Wave Optics
CHAPTER OUTLINE
27.1
27.2
27.3
27.4
27.5
27.6
27.7
27.8
27.9
27.10
Conditions for
Interference
Young’s Double-Slit
Experiment
Light Waves in
Interference
Change of Phase Due to
Reflection
Interference in Thin Films
Diffraction Patterns
Resolution of Single-Slit
and Circular Apertures
The Diffraction Grating
Diffraction of X-Rays by
Crystals
Context
ConnectionHolography
ANSWERS TO QUESTIONS
Q27.1
(a)
Two waves interfere constructively if their path difference is
zero, or an integral multiple of the wavelength, according to
δ = mλ , with m = 0 , 1, 2 , 3 , K.
(b)
Two waves interfere destructively if their path difference is a
half wavelength, or an odd multiple of
FG
H
δ = m+
IJ
K
1
λ , with m = 0 , 1, 2 , 3 , K.
2
λ
2
, described by
Q27.2
The light from the flashlights consists of many different wavelengths (that’s why it’s white) with
random time differences between the light waves. There is no coherence between the two sources.
The light from the two flashlights does not maintain a constant phase relationship over time. These
three equivalent statements mean no possibility of an interference pattern.
Q27.3
Every color produces its own pattern, with a spacing between the maxima that is characteristic of the
wavelength. With several colors, the patterns are superimposed and it can be difficult to pick out a
single maximum. Using monochromatic light can eliminate this problem.
Q27.4
Underwater, the wavelength of the light would decrease, λ water =
Q27.5
The threads that are woven together to make the cloth have small meshes between them. These bits
of space act as pinholes through which the light diffracts. Since the cloth is a grid of such pinholes,
an interference pattern is formed, as when you look through a diffraction grating.
Q27.6
If the oil film is brightest where it is thinnest, then n air < n oil < n water . With this condition, light
reflecting from both the top and the bottom surface of the oil film will undergo phase reversal. Then
these two beams will be in phase with each other where the film is very thin. This is the condition
for constructive interference as the thickness of the oil film decreases toward zero.
λ
air
. Since the positions of light
n water
and dark bands are proportional to λ, (according to Equations 27.2 and 27.3), the underwater fringe
separations will decrease.
745
746
Wave Optics
Q27.7
As water evaporates from the ‘soap’ bubble, the thickness of the bubble wall approaches zero. Since
light reflecting from the front of the water surface is phase-shifted 180° and light reflecting from the
back of the soap film is phase-shifted 0°, the reflected light meets the conditions for a minimum.
Thus the soap film appears black, as in the illustration accompanying textbook Example 27.4,
“Interference in a Wedge-Shaped Film.”
Q27.8
If the film is more than a few wavelengths thick, the interference fringes are so close together that
you cannot resolve them.
Q27.9
Suppose the coating is intermediate in index of refraction between vacuum and the glass. When the
coating is very thin, light reflected from its top and bottom surfaces will interfere constructively, so
you see the surface white and brighter. As the thickness reaches one quarter of the wavelength of
violet light in the coating, destructive interference for violet will make the surface look red or
perhaps orange. Next to interfere destructively are blue, green, yellow, orange, and red, making the
surface look red, purple, and then blue. As the coating gets still thicker, we can get constructive
interference for violet and then for other colors in spectral order. Still thicker coating will give
constructive and destructive interference for several visible wavelengths, so the reflected light will
start to look white again.
Q27.10
The wavelength of light is extremely small in comparison to the dimensions of your hand, so the
diffraction of light around an obstacle the size of your hand is totally negligible. However, sound
waves have wavelengths that are comparable to the dimensions of the hand or even larger.
Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.
Q27.11
Audible sound has wavelengths on the order of meters or centimeters, while visible light has a
wavelength on the order of half a micrometer. In this world of breadbox-sized objects,
λ
is large for
a
sound. When sound goes through a doorway, the sound diffracts around behind the adjoining
walls. But
λ
is a tiny fraction for visible light passing ordinary-size objects or apertures, so light
a
changes its direction by only very small angles when it diffracts.
Another way of phrasing the answer: We can see by a small angle around a small obstacle or
around the edge of a small opening. The side fringes in Figure 27.12 and the Arago spot in the center
of Figure 27.13 show this diffraction. We cannot always hear around corners. Out-of-doors, away
from reflecting surfaces, have someone a few meters distant face away from you and whisper. The
high-frequency, short-wavelength, information-carrying components of the sound do not diffract
around his head enough for you to understand his words.
Q27.12
Yes, but no diffraction effects are observed because the separation distance between adjacent ribs is
so much greater than the wavelength of x-rays. Diffraction does not limit the resolution of an x-ray
image. Diffraction might sometimes limit the resolution of an ultrasonogram.
Q27.13
The intensity of the light coming through the slit decreases, as you would expect. The central
maximum increases in width as the width of the slit decreases. In the condition sin θ =
λ
a
destructive interference on each side of the central maximum, θ increases as a decreases.
for
Chapter 27
Q27.14
747
It is shown in the correct orientation. If the horizontal width of the opening is equal to or less than
the wavelength of the sound, then the equation a sin θ = 1 λ has the solution θ = 90° , or has no
solution. The central diffraction maximum covers the whole seaward side. If the vertical height of
the opening is large compared to the wavelength, then the angle in a sin θ = 1 λ will be small, and
the central diffraction maximum will form a thin horizontal sheet.
af
af
Q27.15
The fine hair blocks off light that would otherwise go through a fine slit and produce a diffraction
pattern on a distant screen. The width of the central maximum in the pattern is inversely
proportional to the distance across the slit. When the hair is in place, it subtracts the same diffraction
pattern from the projected disk of laser light. The hair produces a diffraction minimum that crosses
the bright circle on the screen. The width of the minimum is inversely proportional to the diameter
of the hair. The central minimum is flanked by narrower maxima and minima. Measure the width 2y
of the central minimum between the maxima bracketing it, and use Equation 27.13 in the form
y λ
= to find the width a of the hair.
L a
Q27.16
The condition for constructive interference is that the three radio signals arrive at the city in phase.
We know the speed of the waves (it is the speed of light c), the angular bearing θ of the city east of
north from the broadcast site, and the distance d between adjacent towers. The wave from the
westernmost tower must travel an extra distance 2d sin θ to reach the city, compared to the signal
from the eastern tower. For each cycle of the carrier wave, the western antenna would transmit first,
d sin θ
, and the eastern antenna after an additional equal time
the center antenna after a time delay
c
delay.
SOLUTIONS TO PROBLEMS
Section 27.1 Conditions for Interference
Section 27.2 Young’s Double-Slit Experiment
Section 27.3 Light Waves in Interference
P27.1
y bright =
λL
d
m
For m = 1 ,
P27.2
λ=
e
je
j
3.40 × 10 −3 m 5.00 × 10 −4 m
yd
λ=
=
= 515 nm
L
3.30 m
v 354 m s
=
= 0.177 m
f 2 000 s −1
(a)
d sin θ = mλ
so
a0.300 mf sinθ = 1a0.177 mf
and
θ = 36.2°
(b)
d sin θ = mλ
so
d sin 36.2° = 1 0.030 0 m
b
and
d = 5.08 cm
continued on next page
g
748
Wave Optics
e1.00 × 10
(c)
f=
P27.3
c
λ
=
−6
j
af
m sin 36.2° = 1 λ
3.00 × 10 8 m s
5.90 × 10 −7 m
λ = 590 nm
so
= 508 THz
Note, with the conditions given, the small angle approximation does not
work well. That is, sin θ , tan θ , and θ are significantly different. We treat
the interference as a Fraunhofer pattern.
At the m = 2 maximum, tan θ =
(a)
400 m
= 0.400
1 000 m
400 m
θ = 21.8°
λ=
so
(b)
300 m
a
f
300 m sin 21.8°
d sin θ
=
= 55.7 m .
m
2
The next minimum encountered is the m = 2 minimum;
FG
H
IJ
K
1
λ
2
and at that point,
d sin θ = m +
which becomes
d sin θ =
or
sin θ =
and
θ = 27.7°
so
y = 1 000 m tan 27.7° = 524 m .
5
λ
2
FG
H
IJ
K
5 λ 5 55.7 m
=
= 0.464
2 d 2 300 m
b
g
Therefore, the car must travel an additional 124 m .
If we considered Fresnel interference, we would more precisely find
1
550 2 + 1 000 2 − 250 2 + 1 000 2 = 55.2 m and (b) 123 m.
(a) λ =
2
FH
P27.4
λ=
IK
340 m s
= 0.170 m
2 000 Hz
Maxima are at
d sin θ = mλ :
m=0
gives
θ = 0°
m=1
gives
sin θ =
m=2
gives
sin θ =
m=3
gives
sin θ = 1. 46
continued on next page
λ
d
=
0.170 m
0.350 m
2λ
= 0.971
d
θ = 29.1°
θ = 76.3°
No solution.
1 000 m
FIG. P27.3
Chapter 27
FG
H
749
IJ
K
1
λ:
2
Minima are at
d sin θ = m +
m=0
gives
sin θ =
m=1
gives
sin θ =
m=2
gives
sin θ = 1.21
λ
= 0.243
θ = 14.1°
3λ
= 0.729
2d
θ = 46.8°
2d
No solution.
So we have maxima at 0° , 29.1° , and 76.3° ; minima at 14.1° and 46.8° .
P27.5
d sin θ = m +
The first minimum is described by
m=0
and the tenth by m = 9 :
sin θ =
FG 9 + 1 IJ .
dH
2K
sin θ ≈ tan θ .
Thus,
d=
ja
7.26 × 10
−3
f = 1.54 × 10
m
Source
y
L
but for small θ,
e
y
λ
tan θ =
9.5 5 890 × 10 −10 m 2.00 m
IJ
K
1
λ.
2
L
Also,
d=
*P27.6
FG
H
In the equation
FIG. P27.5
9.5 λ 9.5 λL
=
sin θ
y
−3
m = 1.54 mm .
Location of A = central maximum,
Location of B = first minimum.
P27.7
FG 0 + 1 IJ − 0 = 1 λL = 20.0 m .
2 d
d H
2K
a3.00 mfa150 mf = 11.3 m .
=
So,
∆y = y min − y max =
Thus,
d=
f
2 20.0 m
40.0 m
d sin θ = mλ
At 30.0° ,
e3.20 × 10
a
λL
λL
−4
j
e
j
m sin 30.0° = m 500 × 10 −9 m
so
m = 320
There are 320 maxima to the right, 320 to the left, and one for m = 0 straight ahead.
There are 641 maxima .
d
750
P27.8
Wave Optics
Observe that the pilot must not only home in on the airport, but must be headed in the right
direction when she arrives at the end of the runway.
8
c 3 × 10 m s
=
= 10.0 m
f 30 × 10 6 s −1
(a)
λ=
(b)
The first side maximum is at an angle given by d sin θ = 1 λ .
af
θ = 14.5°
a40 mf sinθ = 10 m
y = L tan θ = b 2 000 mg tan 14.5° = 516 m
(c)
P27.9
tan θ =
y
L
The signal of 10-m wavelength in parts (a) and (b) would show maxima at 0°, 14.5°, 30.0°,
48.6°, and 90°. A signal of wavelength 11.23-m would show maxima at 0°, 16.3°, 34.2°, and
57.3°. The only value in common is 0°. If λ 1 and λ 2 were related by a ratio of small integers
λ
n
(a just musical consonance!) in 1 = 1 , then the equations d sin θ = n 2 λ 1 and d sin θ = n1 λ 2
λ 2 n2
would both be satisfied for the same nonzero angle. The pilot could come flying in with that
inappropriate bearing, and run off the runway immediately after touchdown.
φ=
2π
λ
d sin θ =
2π
d
λ
FG y IJ
H LK
2π
−7
e1.20 × 10
mj
−7
e1.20 × 10
mj
(a)
φ=
(b)
φ=
(c)
If φ = 0.333 rad =
e5.00 × 10
2π
e5.00 × 10
2πd sin θ
λ
j a
f
−4
m sin 0.500° = 13.2 rad
−4
m
× 10 m I
=
jFGH 5.001.20
m JK
−3
θ = sin −1
F λ φ I = sin
GH 2π d JK
6.28 rad
LM e5.00 × 10 mja0.333 radf OP
MN 2π e1.20 × 10 mj PQ
−7
−1
−4
θ = 1.27 × 10 −2 deg .
(d)
If d sin θ =
λ
4
θ = sin −1
FG λ IJ = sin
H 4d K
−1
LM 5 × 10 m OP
MN 4e1.20 × 10 mj PQ
−7
−4
θ = 5.97 × 10 −2 deg .
*P27.10
The path difference between rays 1 and 2 is:
δ = d sin θ 1 − d sin θ 2 .
For constructive interference, this path difference must be equal to an integral number of
wavelengths: d sin θ 1 − d sin θ 2 = mλ , or
b
g
d sin θ 1 − sin θ 2 = mλ .
Chapter 27
P27.11
I av = I max cos 2
FG π d sinθ IJ
H λ K
For small θ,
sin θ ≈
and
I av = 0.750 I max
y=
y
L
λL
I av
cos −1
πd
I max
e6.00 × 10 ja1.20 mf cos
π e 2.50 × 10 mj
−7
y=
P27.12
(a)
−3
I
φ = 2 cos −1
P27.13
−1
0.750 I max
= 48.0 µ m
I max
The intensity of the sum of two coherent waves is described by
I max
I
I max
= cos 2
FG φ IJ . Therefore,
H 2K
= 2 cos −1 0.640 = 1.29 rad .
a
fa
f
486.1 nm 1.29 rad
λφ
=
= 99.6 nm
2π
2π
δ=
(b)
751
F π yd I
GH λ L JK
LM π e6.00 × 10 mje1.80 × 10 mj OP
MN e656.3 × 10 mja0.800 mf PQ =
I = I max cos 2
I
I max
−3
= cos
2
−4
−9
0.968
Section 27.4 Change of Phase Due to Reflection
Section 27.5 Interference in Thin Films
P27.14
Light reflecting from the first surface suffers phase reversal. Light reflecting from the second surface
does not, but passes twice through the thickness t of the film. So, for constructive interference, we
require
λn
2
+ 2t = λ n
where
λn =
Then
2t =
λ
n
λn
2
is the wavelength in the material.
=
λ
2n
a fa
f
λ = 4 nt = 4 1.33 115 nm = 612 nm .
752
P27.15
Wave Optics
(a)
The light reflected from the top of the oil film undergoes phase
reversal. Since 1.45 > 1.33 , the light reflected from the bottom
undergoes no reversal. For constructive interference of reflected
light, we then have
FG
H
IJ
K
1
λ
2
2nt = m +
λm =
or
a fa
f
FIG. P27.15
2 1.45 280 nm
2nt
=
.
m+ 1 2
m+ 1 2
b g
Substituting for m gives:
b g
m=0 ,
λ 0 = 1 620 nm (infrared)
m=1,
λ 1 = 541 nm (green)
m=2,
λ 2 = 325 nm (ultraviolet).
Both infrared and ultraviolet light are invisible to the human eye, so the dominant color in
reflected light is green .
(b)
The dominant wavelengths in the transmitted light are those that produce destructive
interference in the reflected light. The condition for destructive interference upon reflection
is
2nt = mλ
λm =
or
2nt 812 nm
=
.
m
m
Substituting for m gives:
m=1,
λ 1 = 812 nm (near infrared)
m=2,
λ 2 = 406 nm (violet)
m=3 ,
λ 3 = 271 nm (ultraviolet).
Of these, the only wavelength visible to the human eye (and hence the dominate
wavelength observed in the transmitted light) is 406 nm. Thus, the dominant color in the
transmitted light is violet .
P27.16
Treating the anti-reflectance coating like a camera-lens coating,
FG
H
2t = m +
Let m = 0 :
t=
λ
4n
=
IJ
K
1 λ
.
2 n
3.00 cm
= 0.500 cm .
4 1.50
a f
This anti-reflectance coating could be easily countered by changing the wavelength of the radar—to
1.50 cm—now creating maximum reflection!
P27.17
FG
H
2nt = m +
Minimum
IJ
K
1
λ
2
so
FG 1 IJ λ
H 2 K 2n
F 1 I a500 nmf =
t=G J
H 2 K 2a1.30f
t= m+
96.2 nm .
Chapter 27
*P27.18
(a)
For maximum transmission, we want destructive interference in the light reflected from the
front and back surfaces of the film.
If the surrounding glass has refractive index greater than 1.378, light reflected from
the front surface suffers no phase reversal and light reflected from the back does undergo
phase reversal. This effect by itself would produce destructive interference, so we want the
distance down and back to be one whole wavelength in the film: 2t =
t=
λ
=
2n
λ
n
.
656.3 nm
= 238 nm
2 1.378
a
f
(b)
The filter will expand. As t increases in 2nt = λ , so does λ increase .
(c)
Destructive interference for reflected light happens also for λ in 2nt = 2 λ ,
a
f
λ = 1.378 238 nm = 328 nm
or
P27.19
anear ultravioletf .
For destructive interference in the air,
2t = mλ .
For 30 dark fringes, including the one where the plates
meet,
t=
a
f
29 600 nm
= 8.70 × 10 −6 m .
2
Therefore, the radius of the wire is
r=
t 8.70 µ m
=
= 4.35 µ m .
2
2
FIG. P27.19
Section 27.6 Diffraction Patterns
P27.20
753
sin θ =
λ
a
=
6.328 × 10 −7
= 2.11 × 10 −3
3.00 × 10 −4
a
y
= tan θ ≈ sin θ = θ for small θ
1.00 m
2 y = 4.22 mm
f
754
P27.21
Wave Optics
y
mλ
≈ sin θ =
L
a
∆y = 3.00 × 10 −3 nm
∆m = 3 − 1 = 2
and
a=
a=
P27.22
∆ mλ L
∆y
e
ja
f=
2 690 × 10 −9 m 0.500 m
e
The positions of the first-order minima are
3.00 × 10
−3
j
m
2.30 × 10 −4 m
y
λ
≈ sin θ = ± . Thus, the spacing between these two
L
a
FG λ IJ L and the wavelength is
H aK
F ∆y I F a I F 4.10 × 10 mIJ FG 0.550 × 10 mIJ =
λ = G JG J = G
H 2 K H L K H 2 K H 2.06 m K
minima is ∆y = 2
−3
P27.23
−3
547 nm .
For destructive interference,
sin θ = m
and
λ
a
=
λ
a
5.00 cm
= 0.139
36.0 cm
=
θ = 7.98°
d
= tan θ
L
gives
a
f
d = L tan θ = 6.50 m tan 7.98° = 0.912 m
d = 91.2 cm .
P27.24
v 340 m s
=
= 0.523 m .
f
650 s −1
If the speed of sound is 340 m/s,
λ=
Diffraction minima occur at angles described by
a sin θ = mλ
a1.10 mf sinθ
a1.10 mf sinθ
a1.10 mf sinθ
1
2
3
a
f
= 2a0.523 mf
= 3a0.523 mf
θ 1 = 28.4°
= 1 0.523 m
θ 2 = 72.0°
θ 3 nonexistent
Maxima appear straight ahead at 0° and left and right at an angle given approximately by
a1.10 mf sinθ
x
a
f
= 1.5 0.523 m
θ x ≈ 46° .
There is no solution to a sin θ = 2.5 λ , so our answer is already complete, with three sound
maxima.
Chapter 27
*P27.25
755
The rectangular patch on the wall is wider than it is tall. The aperture will be taller than it is wide.
For horizontal spreading we have
y width 0.110 m 2
=
= 0.012 2
L
4.5 m
a width sin θ width = 1λ
tan θ width =
a width =
632.8 × 10 −9 m
= 5.18 × 10 −5 m
0.012 2
For vertical spreading, similarly
0.006 m 2
= 0.000 667
4.5 m
1λ
632.8 × 10 −9 m
=
=
= 9.49 × 10 −4 m
0.000 667
sin θ h
tan θ height =
a height
Section 27.7 Resolution of Single-Slit and Circular Apertures
λ
5.00 × 10 −7 m
= 1.00 × 10 −3 rad
5.00 × 10 −4
P27.26
sin θ =
P27.27
Undergoing diffraction from a circular opening, the beam spreads into a cone of half-angle
a
=
θ min = 1.22
λ
D
= 1.22
F 632.8 × 10 mI = 1.54 × 10
GH 0.005 00 m JK
−9
−4
rad .
The radius of the beam ten kilometers away is, from the definition of radian measure,
e
j
rbeam = θ min 1.00 × 10 4 m = 1.544 m
and its diameter is d beam = 2rbeam = 3.09 m .
*P27.28
λ
d
= 1.22 , where θ min is the smallest angular separation of two
L
D
objects for which they are resolved by an aperture of diameter D, d is the separation of the two
objects, and L is the maximum distance of the aperture from the two objects at which they can be
resolved.
Two objects can be resolved if their angular separation is greater than θ min . Thus, θ min should
be as small as possible. Therefore, the smaller of the two given wavelengths is easier to resolve, i.e.
violet .
By Rayleigh’s criterion: θ min =
L=
e
je
j
5.20 × 10 −3 m 2.80 × 10 −2 m 1.193 × 10 −4 m 2
Dd
=
=
1.22 λ
1.22 λ
λ
Thus L = 186 m for λ = 640 nm, and L = 271 m for λ = 440 nm. The viewer can resolve adjacent tubes
of violet in the range 186 m to 271 m , but cannot resolve adjacent tubes of red in this range.
756
P27.29
Wave Optics
By Rayleigh’s criterion, two dots separated center-to-center by 2.00 mm would overlap
θ min =
when
λ
d
= 1.22 .
L
D
e
je
j
2.00 × 10 −3 m 4.00 × 10 −3 m
dD
Thus, L =
=
= 13.1 m .
1.22λ
1.22 500 × 10 −9 m
P27.30
e
j
The concave mirror of the spy satellite is probably about 2 m in diameter, and is surely not more
than 5 m in diameter. That is the size of the largest piece of glass successfully cast to a precise shape,
for the mirror of the Hale telescope on Mount Palomar. If the spy satellite had a larger mirror, its
manufacture could not be kept secret, and it would be visible from the ground. Outer space is
probably closer than your state capitol, but the satellite is surely above 200-km altitude, for
reasonably low air friction. We find the distance between barely resolvable objects at a distance of
200 km, seen in yellow light through a 5-m aperture:
y
λ
= 1.22
L
D
b
ga fFGH 6 × 105 m mIJK = 3 cm
−7
y = 200 000 m 1.22
(Considering atmospheric seeing caused by variations in air density and temperature, the distance
b
ga
between barely resolvable objects is more like 200 000 m 1 s
1° I F π rad I
fFGH 3 600
G J = 97 cm .) Thus the
s JK H 180° K
snooping spy satellite cannot see the difference between III and II or IV on a license plate. It cannot
count coins spilled on a sidewalk, much less read the date on them.
P27.31
1.22
λ
D
=
d
L
λ=
D = 2.10 m
d = 1.22
c
= 0.020 0 m
f
L = 9 000 m
b0.020 0 mgb9 000 mg =
2.10 m
105 m
Section 27.8 The Diffraction Grating
P27.32
The principal maxima are defined by
d sin θ = mλ
m = 0 , 1, 2 , K .
For m = 1 ,
λ = d sin θ
where θ is the angle between the central ( m = 0 ) and the first order
( m = 1 ) maxima. The value of θ can be determined from the
information given about the distance between maxima and the
grating-to-screen distance. From the figure,
tan θ =
continued on next page
0.488 m
= 0.284
1.72 m
0.488 m
θ
1.72 m
FIG. P27.32
Chapter 27
θ = 15.8°
so
and
757
sin θ = 0.273 .
The distance between grating “slits” equals the reciprocal of the number of grating lines per
centimeter
d=
The wavelength is
P27.33
1
5 310 cm −1
= 1.88 × 10 −4 cm = 1.88 × 10 3 nm .
ja
e
f
λ = d sin θ = 1.88 × 10 3 nm 0.273 = 514 nm .
The grating spacing is
d=
1.00 × 10 −2 m
= 2.22 × 10 −6 m .
4 500
Blue
In the 1st-order spectrum, diffraction angles are given by
sin θ =
λ
d
sin θ 1 =
:
656 × 10 −9 m
2.22 × 10 −6 m
so that for red
θ 1 = 17.17°
and for blue
sin θ 2 =
so that
θ 2 = 11.26° .
434 × 10 −9 m
2.22 × 10 −6 m
= 0.295
FIG. P27.33
= 0.195
The angular separation is in first-order,
∆θ = 17.17°−11.26° = 5.91° .
In the second-order spectrum,
∆θ = sin −1
Again, in the third order,
∆θ = sin −1
FG 2λ
Hd
FG 3λ
Hd
1
1
IJ − sin
K
IJ − sin
K
−1
−1
FG 2λ
Hd
FG 3λ
Hd
2
2
IJ =
K
IJ =
K
13.2° .
26.5° .
Since the red does not appear in the fourth-order spectrum, the answer is complete.
P27.34
sin θ = 0.350 :
d=
λ
sin θ
=
632.8 nm
= 1.81 × 10 3 nm
0.350
Line spacing = 1.81 µ m
P27.35
d=
(a)
1.00 × 10 −3 m mm
= 4.00 × 10 −6 m = 4 000 nm
250 lines mm
d sin θ = mλ ⇒ m =
d sin θ
λ
The number of times a complete order is seen is the same as the number of orders in which
the long wavelength limit is visible.
m max =
or
continued on next page
d sin θ max
λ
=
b4 000 nmg sin 90.0° = 5.71
700 nm
5 orders is the maximum .
758
Wave Optics
(b)
The highest order in which the violet end of the spectrum can be seen is:
m max =
or
*P27.36
sin θ =
d sin θ max
λ
=
b4 000 nmg sin 90.0° = 10.0
400 nm
10 orders in the short-wavelength region .
mλ
d
Therefore, taking the ends of the visible spectrum to be λ v = 400 nm and λ r = 750 nm , the ends the
different order spectra are:
End of second order:
sin θ 2 r =
2 λ r 1 500 nm
=
.
d
d
Start of third order:
sin θ 3 v =
3 λ v 1 200 nm
=
.
d
d
Thus, it is seen that θ 2 r > θ 3 v and these orders must overlap regardless of the value of the grating
spacing d.
*P27.37
The sound has wavelength λ =
343 m s
v
=
= 9.22 × 10 −3 m . Each diffracted beam is described by
f 37.2 × 10 3 s
d sin θ = mλ , m = 0 , 1, 2 , K
The zero-order beam is at m = 0 , θ = 0 . The beams in the first order of interference are to the left and
1λ
9.22 × 10 −3 m
right at θ = sin −1
= sin −1
= sin −1 0.709 = 45.2° . For a second-order beam we would
d
1.3 × 10 −2 m
2λ
need θ = sin −1
= sin −1 2 0.709 = sin −1 1.42 . No angle, smaller or larger than 90° has a sine greater
d
than 1. Then a diffracted beam does not exist for the second order or any higher order. The whole
answer is then, three beams, at 0° and at 45.2° to the right and to the left .
a
f
Section 27.9 Diffraction of X-Rays by Crystals
P27.38
2d sin θ = mλ :
P27.39
2d sin θ = mλ :
and
e
j
−9
2d sin θ 2 0.353 × 10 m sin 7.60°
λ=
=
= 9.34 × 10 −11 m = 0.093 4 nm
m
1
sin θ =
e
e
j
j
−9
mλ 1 0.140 × 10 m
=
= 0.249
2d 2 0.281 × 10 −9 m
θ = 14.4°
Chapter 27
P27.40
759
Figure 27.28 of the text shows the situation.
2d sin θ = mλ
λ=
or
2d sin θ
m
a
f
a
f
a
f
m=1:
λ1 =
2 2.80 m sin 80.0°
= 5.51 m
1
m=2:
λ2 =
2 2.80 m sin 80.0°
= 2.76 m
2
m=3 :
λ3 =
2 2.80 m sin 80.0°
= 1.84 m
3
Section 27.10 Context Connection
Holography
P27.41
(a)
The several narrow parallel slits make a diffraction
grating. The zeroth- and first- order maxima are
separated according to
af
d sin θ = 1 λ
sin θ =
b
λ
d
=
632.8 × 10 −9 m
1.2 × 10 −3 m
g
θ = sin −1 0.000 527 = 0.000 527 rad
a
fb
g
y = L tan θ = 1. 40 m 0.000 527 = 0.738 mm .
(b)
FIG. P27.41
Many equally spaced transparent lines appear on the film. It is itself a diffraction grating.
When the same light is sent through the film, it produces interference maxima separated
according to
af
d sin θ = 1 λ
sin θ =
a
fb
λ
d
=
632.8 × 10 −9 m
0.738 × 10 −3 m
= 0.000 857
g
y = L tan θ = 1. 40 m 0.000 857 = 1.20 mm
An image of the original set of slits appears on the screen. If the screen is removed, light
diverges from the real images with the same wave fronts reconstructed as the original slits
produced. Reasoning from the mathematics of Fourier transforms, Gabor showed that light
diverging from any object, not just a set of slits, could be used. In the picture, the slits or
maxima on the left are separated by 1.20 mm. The slits or maxima on the right are separated
by 0.738 mm. The length difference between any pair of lines is an integer number of
wavelengths. Light can be sent through equally well toward the right or toward the left.
760
P27.42
Wave Optics
(a)
The light in the cavity is incident perpendicularly on the
mirrors, although the diagram shows a large angle of incidence
for clarity. We ignore the variation of the index of refraction
with wavelength. To minimize reflection at a vacuum
wavelength of 632.8 nm, the net phase difference between rays
(1) and (2) should be 180°. There is automatically a 180° shift in
one of the two rays upon reflection, so the extra distance
traveled by ray (2) should be one whole wavelength:
2t =
t=
(b)
FIG. P27.42
λ
n
λ
2n
=
632.8 nm
= 217 nm
2 1.458
a
f
The total phase difference should be 360°, including contributions of 180° by reflection and
180° by extra distance traveled:
2t =
t=
λ
2n
λ
4n
=
543 nm
= 93.1 nm
4 1.458
a
f
Additional Problems
*P27.43
Along the line of length d joining the source, two identical waves
moving in opposite directions add to give a standing wave. An
d λ
antinode is halfway between the sources. If > , there is space for
2 2
d
two more antinodes for a total of three. If > λ , there will be at least
2
d
five antinodes, and so on. To repeat, if > 0 , the number of antinodes
is 1 or more. If
d
λ
λ
> 1, the number of antinodes is 3 or more. If
number of antinodes is 5 or more. In general,
d
λ
s
N
A
N
s
λ
4
FIG. P27.43
> 2 , the
d
The number of antinodes is 1 plus 2 times the greatest integer less than or equal to
λ
.
d λ
d λ
< , there will be no nodes. If > , there will be space for at least two nodes, as shown in the
2 4
2 4
d 3λ
d 5λ
picture. If >
, there will be at least four nodes. If >
six or more nodes will fit in, and so on.
2
4
2
4
To repeat, if 2 d < λ the number of nodes is 0. If 2d > λ the number of nodes is 2 or more. If 2d > 3 λ
d 1
+
> 1,
the number of nodes is 4 or more. If 2d > 5 λ the number of nodes is 6 or more. Again, if
λ 2
d 1
d 1
+
> 2 , the number of nodes is at least 4. If
+
> 3 , the
the number of nodes is at least 2. If
λ 2
λ 2
number of nodes is at least 6. In general,
If
FG
H
continued on next page
IJ
K
FG
H
FG
H
IJ
K
IJ
K
Chapter 27
the number of nodes is 2 times the greatest nonzero integer less than
761
FG d + 1 IJ .
H λ 2K
Next, we enumerate the zones of constructive interference. They are described by d sin θ = mλ ,
m = 0 , 1, 2 , K with θ counted as positive both left and right of the maximum at θ = 0 in the center.
d
The number of side maxima on each side is the greatest integer satisfying sin θ ≤ 1, d1 ≥ mλ , m ≤ .
λ
So the total
number of bright fringes is one plus 2 times the greatest integer less than or equal to
d
λ
.
It is equal to the number of antinodes on the line joining the sources.
FG
H
The interference minima are to the left and right at angles described by d sin θ = m +
FG
H
IJ
K
IJ
K
1
λ,
2
d 1
d 1
1
λ , m max < − or m max + 1 < + . Let n = 1, 2 , 3 , K .
λ 2
λ 2
2
d 1
Then the number of side minima is the greatest integer n less than + . Counting both left and
λ 2
d 1
+ . It is
right, the number of dark fringes is two times the greatest positive integer less than
λ 2
equal to the number of nodes in the standing wave between the sources.
m = 0 , 1, 2 , K. With sin θ < 1, d1 > m max +
FG
H
P27.44
IJ
K
My middle finger has width d = 2 cm .
(a)
Two adjacent directions of constructive interference for 600-nm light are described by
d sin θ = mλ
θ0 = 0
e2 × 10
(b)
−2
j
e
Thus,
θ 1 = 2 × 10 −3 degree
and
θ 1 − θ 0 ~ 10 −3 degree .
Choose
θ 1 = 20°
e2 × 10
−2
j
af
m sin 20° = 1 λ
λ = 7 mm
Millimeter waves are microwaves .
f=
c
λ
:
f=
j
m sin θ 1 = 1 6 × 10 −7 m
3 × 10 8 m s
7 × 10 −3 m
~ 10 11 Hz
762
P27.45
Wave Optics
No phase shift occurs upon reflection from the upper surface (glass to air) of the film, but there will
λ
due to the reflection at the lower surface of the film (air to metal). The total phase
2
difference in the two reflected beams is
be a shift of
then
δ = 2nt +
λ
2
.
For constructive interference, δ = mλ
or
a f
2 1.00 t +
λ
= mλ .
2
Thus, the film thickness for the m th order bright fringe is
FG
H
tm = m −
FG IJ
H K
IJ
K
1 λ
λ
λ
=m
−
2 2
2
4
and the thickness for the m − 1 bright fringe is:
a
t m −1 = m − 1
fFGH λ2 IJK − λ4 .
Therefore, the change in thickness required to go from one bright fringe to the next is
λ
∆t = t m − t m −1 =
2
.
To go through 200 bright fringes, the change in thickness of the air film must be:
200
FG λ IJ = 100λ .
H 2K
Thus, the increase in the length of the rod is
e
j
∆L = 100 λ = 100 5.00 × 10 −7 m = 5.00 × 10 −5 m .
*P27.46
From
∆L = L i α ∆ T
we have:
α=
∆L
5.00 × 10 −5 m
=
= 20.0 × 10 −6 ° C −1 .
Li ∆T
0.100 m 25.0° C
a
fa
f
The central bright fringe is wider than the side bright fringes, so the light must have been diffracted
by a single slit . For precision, we measure to the third minimum from the center
y = 4.0 cm
y 0.04 m
=
= 0.015 4
L 2.6 m
θ = 0.881° = 0.015 4 rad
tan θ =
a sin θ = mλ
a=
e
j e
j
3 632.8 × 10 −9 m 3 632.8 × 10 −9 m
mλ
=
=
= 1.23 × 10 −4 m
sin θ
sin 0.881°
0.015 4
Chapter 27
P27.47
763
For destructive interference, the path length must differ by mλ . We may treat this problem as a
double slit experiment if we remember the light undergoes a
π
-phase shift at the mirror. The
2
second slit is the mirror image of the source, 1.00 cm below the mirror plane. Modifying the equation
dy
= mλ we have
for Young’s experiment,
L
y dark
P27.48
e
ja
f
−7
mλ L 1 5.00 × 10 m 100 m
=
=
= 2.50 mm .
d
2.00 × 10 −2 m
e
j
a15.0 kmf + h = 30.175 km
a15.0 kmf + h = 227.63
2
2
2
2
2
h = 1.62 km
P27.49
FIG. P27.48
2nt = mλ
For dark fringes,
a
2nt = mλ
When submerged in water,
m=
so
P27.50
(a)
f
84 500 nm
.
2
and at the edge of the wedge, t =
a fa fa
f
2 1.33 42 500 nm
500 nm
FIG. P27.49
m + 1 = 113 dark fringes .
Minimum:
2nt = mλ 2
Maximum:
2nt = m ′ +
for λ 1 > λ 2 ,
FG m′ + 1 IJ < m
H 2K
so
m′ = m − 1 .
Then
2nt = mλ 2 = m −
FG
H
for m = 0 , 1, 2 , K
IJ
K
1
λ1
2
FG
H
for m ′ = 0 , 1, 2 , K
IJ
K
1
λ1
2
2mλ 2 = 2mλ 1 − λ 1
so
continued on next page
m=
b
λ1
2 λ1 − λ 2
g
.
764
Wave Optics
(b)
m=
500
= 1.92 → 2 (wavelengths measured to ±5 nm )
2 500 − 370
a
Minimum:
f
2nt = mλ 2
a f a
f
1I
F
2nt = G m − 1 + J λ = 1.5 λ
H
2K
2a1.40ft = 1.5a500 nmf
2 1.40 t = 2 370 nm
Maximum:
t = 264 nm
t = 268 nm
Film thickness = 266 nm .
P27.51
The shift between the two reflected waves is δ = 2na − b −
λ
2
where a and b are as shown in the ray diagram, n is the index of
λ
is due to phase reversal at the top
2
surface. For constructive interference, δ = mλ where m has
integer values. This condition becomes
refraction, and the term
FG
H
2na − b = m +
From the figure’s geometry,
a=
IJ
K
1
λ
2
FIG. P27.51
(1)
t
cos θ 2
t sin θ 2
cos θ 2
2t sin θ 2
sin φ 1
b = 2 c sin φ 1 =
cos θ 2
c = a sin θ 2 =
Also, from Snell’s law,
sin φ 1 = n sin θ 2 .
Thus,
b=
2nt sin 2 θ 2
.
cos θ 2
With these results, the condition for constructive interference given in Equation (1) becomes:
FG t IJ − 2nt sin θ = 2nt e1 − sin θ j = FG m + 1 IJ λ
H 2K
H cos θ K cos θ cosθ
F 1I
2nt cos θ = G m + J λ .
H 2K
2
2n
2
2
or
2
2
2
2
2
Chapter 27
P27.52
(a)
(b)
2nt = m +
Hence, the first bright band ( m = 0 ) corresponds to
nt =
Since
x 1 t1
=
x2 t2
we have
x 2 = x1
t1 =
t2 =
(c)
P27.53
λ1
4n
λ2
4n
=
420 nm
= 78.9 nm
4 1.33
=
680 nm
= 128 nm
4 1.33
λ
4
.
FG t IJ = x FG λ IJ = a3.00 cmfFG 680 nmIJ =
H 420 nm K
Ht K Hλ K
2
1
1
2
IJ
K
1
λ.
2
4.86 cm .
1
a f
θ ≈ tan θ =
a f
t1 78.9 nm
=
= 2.63 × 10 −6 rad
x1 3.00 cm
af
a sin θ = 1 λ .
The first minimum is at
λ
This has no solution if
a
>1.
a < λ = 632.8 nm .
or if
P27.54
FG
H
Bright bands are observed when
765
Consider vocal sound moving at 340 m/s and of frequency 3 000 Hz. Its wavelength is
λ=
v 340 m s
=
= 0.113 m .
f 3 000 Hz
If your mouth, for horizontal dispersion, behaves similarly to a slit 6.00 cm wide, then a sin θ = mλ
predicts no diffraction minima. You are a nearly isotropic source of this sound. It spreads out from
you nearly equally in all directions. On the other hand, if you use a megaphone with width 60.0 cm
at its wide end, then a sin θ = mλ predicts the first diffraction minimum at
θ = sin −1
FG mλ IJ = sin FG 0.113 mIJ = 10.9° .
H 0.600 m K
HaK
−1
This suggests that the sound is radiated mostly toward the front into a diverging beam of angular
diameter only about 20°. With less sound energy wasted in other directions, more is available for
your intended auditors. We could check that a distant observer to the side or behind you receives
less sound when a megaphone is used.
766
P27.55
Wave Optics
d=
1
400 mm
−1
= 2.50 × 10 −6 m
(a)
d sin θ = mλ
(b)
λ=
(c)
d sin θ a = 2 λ
θ a = sin −1
541 × 10 −9 m
= 4.07 × 10 −7 m
1.33
θ b = sin −1
d sin θ b =
F 2 × 541 × 10 mI = 25.6°
GH 2.50 × 10 m JK
F 2 × 4.07 × 10 mI = 19.0°
GH 2.50 × 10 m JK
−9
−6
−7
−6
2λ
n
af
n sin θ b = 1 sin θ a
P27.56
(a)
λ=
v
:
f
λ=
θ min = 1.22
λ
D
3.00 × 10 8 m s
d
:
L
= 0.214 m
FG 0.214 m IJ = 7.26 µ rad
H 3.60 × 10 mK
F 180 × 60 × 60 s IJ = 1.50 arc seconds
= 7.26 µ radG
H π K
θ min = 1.22
:
θ min
P27.57
1.40 × 10 9 s −1
4
g
e
jb
F 500 × 10 m I = 50.8 µ rad a10.5 seconds of arcf
θ
= 1.22G
H 12.0 × 10 m JK
radja30.0 mf = 1.52 × 10 m = 1.52 mm
d = θ min L = 7.26 × 10 −6 rad 26 000 ly = 0.189 ly
(b)
θ min =
(c)
θ min = 1.22
(d)
d = θ min L = 50.8 × 10 −6
(a)
d sin θ = mλ
−9
λ
min
D
e
e
−3
−3
j
3 500 × 10 −9 m
mλ
= 2.83 µm
or d =
=
sin θ
sin 32.0°
Therefore, lines per unit length =
1
1
=
d 2.83 × 10 −6 m
or lines per unit length = 3.53 × 10 5 m −1 = 3.53 × 10 3 cm −1 .
continued on next page
Chapter 27
sin θ =
(b)
*P27.58
e
j a
−9
mλ m 500 × 10 m
=
= m 0.177
d
2.83 × 10 −6 m
f
a
f
For sin θ ≤ 1.00 , we must have
m 0.177 ≤ 1.00
or
m ≤ 5.66 .
Therefore, the highest order observed is
m=5 .
Total number of primary maxima observed is
2m + 1 = 11 .
Bragg’s law applies to the space lattice of melanin rods. Consider the planes d = 0.25 µm
apart. For light at near-normal incidence, strong reflection happens for the wavelength
given by 2d sin θ = mλ . The longest wavelength reflected strongly corresponds to m = 1:
(a)
e
j
2 0.25 × 10 −6 m sin 90° = 1λ
*P27.59
λ = 500 nm. This is the blue-green color.
(b)
For light incident at grazing angle 60°, 2d sin θ = mλ gives
1λ = 2 0.25 × 10 −6 m sin 60° = 433 nm . This is violet.
(c)
Your two eyes receive light reflected from the feather at different angles, so they receive
light incident at different angles and containing different colors reinforced by constructive
interference.
(d)
The longest wavelength that can be reflected with extra strength by these melanin rods is
the one we computed first, 500 nm blue-green.
(e)
If the melanin rods were farther apart (say 0.32 µm ) they could reflect red with constructive
interference.
e
j
A central maximum and side maxima in seven orders of
interference appear. If the seventh order is just at 90°,
e
j
d1 = 7 654 × 10 −9 m
d sin θ = mλ
d = 4.58 µ m .
If the seventh order is at less than 90°, the eighth order
might be nearly ready to appear according to
e
j
d1 = 8 654 × 10 −9 m
d = 5.23 µ m .
FIG. P27.59
Thus 4.58 µ m < d < 5.23 µ m .
P27.60
767
(a)
We require
λ
D
=
radius of diffraction disk D
=
.
L
2L
D 2 = 2.44λ L .
Then
(b)
θ min = 1.22
e
ja
f
D = 2.44 500 × 10 −9 m 0.150 m = 428 µ m
768
P27.61
Wave Optics
d sin θ = mλ
a
f
and, differentiating,
d cos θ dθ = mdλ
or
d 1 − sin 2 θ ∆θ ≈ m∆λ
d 1−
so
∆θ ≈
m 2 λ2
∆θ ≈ m∆λ
d2
ed
∆λ
2
j
m 2 − λ2
.
ANSWERS TO EVEN PROBLEMS
P27.2
(a) 36.2°; (b) 5.08 cm; (c) 508 THz
P27.26
1.00 mrad
P27.4
maxima at 0°, 29.1°, and 76.3°; minima at
14.1° and 46.8°
P27.28
she can resolve only the violet tubes if she
is between 186 m and 271 m away
P27.6
11.3 m
P27.30
P27.8
(a) 10.0 m;
(b) 516 m;
(c) Only the runway centerline is a
maximum for the interference patterns for
both frequencies. If the frequencies were
k
related by a ratio of small integers , the
l
plane could by mistake fly along the k-th
side maximum of one signal where it
coincides with the l-th side maximum of
the other.
neither, it can resolve objects no closer
than several centimeters apart
P27.32
514 nm
P27.34
1.81 µ m 1.81 m
P27.36
see the solution
P27.38
93.4 pm
P27.40
5.51 m, 2.76 m, 1.84 m
P27.42
(a) 217 nm; (b) 93.1 nm
P27.44
(a) ~ 10 −3 degree ;
(b) ~ 10 11 Hz , microwave
P27.10
see the solution
P27.12
(a) 1.29 rad; (b) 99.6 nm
P27.14
612 nm
P27.46
one slit 0.123 mm wide
P27.16
0.500 cm
P27.48
1.62 km
P27.18
(a) 238 nm; (b) λ increase; (c) 328 nm
P27.50
(a) see the solution; (b) 266 nm
P27.20
4.22 mm
P27.52
P27.22
547 nm
(a) 4.86 cm from the top;
(b) 78.9 nm and 128 nm;
(c) 2.63 × 10 −6 rad
P27.24
three maxima, at 0° and near 46° on both
sides
P27.54
see the solution
Chapter 27
P27.56
(a) 7.26 µ rad (1.50 seconds of arc);
(b) 0.189 ly;
(c) 50.8 µ rad (10.5 seconds of arc);
(d) 1.52 mm
P27.58
see the solution
P27.60
(a) D 2 = 2.44λ L ; (b) 0.428 mm
769
CONTEXT 8 CONCLUSION SOLUTIONS TO PROBLEMS
CC8.1
Depth = one-quarter of the wavelength in plastic.
t=
CC8.2
λ
4n
=
780 nm
= 130 nm
4 1.50
a f
For a side maximum,
tan θ =
y 0.4 µ m
=
L 6.9 µ m
θ = 3.32°
d sin θ = mλ
a1fe780 × 10
d=
−9
j = 13.5 µ m .
m
sin 3.32°
The number of grooves per millimeter =
1 × 10 −3 m
= 74.2 .
13.5 × 10 -6 m
FIG. CC8.2
*CC8.3
In one second, a 1.3-m length of audio track passes the laser. This length includes 705 600 bits of
1.3 m
information. The average length per bit is
= 1.8 × 10 −6 m bit = 1.8 µ m bit . The average
705 600 bits
length per bit of total information on a CD is smaller than this because the disc includes additional
information, such as error correction codes, song numbers, and timing codes. The shortest length
per bit of total information is actually about 0.8 µ m bit .
*CC8.4
We convert each bit to a power of 2 and add them up:
1 × 2 15
1 × 2 11
1 × 27
1 × 23
= 32 768 0 × 2 14 = 0 1 × 2 13 = 8 192 1 × 2 12 = 4 096
= 2 048
= 128
=8
0 × 2 12 = 0
0 × 26 = 0
0 × 22 = 0
1 × 2 9 = 512
1 × 2 5 = 32
1 × 21 = 2
1 × 2 8 = 256
1 × 2 4 = 16
1 × 20 = 1
sum = 48 059
The CD player converts this number into a voltage, representing one of the 44 100 values that will be
used to build one second of the electronic waveform that represents the recorded sound.
CC8.5
e
j e10 mj ≈ 3 × 10 m
kg m je3 × 10
m j ≈ 6 × 10
The volume of the cylinder is
π r 2 h = π 10 −6 m
The mass of the magnetic material is
ρ V = 2 × 10 3
e
2
−6
3
−18
−18
3
3
−15
kg
We model the disk as absorbing all of the incident light. To identify the amount of energy transfer
TER by electromagnetic radiation, we compute the heat that would have the same effect:
continued on next page
770
Wave Optics
e
“Q ” = m c∆T = 6 × 10 −15 kg
jFGH kg300⋅° CJ IJK a600 K − 300 K fFGH 11°KC IJK
“Q ” ≈ 6 × 10 −10 J
The pit moves past in a time interval of
So, the intensity is I =
P
A
=
“Q ”
:
A ∆t
∆t =
I≈
2 × 10 −6 m
= 2 × 10 −6 s.
1 ms
LMπ e10
N
6 × 10 −10 J
−6
= 9 × 10
mj Oe 2 × 10 sj
PQ
2
−6
7
W m2
I ~ 10 8 W m 2
ANSWERS TO EVEN CONTEXT 8 CONCLUSION PROBLEMS
CC8.2
74.2 grooves mm
CC8.4
48 059