maximum for wavelength О» is specified by
d sin Оё = mО» (Equation 33-26). Compute the differential of each side of this
equation to show that angular separation of the mth-order maximum for two
m dО»
. (e) According
nearly equal wavelengths differing by dО» is given by dОё =
d cos Оё
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 R = О» О”О» = mN (Equation 3327) for the resolving power of a grating.
Picture the Problem We can follow the procedure outlined in the problem
statement to obtain R = О» О”О» = mN .
(a) Express the relationship between
the phase difference П† and the path
difference О”r:
Because О”r = dsinОё :
(b) Differentiate this expression with
respect to Оё to obtain:
Solve for dφ:
(c) From Part (b):
2ПЂО”r
П†
О”r
в‡’П†=
=
2ПЂ
О»
О»
П†=
2ПЂd
О»
sin Оё
dφ
d вЋЎ 2ПЂd
вЋ¤ 2ПЂd
=
sin Оё вЋҐ =
cos Оё
вЋў
dОё d Оё вЋЈ О»
О»
вЋ¦
dφ =
dОё =
2ПЂd
О»
cos Оё dОё
λ dφ
2ПЂd cos Оё
3104
Chapter 33
Substitute 2π/N for dφ to obtain:
(d) Equation 33-26 is:
Differentiate this expression
implicitly with respect to О» to obtain:
Solve for dОё to obtain:
(e) Equate the two expressions for
dОё obtained in (c) and (d):
Approximating dО» by О”О» and
allowing for the possibility that
О”О» < 0 yields:
Solving for R = О»/О”О» yields:
dОё =
О»
Nd cos Оё
d sin Оё = mО», m = 0, 1, 2, ...
d
[d sin Оё ] = d [mО» ]
dО»
dО»
or
dОё
d cos Оё
=m
dО»
dОё =
mdО»
d cos Оё
О»
Nd cos Оё
О»
Nd cosОё
R=
=
=
mdО»
d cos Оё
m О”О»
d cosОё
О»
= mN
О”О»
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 10Вє. 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 sin Оё = 1.22О» D to relate the diameter D of the
opaque-disk water droplets to the angular diameter Оё of a coronal ring and to the
wavelength of light. We’ll assume a wavelength of 500 nm.
Interference and Diffraction 3105
The angle Оё subtended by the first
diffraction minimum is related to the
wavelength О» of light and the
diameter D of the opaque-disk water
droplet:
О»
sin Оё = 1.22
Because of the great distance to the
cloud of water droplets, Оё << 1 and:
θ ≈ 1.22
Substitute numerical values and
evaluate D:
D=
О»
D
D
в‡’D =
1.22О»
Оё
1.22(500 nm )
= 3.5 Ојm
ПЂ rad
10В° Г—
180В°
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-Ојm-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 sin Оё = 1.22О»n D to relate the diameter D of a
microsphere to the angular diameter Оё of a coronal ring and to the wavelength of
light in water.
The angle Оё subtended by the first
diffraction minimum is related to the
wavelength О»n of light in water and
the diameter D of the microspheres:
Because Оё << 1:
Substitute numerical values and
evaluate Оё :
sin Оё = 1.22
θ ≈ 1.22
θ≈
О»n
D
= 1.22
О»
nD
О»
nD
(1.22)(632.8 nm ) = 0.116 rad
(1.33)(5.00 Ојm )
= 6.65В°
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 Ојm. 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?
3106
Chapter 33
Picture the Problem We can use sin Оё = 1.22О» D to relate the diameter D of a
pollen grain to the angular diameter Оё 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.
The angleО± subtended by the first
diffraction intensity minima is
related to the wavelength О» of light
and to the diameter D of the
microspheres:
sin О± = 1.22
О»
D
or, because О± = 12 Оё where Оё is the
angular diameter of the coronal ring,
О»
sin 12 Оё = 1.22
D
Because θ << 1, sinθ ≈ θ and:
1
2
Substitute numerical values and
evaluate Оё for red light:
θ ≈ 1.22
θ red ≈
О»
D
⇒ θ ≈ 2.44
О»
D
2.44(650 nm )
= 6.344 Г— 10 в€’ 2 rad
25 Ојm
= 3.6В°
Substitute numerical values and
evaluate Оё for blue light:
θ blue ≈
2.44(450 nm )
= 4.392 Г— 10 в€’ 2 rad
25 Ојm
= 2.5В°
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 О”y of the first diffraction
peak from the center. We can use the geometry of the experiment to relate О”y to L
and a and the condition for diffraction maxima to express Оё1 in terms of the
diameter of the hair and the wavelength of the light illuminating the hair.
a
Оё1
О”y
L
Interference and Diffraction 3107
Relate Оё to О”y:
Diffraction maxima occur where:
tan Оё =
1
2
О”y
вЋ› О”y вЋћ
в‡’ Оё = tan в€’1 вЋњ вЋџ
L
вЋќ 2L вЋ a sin Оё = (m + 12 )О» в‡’ a =
(m + 12 )О»
sin Оё
where m = 1, 2, 3, …
Substituting for Оё yields:
Substitute numerical values and
evaluate a for m = 1:
a=
a=
(m + 12 )О»
вЋЎ
вЋ› О”y вЋћвЋ¤
sin вЋў tan в€’1 вЋњ вЋџвЋҐ
вЋќ 2 L вЋ вЋ¦
вЋЈ
(1 + 12 )(632.8 nm )
вЋЎ
вЋ› 14.6 cm вЋћвЋ¤
вЋџвЋџвЋҐ
sin вЋў tan в€’1 вЋњвЋњ
вЋќ 2(7.5 m ) вЋ вЋ¦
вЋЈ
= 98 Ојm
79 •
[SSM] A long, narrow horizontal slit lies 1.00 Ојm 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:
Express the condition for
constructive interference:
y0 = LОё 0
(1)
d sin Оё = (m + 12 )О»
where m = 0, 1, 2, …
Solving for Оё yields:
О»вЋ¤
вЋЎ
Оё = sin в€’1 вЋў(m + 12 ) вЋҐ
dвЋ¦
вЋЈ
For the first maximum, m = 0 and:
Оё 0 = sin в€’1 вЋў
Substitute in equation (1) to obtain:
вЋЎО» вЋ¤
y 0 = L sin в€’1 вЋў вЋҐ
вЋЈ 2d вЋ¦
вЋЎО» вЋ¤
вЋЈ 2d вЋҐвЋ¦
3108
Chapter 33
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 Ојm.
Substitute numerical values and
evaluate y0:
вЋЎ 600 nm вЋ¤
y 0 = (1.00 m )sin в€’1 вЋў
вЋҐ
вЋЈ 2(2.00 Ојm ) вЋ¦
(b) The separation of the fringes on
the screen is given by:
О”y =
The number of dark bands per unit
length is the reciprocal of the fringe
separation:
n=
1
d
=
О”y О» L
Substitute numerical values and
evaluate n:
n=
2.00 Ојm
= 3.33 m в€’1
(600 nm )(1.00 m )
= 15.1cm
О»L
d
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 180Вє 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 180В°, 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.
Radio waves directly from galaxy
Оё
m
l
ed
ect
f
s re
ave
w
О”r
dio
a
R
Оё
fro
Telescope
P
Оё
e
lak
e
h
t
d
Interference and Diffraction 3109
Because the reflected radio waves are
phase shifted by 180В°, the condition
for constructive interference at point
P is:
Referring to the figure, note that:
Substitute for О”r to obtain:
Noting that m = 0 for the first
interference maximum, substitute
numerical values and evaluate Оё0:
О”r = (m + 12 )О»
where m = 0, 1, 2, …
sin θ ≈
О”r
вЋЎ О”r вЋ¤
в‡’ Оё = sin в€’1 вЋў вЋҐ
d
вЋЈd вЋ¦
вЋЎ (m + 12 )О» вЋ¤
вЋҐ
вЋЈ d
вЋ¦
Оё = sin в€’1 вЋў
вЋЎ 1 (20 cm ) вЋ¤
= 5.00 Г— 10 в€’3 rad
вЋҐ
вЋЈ 20 m вЋ¦
Оё 0 = sin в€’1 вЋў 2
= 0.29В°
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:
Substitute numerical values and
evaluate О±c:
О± c = 1.22
О»
D
вЋ› 3.2 cm вЋћ
вЋџвЋџ = 0.13 mrad
вЋќ 300 m вЋ О± c = 1.22вЋњвЋњ
82 ••
A thin layer of a transparent material that has an index of refraction of
1.30 is used as a nonreflective coating 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 ПЂ rad. We can apply the
condition for destructive interference to
find the thickness t of the nonreflective
coating.
О» = 600 nm
ПЂ
Air
t
ПЂ
Coating
Glass
3110
Chapter 33
The condition for destructive
interference is:
2t = (m + 12 )О»coating = (m +
Solve for t to obtain:
t = (m +
Evaluate t for m = 0:
t = ( 12 )
1
2
)
1
2
)
О»air
ncoating
О»air
2ncoating
600 nm
= 115 nm
2(1.30)
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 Оё, the transmitted light will
have maximum intensity when 2a = mО»/cos Оё.
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.
Express the path difference between
the two rays that emerge from the
interferometer:
For constructive interference we
require that:
a
Оё
О”r =
2a
cos Оё
О”r = mО», m = 0, 1, 2, ...
2a
cos Оё
Equate these expressions to
obtain:
mО» =
Solve for 2a to obtain:
2a = mО» cosОё
84 ••
A mica sheet 1.20 Ојm 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 ПЂ
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.
Because there is a ПЂ rad phase shift
at the first air-mica interface, the
condition for destructive interference
is:
Solving for n yields:
Mica
Air
Air
t
ПЂ
n
2t = mО»mica = m
n=m
О»air
n
, m = 1, 2, 3, ...
О»air
2t
For О» = 474 nm:
2t = (474 nm) m
For О» = 421 nm:
2t = (421nm)(m + 1)
Equate these two expressions and
solve for m to obtain:
m = 8 for О» = 474 nm
Substitute numerical values in
equation (1) and evaluate n:
вЋ› 474 nm вЋћ
вЋџвЋџ = 1.58
n = 8вЋњвЋњ
вЋќ 2(1.20 Ојm ) вЋ (1)
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.
3112
Chapter 33
Picture the Problem Note that the
light reflected at both the air-film and
film-lens interfaces undergoes a ПЂ 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 I в€ќ cos 2 12 Оґ .
(a) Express the condition for
destructive interference between the
light reflected from the air-film
interface and the film-lens interface:
Solving for t gives:
Evaluate t for m = 0:
(b) Solve equation (1) for О»air to
obtain:
Evaluate О»air for m = 1:
Air
Film
Lens
t
ПЂ
ПЂ
n
2t = (m +
1
2
)О»film = (m + 12 ) О»air
(1)
n
where m = 0, 1, 2, …
t = (m +
1
2
) О»air
2n
вЋ› 1 вЋћ 540 nm
t =вЋњ вЋџ
= 97.83 nm = 97.8 nm
вЋќ 2 вЋ 2(1.38)
О»air =
2tn
m + 12
О»air =
2(97.8 nm )(1.38)
= 180 nm
1 + 12
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 Оґ
between the two reflected waves:
f = cos 2 12 Оґ
Relate the phase difference to the
path difference О”r:
вЋ› О”r вЋћ
Оґ
О”r
вЋџвЋџ
=
в‡’ Оґ = 2ПЂ вЋњвЋњ
2ПЂ О»film
вЋќ О»film вЋ Because О”r = 2t:
вЋ› 2t вЋћ
вЋџвЋџ
вЋќ О»film вЋ Оґ = 2ПЂ вЋњвЋњ
(2)
Interference and Diffraction 3113
Substitute in equation (2) to obtain:
вЋЎ
вЋ› 2t вЋћвЋ¤
вЋЎ 2ПЂ t вЋ¤
вЋџвЋџвЋҐ = cos 2 вЋў
f = cos 2 вЋў 12 2ПЂ вЋњвЋњ
вЋҐ
вЋќ О»film вЋ вЋ¦
вЋЈ О»film вЋ¦
вЋЈ
вЋЎ 2ПЂ nt вЋ¤
= cos 2 вЋў
вЋҐ
вЋЈ О»air вЋ¦
Evaluate f for О» = 400 nm:
вЋЎ 2ПЂ (1.38)(97.83 nm ) вЋ¤
f 400 = cos 2 вЋў
вЋҐ
400 nm
вЋЈ
вЋ¦
= 0.273
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.
Film
Object
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.
D
2Оё
L
D
D
в‡’Оё=
L
2L
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:
2Оё =
Using Rayleigh’s criterion, express
the angular width of the diffraction
pattern:
Оё diffraction = 1.22
Equate these two expressions to
obtain:
О»
D
= 1.22 в‡’ D = 2.44О»L
2L
D
О»
D
3114
Chapter 33
D = 2.44(550 nm )(10.0 cm )
Substitute numerical values and
evaluate D:
= 0.366 mm
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
d
Оё
О±c
L
d
L
Referring to the diagram, express
the angle subtended by the adjacent
dots:
θ≈
Letting the diameter of the pupil of
the eye be D, apply Rayleigh’s
criterion to obtain:
О± c = 1.22
Set Оё = О±c to obtain:
О»
d
Dd
= 1.22 в‡’ L =
L
D
1.22О»
Evaluate L for the shortest
wavelength light in the visible
portion of the spectrum:
L=
О»
D
(3.0 mm )(2.0 mm) =
(1.22)(400 nm )
12 m
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