Chapter 22
Reflection and
Refraction of Light
Quick Quizzes
1.
(a). In part (a), you can see clear reflections of the headlights and the lights on the top of
the truck. The reflection is specular. In part (b), although bright areas appear on the
roadway in front of the headlights, the reflection is not as clear, and no separate reflection
of the lights from the top of the truck is visible. The reflection in part (b) is mostly diffuse.
2.
Beams 2 and 4 are reflected; beams 3 and 5 are refracted.
3.
(b). When light goes from one material into one having a higher index of refraction, it
refracts toward the normal line of the boundary between the two materials. If, as the light
travels through the new material, the index of refraction continues to increase, the light
ray will refract more and more toward the normal line.
4.
(c). Both the wave speed and the wavelength decrease as the index of refraction increases.
The frequency is unchanged.
233
234
CHAPTER 22
Answers to Even Numbered Conceptual Questions
2.
Ceilings are generally painted a light color so they will reflect more light, making the
room brighter. Textured materials are often used on the ceiling to diffuse the reflected
light and reduce glare (specular reflections).
4.
At the altitude of the plane the surface of Earth does not block off the lower half of the
rainbow. Thus, the full circle can be seen. You can see such a rainbow by climbing on a
stepladder above a garden sprinkler in the middle of a sunny day.
6.
The spectrum of the light sent back to you from a drop at the top of the rainbow arrives
such that the red light (deviated by an angle of 42°) strikes the eye while the violet light
(deviated by 40°) passes over your head. Thus, the top of the rainbow looks red. At the
bottom of the bow, violet light arrives at your eye and red light is deviated toward the
ground. Thus, the bottom part of the bow appears violet.
8.
A mirage occurs when light changes direction as it moves between batches of air having
different indices of refraction. The different indices of refraction occur because the air has
different densities at different temperatures. Two images are seen; One from a direct path
from the object to you, and the second arriving by rays originally heading toward Earth
but refracted to your eye. On a hot day, the Sun makes the surface of blacktop hot, so the
air is hot directly above it, becoming cooler as one moves higher into the sky. The “water”
we see far in front of us is an image of the blue sky. Adding to the effect is the fact that the
image shimmers as the air changes in temperature, giving the appearance of moving
water.
10.
The upright image of the hill is formed by light that has followed a direct path from the
hill to the eye of the observer. The second image is a result of refraction in the atmosphere.
Some light is reflected from the hill toward the water. As this light passes through warmer
layers of air directly above the water, it is refracted back up toward the eye of the
observer, resulting in the observation of an inverted image of the hill directly below the
upright image.
12.
The color traveling slowest is bent the most. Thus, X travels more slowly in the glass
prism.
14.
Total internal reflection occurs only when light attempts to move from a medium of high
index of refraction to a medium of lower index of refraction. Thus, light moving from air
(n = 1) to water (n = 1.33) cannot undergo total internal reflection.
16.
Objects beneath the surface of water appear to be raised toward the surface by refraction.
Thus, the bottom of the oar appears to be closer to the surface than it really is, and the oar
looks to be bent.
Reflection and Refraction of Light
18.
235
The cross section can be visualized by considering just the two rays of light on the edges
of the beam. If the beam of light enters a new medium with a higher index of refraction,
the rays bend toward the normal, and the cross section of the refracted beam will be larger
than that of the incident beam as suggested by Fig. CQ22.18a. If the new index of
refraction is lower, the rays bend away from the normal, and the cross section of the beam
is reduced, as shown in Fig. CQ22.18b.
1
1
2
2
n2 > n1
n2 < n1
(b)
(a)
Figure CQ22.18
236
CHAPTER 22
Answers to Even Numbered Problems
2.
2.97 × 108 m s
4.
(a)
536 rev s
(b)
1.07 × 10 3 rev s
6.
(a)
1.94 m
(b)
50.0° above horizontal (parallel to incident ray)
8.
2.09 × 10 −11 s
The longer the wavelength, the less it is deviated (or refracted) from the original
path.
(b) Using data from Figure 22.14, the angles of refraction are:
(400 nm) θ 2 = 16.0° , (500 nm) θ 2 = 16.1° , (650 nm) θ 2 = 16.3°
10.
(a)
12.
(a)
14.
67.4°
16.
53.4°
18.
First surface: θ i = 30.0°, θ r = 19.5°
Second surface: θ i = 19.5°, θ r = 30.0°
20.
1.06 × 10 −10 s
22.
107 m
24.
6.30 cm
26.
23.1°
28.
2.5 m
30.
0.40°
32.
4.6°
34.
(a)
36.
48.5°
38.
67.2°
40.
4.54 m
327 nm
24.4°
(b)
(b)
287 nm
37.0°
Reflection and Refraction of Light
44.
(a) θ1′ = 30.0°, θ 2 = 18.8°
(c) See solution.
(b)
(d)
θ1′ = 30.0°, θ 2 = 50.8°
See solution.
46.
(a) Any angle of incidence ≤ 90°
(b) 30.0°
(c) not possible since npolystyrene < ncarbon disulfide
48.
(a) 0.172 mm s
(b) 0.345 mm s
(c) and (d) Northward at 50.0° below horizontal.
50.
77.5°
52.
(a)
54.
7.91°
56.
82
58.
62.2% of a circle
60.
The graph is a straight line passing through the origin. From the slope of the graph,
nwater = 1.33 .
62.
(a)
R ≥ nd ( n − 1)
(
(b)
)
n = 1 + 4t d 2 


12
(b)
237
yes; yes; yes
2.10 cm
(c)
(c)
350 µ m
violet
238
CHAPTER 22
Problem Solutions
22.1
The total distance the light travels is


∆d = 2  Dcenter to − REarth − RMoon 
 center

= 2 ( 3.84 × 10 8 − 6.38 × 10 6 − 1.76 × 10 6 ) m = 7.52 × 10 8 m
Therefore,
22.2
v=
∆d 7.52 × 10 8 m
=
= 3.00 × 10 8 m s
∆t
2.51 s
If the wheel has 360 teeth, it turns through an angle of 1 720 rev in the time it takes the
light to make its round trip. From the definition of angular velocity, we see that the time
is
t=
θ ( 1 720 ) rev
=
= 5.05 × 10 −5 s
ω 27.5 rev s
Hence, the speed of light is c =
22.3
2d 2 ( 7 500 m )
=
= 2.97 × 108 m s
t
5.05 × 10 −5 s
The experiment is most convincing if the wheel turns fast enough to pass outgoing light
through one notch and returning light through the next. Then,
2π
 1

rev  ( 2π rad rev ) =
rad
∆θ = 
720
 720

and
8
c ( ∆θ ) ( 2.998 × 10 m s )  2π
∆θ
∆θ

rad  = 114 rad s
ω=
=
=
=

3
2d
∆t 2 d c
2 ( 11.45 × 10 m )  720

Reflection and Refraction of Light
22.4
239
(a) The time for the light to travel to the stationary mirror and back is
3
2 d 2 ( 35.0 × 10 m )
∆t =
=
= 2.33 × 10 −4 s
8
c
3.00 × 10 m s
At the lowest angular speed, the octagonal mirror will have rotated 1 8 rev in this
time, so
ωmin =
1 8 rev
∆θ
=
= 536 rev s
∆t 2.33 × 10 −4 s
(b) At the next higher angular speed, the mirror will have rotated 2 8 rev in the
elapsed time, or
ω2 = 2 ωmin = 2 ( 536 rev s ) = 1.07 × 10 3 rev s
22.5
(a) For the light beam to make it through both slots, the time for the light to travel
distance d must equal the time for the disks to rotate through angle θ. Therefore, if c
is the speed of light,
t=
d θ
= , or
c ω
c=
ωd
θ

1 rev
 1
  1 rev
, and ω = 5 555 rev s
(b) If d = 2.500 m , θ = 
degree  
=
 60
  360 degree  ( 60 )( 360 )
rev   ( 60 )( 360 ) 

8
c = ( 2.500 m )  5 555
 = 3.000 × 10 m s

s   1 rev 

240
CHAPTER 22
22.6
Mirror 2
Light
beam
Mirror
2
50°
40.0°
i2 = 50°
40°
P
1.25 m
d
40°
Mirror
1
i1 = 40°
50°
50°
Mirror 1
1.25 m
(a) From geometry, 1.25 m = d sin 40.0° , so d = 1.94 m
(b)
22.7
50.0° above horizontal , or parallel to the incident ray
n1 sin θ1 = n2 sin θ 2
q1
n1 = 1.00
sin θ1 = 1.333 sin 45.0°
q2
sin θ1 = (1.333)(0.707) = 0.943
n2 = 1.333
θ1 = 70.5° →
22.8
19.5° above the horizontal
c
The speed of light in water is vwater =
, and in Lucite vLucite =
®
nwater
time required to transverse the double layer is
∆t1 =
c
nLucite
. Thus, the total
dwater dLucite dwater nwater + dLucite nLucite
+
=
vwater vLucite
c
The time to travel the same distance in air is ∆t2 =
dwater + dLucite
, so the additional time
c
required for the double layer is
∆t = ∆t1 − ∆t2 =
=
(1.00 × 10
−2
dwater ( nwater − 1) + dLucite ( nLucite − 1)
c
m ) ( 1.333 − 1) + ( 0.500 × 10 −2 m ) ( 1.59 − 1)
3.00 × 10 m s
8
= 2.09 × 10 −11 s
Reflection and Refraction of Light
22.9
(a) From Snell’s law, n2 =
(b) λ2 =
(c)
f=
λ0
n2
c
λ0
=
=
241
n1 sin θ1 ( 1.00 ) sin 30.0°
=
= 1.52
sin θ 2
sin19.24°
632.8 nm
= 417 nm
1.52
3.00 × 108 m s
= 4.74 × 1014 Hz in air and in syrup
632.8 × 10 −9 m
c 3.00 × 108 m s
(d) v2 =
=
= 1.98 × 108 m s
n2
1.52
22.10
(a) When light refracts from air ( n1 = 1.00 ) into the
Crown glass, Snell’s law gives the angle of
refraction as

n
1.54

θ 2 = sin −1  sin 25.0° nCrown 

glass
Crown glass
1.52

1.50
For first quadrant angles, the sine of the angle
increases as the angle increases. Thus, from the
above equation, note that θ 2 will increase when
the index of refraction of the Crown glass
decreases. From Figure 22.14, this means that
the longer wavelengths have the largest angles
of refraction, and deviate the least from the
Acrylic
1.48
Fused quartz
1.46
400
500
600
700
l, nm
original path.
(b) From Figure 22.14, observe that the index of refraction of Crown glass for the given
wavelengths is:
and
λ = 400 nm:
λ = 650 nm:
nCrown glass = 1.53 ;
λ = 500 nm:
nCrown glass = 1.52 ;
nCrown glass = 1.51
Thus, Snell’s law gives: λ = 400 nm:
θ 2 = sin −1 ( sin 25.0° 1.53 ) = 16.0°
λ = 500 nm:
θ 2 = sin −1 ( sin 25.0° 1.52 ) = 16.1°
λ = 650 nm:
θ 2 = sin −1 ( sin 25.0° 1.51) = 16.3°
242
CHAPTER 22
22.11
(a)
λwater =
λ0
nwater
, so λ0 = nwater λwater = ( 1.333 )( 438 nm ) = 584 nm
(b) λ0 = nwater λwater = nbenzene λbenzene
nbenzene
λ
438 nm
= water =
= 1.12
nwater
λbenzene 390 nm
and
22.12
(a)
λwater =
(b) λglass =
λ0
nwater
λ0
ncrown
=
436 nm
= 327 nm
1.333
=
436 nm
= 287 nm
1.52
glass
22.13
From Snell’s law,
40.0° f
 n1 sin θ1 
−1  ( 1.00 ) sin 40.0° 
 = sin 
 = 29.4°
1.309


 n2 
θ 2 = sin −1 
air
ice
q2
and from the law of reflection, φ = θ1 = 40.0°
Hence, the angle between the reflected and refracted rays is
α = 180° − θ 2 − φ = 180° − 29.4° − 40.0° = 111°
22.14
 v
Thus, θ 2 = sin −1  2
 v1
22.15
sin θ 2 n1 c v1 v2
=
=
=
sin θ1 n2 c v2 v1
Snell’s law may be written as:




−1  1 510 m s 
 sin12.0° = 67.4°
 sin θ1  = sin 


 340 m s 

n = 1.923
The index of refraction of zircon is
(a)
v=
c 3.00 × 10 8 m s
=
= 1.56 × 108 m s
1.923
n
λn =
(b) The wavelength in the zircon is
(c) The frequency is
f=
v
λn
=
c
λ0
=
a
λ0
n
=
632.8 nm
= 329.1 nm
1.923
3.00 × 10 8 m s
= 4.74 × 1014 m s
632.8 × 10 -9 m
Reflection and Refraction of Light
22.16
The angle of incidence is
air
water
4.00 m
 2.00 m 
θ1 = tan −1 
 = 26.6°
 4.00 m 
Therefore, Snell’s law gives
2.00
m
q2
243
f
q1 q1
 n1 sin θ1 

 n2 
θ 2 = sin −1 
 ( 1.333 ) sin 26.6° 
= sin −1 
 = 36.6°
1.00


and the angle the refracted ray makes with the surface is
φ = 90.0° − θ 2 = 90.0° − 36.6° = 53.4°
The incident light reaches the left-hand mirror at
distance
Mirror
Mirror
( 1.00 m ) tan 5.00° = 0.087 5 m
above its bottom edge. The reflected light first
reaches the right-hand mirror at height
2 ( 0.087 5 m ) = 0.175 m
It bounces between the mirrors with this distance
between points of contact with either.
Since
1.00 m
22.17
reflected beam
5.00°
1.00 m
= 5.72 , the light reflects
0.175 m
five times from the right-hand mirror and six times from the left
1.00 m
244
CHAPTER 22
22.18
At the first surface, the angle of incidence is θ1 = 30.0° , and Snell’s law gives
 nair sin θ1 
−1  ( 1.00 ) sin 30.0° 
 = sin 
 = 19.5°
1.50


 nglass 
θ 2 = sin −1 
Since the second surface is parallel to the first, the angle of incidence at the second
surface is θ1 = 19.5° and the angle of refraction is
 nglass sin θ glass 
−1  ( 1.50 ) sin19.5° 
 = sin 
 = 30.0°
1.00
nair




θ 2 = sin −1 
Thus, the light emerges traveling parallel to the incident beam.
The angle of refraction at the first surface is θ 2 = 19.5° (see
Problem 18). Let h represent the distance from point a to c
(that is, the hypotenuse of triangle abc). Then,
a
2.00 cm 2.00 cm
=
= 2.12 cm
cosθ 2
cos19.5°
Also, α = θ1 − θ 2 = 30.0° − 19.5° = 10.5° , so
b
h=
d = h sin α = ( 2.12 cm ) sin10.5° = 0.388 cm
22.20
q1
2.00 cm
22.19
The distance h the light travels in the glass is
h=
2.00 cm
= 2.12 cm
cos 19.5°
The speed of light in the glass is
v=
c
nglass
Therefore, t =
=
3.00 × 108 m s
= 2.00 × 108 m s
1.50
h 2.12 × 10 − 2 m
=
= 1.06 × 10 −10 s
8
v 2.00 × 10 m s
q1
q2
a
c d
245
Reflection and Refraction of Light
22.21
From Snell’s law, the angle of incidence at the air-oil interface is
 noil sin θ oil 

 nair

θ = sin −1 
 ( 1.48 ) sin 20.0° 
= sin −1 
 = 30.4°
1.00


and the angle of refraction as the light enters the water is
 noil sin θ oil 
−1  ( 1.48 ) sin 20.0° 
 = sin 
 = 22.3°
1.333


 nwater 
θ ′ = sin −1 
22.22
The angle of incidence at the water
surface is
h
θ1 = tan −1 
90.0 m 
 = 42.0°
 100 m 
q2
q2
air
n = 1.00
90.0 m
Then, Snell’s law gives the angle of
refraction as
water
n = 1.333
 nwater sin θ1 
−1  ( 1.333 ) sin 42.0° 
 = sin 
 = 63.1°
nair
1.00




22.23
210 m
210 m
=
= 107 m
tanθ 2 tan63.1°
∆t = ( time to travel 6.20 m in ice ) − ( time to travel 6.20 m in air )
∆t =
6.20 m 6.20 m
−
vice
c
Since the speed of light in a medium of refractive index n is v =
q1
Submarine
θ 2 = sin −1 
so the height of the building is h =
q1
c
n
 1.309 1  ( 6.20 m )( 0.309 )
∆t = ( 6.20 m ) 
− =
= 6.39 × 10 −9 s = 6.39 ns
8
c  3.00 × 10 m s
 c
100 m
210 m
246
CHAPTER 22
22.24
n
From Snell’s law, sin θ =  medium
 nliver
But,
so,

 sin 50.0°

12.0 cm
50.0°
nmedium c vmedium
v
=
= liver = 0.900
nliver
c vliver
vmedium
h
d
nmedium
q
nliver
q
Tumor
θ = sin −1 ( 0.900 ) sin 50.0° = 43.6°
From the law of reflection,
d=
22.25
12.0 cm
= 6.00 cm , and
2
h=
d
6.00 cm
=
= 6.30 cm
tanθ tan ( 43.6° )
As shown at the right, θ1 + β + θ 2 = 180°
incident
ray
When β = 90° , this gives θ 2 = 90° − θ1
Air,
n = 1.00
glass, ng
Then, from Snell’s law
sin θ1 =
q1
nair
= ng sin ( 90° − θ1 ) = ng cos θ1
sin θ1
= tan θ1 = ng or θ1 = tan −1 ng
cosθ1
( )
22.26
Given Conditions and Observed Results
26.5° Sheet 1
n1
b
q2 refracted
ray
ng sin θ 2
Thus, when β = 90° ,
q1
reflected
ray
26.5° Sheet 3
n3
Sheet 2
n2
31.7°
Sheet 2
n2
36.7°
Case 1
Case 2
For the first placement, Snell’s law gives,
26.5° Sheet 1
n1
Sheet 3
n3
qR
Case 3
n2 =
n1 sin 26.5°
sin 31.7°
In the second placement, application of Snell’s law yields
n1 sin 36.7°
 n sin 26.5° 
n3 sin 26.5° = n2 sin 36.7° =  1
 sin 36.7° , or n3 =
sin 31.7°
 sin 31.7° 
Reflection and Refraction of Light
247
Finally, using Snell’s law in the third placement gives
sin θ R =
and
22.27
 sin 31.7° 
n1 sin 26.5°
= ( n1 sin 26.5° ) 
 = 0.392
n3
 n1 sin 36.7° 
θ R = 23.1°
When the Sun is 28.0° above the horizon, the angle of
incidence for sunlight at the air-water boundary is
q1
air, n = 1.00
28.0°
θ1 = 90.0° − 28.0° = 62.0°
Thus, the angle of refraction is
h
q2
 nair sin θ1 

 nwater 
water
n = 1.333
θ 2 = sin −1 
3.00 m
 ( 1.00 ) sin 62.0° 
= sin −1 
 = 41.5°
1.333


The depth of the tank is then h =
22.28
3.00 m
3.00 m
=
= 3.39 m
tanθ 2
tan ( 41.5° )
From the drawing, observe that
d
q2
h2
R
−1  1.5 m 
 = 37°
 = tan 
 h1 
 2.0 m 
θ1 = tan −1 
h1
Applying Snell’s law to the ray shown gives
 nliquid sin θ1 
−1  1.5sin 37° 
 = sin 
 = 64°
1.0
nair




θ 2 = sin −1 
Thus, the distance of the girl from the cistern is
d = h2 tan θ 2 = ( 1.2 m ) tan 64° = 2.5 m
air
n = 1.0
liquid
n = 1.5
q1
R
248
CHAPTER 22
22.29
Using Snell’s law gives
 nair sin θ i
 nred

−1  ( 1.000 ) sin 83.00° 
 = sin 
 = 48.22°
1.331



 nair sin θ i
 nblue

−1  ( 1.000 ) sin 83.00° 
 = sin 
 = 47.79°
1.340



θ red = sin −1 
and
22.30
θ blue = sin −1 
The angles of refraction for the two wavelengths are
 1.00 0 ) sin 30.00° 
 nair sin θ i 
−1 (
 = 18.04°
 = sin 
1.615
 nred 


θ red = sin −1 
and
 nair sin θ i
 nblue
θ blue = sin −1 
 1.00 0 ) sin 30.00° 

−1 (
 = 17.64°
 = sin 
1.650



Thus, the angle between the two refracted rays is
∆θ = θ red − θ blue = 18.04° − 17.64° = 0.40°
22.31
(a) The angle of incidence at the first surface is
θ1i = 30° , and the angle of refraction is
 nair sin θ1i
 nglass

θ1r = sin −1 




nair = 1.0
60°
q1i = 30°
a
q1r
 1.0 sin 30° 
= sin −1 
 = 19°
1.5


b
q2i
q2r
nglass = 1.5
Also, α = 90° − θ1r = 71° and β = 180° − 60° − α = 49°
Therefore, the angle of incidence at the second surface is θ 2i = 90° − β = 41° . The
angle of refraction at this surface is
 nglass sin θ 2i
nair

θ 2r = sin −1 

−1  1.5sin 41° 
 = sin 
 = 77°
1.0



Reflection and Refraction of Light
249
(b) The angle of reflection at each surface equals the angle of incidence at that surface.
Thus,
(θ1 )reflection = θ1i =
22.32
30° , and (θ 2 )reflection = θ 2i = 41°
For the violet light, nglass = 1.66 , and
 n sin θ1i
θ1r = sin −1  air
 nglass

nair = 1.00
60.0°




q1i = 50.0°
 1.00 sin 50.0° 
= sin −1 
 = 27.5°
1.66


a
b
q1r q2i
nglass
α = 90° − θ1r = 62.5°, β = 180.0° − 60.0° − α = 57.5°,
and
θ 2i = 90.0° − β = 32.5° . The final angle of refraction of the violet light is
 nglass sin θ 2i
nair

θ 2r = sin −1 

−1  1.66 sin 32.5° 
 = sin 
 = 63.2°
1.00



(
)
Following the same steps for the red light nglass = 1.62 gives
θ1r = 28.2°, α = 61.8°, β = 58.2°, θ 2i = 31.8°, and θ 2r = 58.6°
Thus, the angular dispersion of the emerging light is
Dispersion = θ 2r
22.33
violet
− θ 2r
red
= 63.2° − 58.6° = 4.6°
When surrounded by air ( n2 = 1.00 ) , the critical angle of a material is
 n2 
1 
−1 

 = sin 
 n1 
 nmaterial 
θ c = sin −1 
 1 
(a) For Zircon, n = 1.923 , and θ c = sin −1 
 = 31.3°
 1.923 
 1 
(b) For fluorite, n = 1.434 , and θ c = sin −1 
 = 44.2°
 1.434 
 1 
(c) For ice, n = 1.309 , and θ c = sin −1 
 = 49.8°
 1.309 
q2r
250
CHAPTER 22
22.34
When surrounded by air ( n2 = 1.00 ) , the critical angle of a material is
 n2
 n1
θ c = sin −1 

1 
−1 

 = sin 

 nmaterial 
 1 
(a) For diamond, θ c = sin −1 
 = 24.4°
 2.419 
 1 
(b) For flint glass, θ c = sin −1 
 = 37.0°
 1.66 
22.35
When surrounded by water ( n2 = 1.333 ) , the critical angle of a material is
 n2
 n1
θ c = sin −1 

−1  1.333 

 = sin 

 nmaterial 
 1.333 
(a) For diamond, θ c = sin −1 
 = 33.4°
 2.419 
 1.333 
(b) For flint glass, θ c = sin −1 
 = 53.4°
 1.66 
22.36
Using Snell’s law, the index of refraction of the liquid is found to be
nliquid =
nair sin θ i ( 1.00 ) sin 30.0°
=
= 1.33
sin θ r
sin 22.0°
 n
Thus, θ c = sin −1  air
 nliquid

22.37

 1.00 
 = sin −1 
 = 48.5°

1.33



When light attempts to cross a boundary from one medium of refractive index n1 into a
new medium of refractive index n2 < n1 , total internal reflection will occur if the angle of
incidence exceeds the critical angle given by θ c = sin −1 ( n2 n1 ) .
(a) If n1 = 1.53 and n2 = nair = 1.00, then
(b) If n1 = 1.53 and n2 = nwater = 1.333, then
θ c = sin −1 
1.00 
 = 40.8°
 1.53 
θ c = sin −1 
1.333 
 = 60.6°
 1.53 
251
Reflection and Refraction of Light
22.38
The critical angle for this material in air is
 nair
 npipe

θ c = sin −1 
q
qr c

 1.00 
 = sin −1 
 = 47.3°

 1.36 

qi
Thus, θ r = 90.0° − θ c = 42.7° and from Snell’s law,
 npipe sin θ r 
−1  ( 1.36 ) sin 42.7° 
 = sin 
 = 67.2°
nair
1.00




θ i = sin −1 
22.39
The light must be totally reflecting from the surface of a hot air layer just above the road
surface. The angle of reflection, and hence the critical angle, is θ c = 90.0° − 1.20° = 88.8° .
Thus, from sin θ c =
n2
, we find
n1
n2 = n1 sin θ c = ( 1.000 3 ) sin 88.8° = 1.000 08
22.40
The circular raft must cover the area of the surface
through which light from the diamond could emerge.
Thus, it must form the base of a cone (with apex at the
diamond) whose half angle is θ, where θ is greater than
or equal to the critical angle.
q
The critical angle at the water-air boundary is
 nair 
−1  1.00 
 = sin 
 = 48.6°
 1.333 
 nwater 
θ c = sin −1 
Thus, the minimum diameter of the raft is
2 rmin = 2 h tan θ min = 2 h tan θ c = 2 ( 2.00 m ) tan 48.6° = 4.54 m
q
h
rmin
q
q
252
CHAPTER 22
22.41
(a) Snell’s law can be written as
sin θ1 v1
= . At the critical angle of incidence (θ1 = θ c ) ,
sin θ 2 v2
v
the angle of refraction is 90° and Snell’s law becomes sin θ c = 1 . At the concrete-air
v2
boundary,
 v1 
−1  343 m s 
 = 10.7°
 = sin 
 v2 
 1 850 m s 
θ c = sin −1 
(b) Sound can be totally reflected only if it is initially traveling in the slower medium.
Hence, at the concrete-air boundary, the sound must be traveling in air .
(c)
Sound in air falling on the wall from most directions is 100% reflected , so the wall
is a good mirror.
22.42
The sketch at the right shows a light ray entering at the painted
corner of the cube and striking the center of one of the three
unpainted faces of the cube. The angle of incidence at this face
is the angle θ1 in the triangle shown. Note that one side of this
triangle is half the diagonal of a face and is given by
d
=
2
l2 +l2
l
=
2
2
l
L q1
d¤2
2
Also, the hypotenuse of this triangle is
Thus, sin θ1 =
l2
3
d
L = l2 +   = l2 +
=l
2
2
2
d2
l 2
1
=
=
L
3
l 3 2
(
)
For total internal reflection at this face, it is necessary that
sin θ1 ≥ sin θ c =
nair
ncube
or
1 1.00
≥
n
3
giving
n≥ 3
l
l
l
Reflection and Refraction of Light
22.43
If θ c = 42.0° at the boundary between the prism glass
n
and the surrounding medium, then sin θ c = 2 gives
n1
q1
nglass
nm
=
Surrounding
Medium, nm
b
nglass
60.0°
qr
nm
= sin 42.0°
nglass
or
253
a
qc = 42.0°
Surface 2
1
= 1.494
sin 42.0°
From the geometry shown in the above figure,
α = 90.0° − 42.0° = 48.0°, β = 180° − 60.0° − α = 72.0°
and θ r = 90.0° − β = 18.0° . Thus, applying Snell’s law at the first surface gives
 nglass sin θ r 
−1
 = sin ( 1.494 sin18.0° ) = 27.5°
nm


θ1 = sin −1 
22.44
(a) θ1′ = θ1 = 30.0°
 n1 sin θ1 

 n2 
θ 2 = sin −1 
 ( 1.00 ) sin 30.0° 
= sin −1 

1.55


air
q1 q1¢
air
q2
glass
glass
q2
Parts (a) and (c)
= 18.8°
 n sin θ1 
−1  ( 1.55 ) sin 30.0° 
(b) θ1′ = θ1 = 30.0° , θ 2 = sin −1  1
 = sin 
 = 50.8°
1.00


 n2 
q1 q1¢
Parts (b) and (d)
254
CHAPTER 22
(c) and (d) The other entries are computed similarly and are shown in the table below.
22.45
(c) air into glass, angles in degrees
(d) glass into air, angles in degrees
incidence
0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
incidence
reflection
0
0
10.0
10.0
20.0
20.0
30.0
30.0
40.0
40.0
50.0
50.0
60.0
60.0
70.0
70.0
80.0
80.0
90.0
90.0
*total internal reflection
reflection
0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
refraction
0
6.43
12.7
18.8
24.5
29.6
34.0
37.3
39.4
40.2
refraction
0
15.6
32.0
50.8
85.1
none*
none*
none*
none*
none*
At the air-ice boundary, Snell’s law gives the angle of refraction in the ice as
 nair sin θ1i
nice

θ1r = sin −1 

−1  ( 1.00 ) sin 30.0° 
 = sin 
 = 22.5°
1.309



Since the sides of the ice layer are parallel, the angle of incidence at the ice-water
boundary is θ 2i = θ1r = 22.5° . Then, from Snell’s law, the angle of refraction in the water
is
 nice sin θ 2i 
−1  ( 1.309 ) sin 22.5° 
 = sin 
 = 22.0°
1.333


 nwater 
θ 2r = sin −1 
22.46
(a) For polystyrene surrounded by air, total internal
reflection at the left vertical face requires that
 nair
 np

θ 3 ≥ θ c = sin −1 

 1.00 
 = sin −1 
 = 42.2°

 1.49 

From the geometry shown in the figure at the right,
θ 2 = 90.0° − θ 3 ≤ 90.0° − 42.2° = 47.8°
q1
q2
q2
q3
polystyrene
np = 1.49
q3
Thus, use of Snell’s law at the upper surface gives
sin θ1 =
np sin θ 2
nair
≤
( 1.49 ) sin 47.8°
1.00
= 1.10
so it is seen that any angle of incidence ≤ 90° at the upper surface will yield total
internal reflection at the left vertical face.
Reflection and Refraction of Light
255
(b) Repeating the steps of part (a) with the index of refraction of air replaced by that of
water yields θ 3 ≥ 63.5° , θ 2 ≤ 26.5° , sin θ1 ≤ 0.499 , and θ1 ≤ 30.0° .
(c) Total internal reflection is not possible since npolystyrene < ncarbon disulfide
22.47
Applying Snell’s law at points A, B, and C, gives
and
1.40 sin α = 1.60 sin θ1
(1)
1.20 sin β = 1.40 sin α
(2)
1.00 sin θ 2 = 1.20 sin β
(3)
n = 1.60
A
a
a
n = 1.40
B
b
n = 1.00
Combining equations (1), (2), and (3) yields
sin θ 2 = 1.60 sin θ1
q1
b
n = 1.20
C
q2
(4)
Note that equation (4) is exactly what Snell’s law would yield if the second and third
layers of this “sandwich” were ignored. This will always be true if the surfaces of all the
layers are parallel to each other.
(a) If θ1 = 30.0° , then equation (4) gives θ 2 = sin −1 ( 1.60 sin 30.0° ) = 53.1°
(b) At the critical angle of incidence on the lowest surface, θ 2 = 90.0° . Then, equation (4)
gives
sin θ 2
 1.60
θ1 = sin −1 
22.48

−1  sin 90.0° 
 = 38.7°
 = sin 
 1.60 

(a) We see the Sun swinging around a circle in the extended plane of our parallel of
latitude. Its angular speed is
ω=
∆θ
2π rad
=
= 7.27 × 10 −5 rad s
∆t 86 400 s
The direction of sunlight crossing the cell from the window changes at this rate,
moving on the opposite wall at speed
v = rω = ( 2.37 m ) ( 7.27 × 10 − 5 rad s ) = 1.72 × 10 − 4 m s = 0.172 mm s
256
CHAPTER 22
(b) The mirror folds into the cell the motion that would occur in a room twice as wide.
Thus,
v = ( 2r ) ω =  2 ( 2.37 m )  ( 7.27 × 10 −5 rad s )
= 3.45 × 10 −4 m s = 0.345 mm s
(c) and (d) As the Sun moves southward and upward at 50.0°, we may regard the
corner of the window as fixed, and both patches of light move
northward at 50.0° below the horizontal
22.49
(a) From the geometry of the figure at
the right, observe that θ1 = 60.0° .
Also, from the law of reflection,
θ 2 = θ1 = 60.0° . Therefore,
α = 90.0° − θ 2 = 30.0° , and
θ 3 + 90.0° = 180 − α − 30.0° or
θ 3 = 30.0° .
60.0°
n2
P
q1
q2
Flint Glass
nglass = 1.66
Then, since the prism is immersed in
water ( n2 = 1.333 ) , Snell’s law gives
n2
 nglass sin θ 3 
−1  ( 1.66 ) sin 30.0° 
 = sin 
 = 38.5°
1.333
n2




(b) For refraction to occur at point P, it is necessary that θ c > θ1 .
 n2
 nglass

θ c = sin −1 

 > θ1 , which gives


n2 > nglass sin θ1 = ( 1.66 ) sin 60.0° = 1.44
q3
30.0°
θ 4 = sin −1 
Thus,
a
n2
q4
257
Reflection and Refraction of Light
22.50
nglass sin θ 2 = nair sin θ1
Applying Snell’s law to this refraction gives
If θ1 = 2θ 2 , this becomes
nglass sin θ 2 = sin ( 2θ 2 ) = 2sin θ 2 cosθ 2 or cosθ 2 =
nglass
2
Then, the angle of incidence is
 nglass
 2
θ1 = 2θ 2 = 2cos −1 
22.51

−1  1.56 
 = 2cos 
 = 77.5°
 2 

In the Figure at the right, observe that
β = 90° − θ1 and α = 90° − θ1 . Thus,
β =α .
Similarly, on the right side of the
prism, δ = 90° − θ 2 and γ = 90° − θ 2 ,
giving δ = γ .
q2
q1
b
q2
A
q1
a
g
q2
q1
Next, observe that the angle between
the reflected rays is B = (α + β ) + ( γ + δ ) ,
d
B
so B = 2 (α + γ ) . Finally, observe that the
left side of the prism is sloped at angle α from the vertical, and the right side is sloped at
angle γ. Thus, the angle between the two sides is A = α + γ , and we obtain the result
B = 2 (α + γ ) = 2 A
22.52
(a) Observe in the sketch at the right that a ray originally
traveling along the inner edge will have the smallest angle
of incidence when it strikes the outer edge of the fiber
in the curve. Thus, if this ray is totally internally reflected,
all of the others are also totally reflected.
For this ray to be totally internally reflected it is necessary
that
n
1
θ ≥ θc
or
sin θ ≥ sin θ c = air =
npipe n
R−d
,
so we must have
R
which simplifies to
R ≥ nd ( n − 1)
But,
sin θ =
R−d 1
≥
R
n
d
q
R
R–d
258
CHAPTER 22
(b) As d → 0, R → 0 . This is reasonable behavior.
As n increases, Rmin =
nd
d
=
decreases. This is reasonable behavior.
n −1 1−1 n
As n → 1 , Rmin increases. This is reasonable behavior.
(c)
22.53
Rmin =
( 1.40 )( 100 µ m )
nd
=
= 350 µ m
1.40 − 1
n−1
Consider light from the bottom end of the wire that
happens to be headed up along the surface of the wire
before and after refraction with angle of incidence θ1 and
angle of refraction θ 2 = 60.0° . Then, from Snell’s law, the
angle of incidence was
q2
30.0°
Air
benzene
 nair sin θ 2 

 nbenzene 
θ1 = sin −1 
 ( 1.00 ) sin 60.0° 
= sin −1 
 = 35.3°
1.50


Thus, the wire is bent by angle θ = 60.0° − θ1 = 60.0° − 35.3° = 24.7°
q1
q
Reflection and Refraction of Light
22.54
259
f
From the sketch at the right, observe that
the angle of incidence at A is the same as
the prism angle at point O. Given that
θ = 60.0° , application of Snell’s law at
point A gives
q
C
A
q
1.50 sin β = ( 1.00 ) sin 60.0° or β = 35.3°
b
g g
O
d
90.0° – q
B
Q
From triangle AOB, we calculate the
angle of incidence and reflection, γ , at
point B:
θ + ( 90.0° − β ) + ( 90.0° − γ ) = 180°
or
γ = θ − β = 60.0° − 35.3° = 24.7°
Now, we find the angle of incidence at point C using triangle BCQ:
( 90.0° − γ ) + ( 90.0° − δ ) + ( 90.0° − θ ) = 180°
or
δ = 90.0° − (θ + γ ) = 90.0° − 84.7° = 5.26°
Finally, application of Snell’s law at point C gives
or
22.55
( 1.00 ) sin φ = ( 1.50 ) sin ( 5.26° )
φ = sin −1 ( 1.50 sin 5.26° ) = 7.91°
The path of a light ray during a reflection
and/or refraction process is always
reversible. Thus, if the emerging ray is
parallel to the incident ray, the path which the
light follows through this cylinder must be
symmetric about the center line as shown at
the right.
air
q1
d/2
d/2
A
ncylinder
R a
d/2 q1
C
b g
R
 1.00 m 
d 2
= sin −1 
Thus, θ1 = sin −1 
 = 30.0°

 R 
 2.00 m 
Triangle ABC is isosceles, so γ = α and β = 180° − α − γ = 180° − 2α . Also, β = 180° − θ1
which gives α = θ1 2 = 15.0° . Then, from applying Snell’s law at point A,
ncylinder =
nair sin θ1 ( 1.00 ) sin 30.0°
=
= 1.93
sin α
sin15.0°
B
260
CHAPTER 22
22.56
N=1
q1 = 50.0°
q2
d
N=3
x = 3d
x=d
x = 2d
0.310 cm
x = 4d
x = 5d
N=2
The angle of refraction as the light enters the left end of the slab is
 nair sin θ1 
−1  ( 1.00 ) sin 50.0° 
 = sin 
 = 31.2°
1.48


 nslab 
θ 2 = sin −1 
Observe from the figure that the first reflection occurs at x = d, the second reflection is at
th
x = 3d, the third is at x = 5d, and so forth. In general, the N reflection occurs at
x = ( 2N − 1) d where
d=
( 0.310 cm 2 ) =
tan θ 2
0.310 cm
= 0.256 cm
2 tan 31.2°
Therefore, the number of reflections made before reaching the other end of the slab at
x = L = 42 cm is found from L = ( 2N − 1) d to be
1 L

N =  + 1 =
2 d

1  42 cm

+ 1  = 82.5 or 82 complete reflections

2  0.256 cm

Reflection and Refraction of Light
22.57
b
Refer to the figure given at the right.
δ min = α + ε = (θ1 − θ 2 ) + (θ 4 − θ 3 )
q1
Note that triangles abc and bcd are
congruent. Therefore, β = γ and their
complementary angles are also equal,
or θ 2 = θ 3 .
From Snell’s law at point a,
sin θ1 = n sin θ 2
and at point d,
sin θ 4 = n sin θ 3
q2
a
a
a
q2
b
A
A —
—
2 2
c
g
d
q3
Since θ 2 = θ 3 , comparison of these results shows that θ 4 = θ1 , so δ min = 2 (θ1 − θ 2 ) .
But, θ 2 = 90° − β = 90° − ( 90° − A 2 ) , or θ 2 = A 2
A + δ min
A

Thus, δ min = 2  θ1 −  , or θ1 =
and equation (1) then yields
2
2

n=
1
sin θ1 sin  2 ( A + δ min ) 
=
sin θ 2
sin ( 12 A )
261
q4 dmin
e
(1)
262
CHAPTER 22
22.58
Horizontal light rays from the setting Sun pass
above the hiker. Those light rays that encounter
raindrops at 40.0° to 42.0° from the hiker’s shadow
are twice refracted and once reflected and reach
the hiker as the rainbow.
Sunlight
R
V
40.0°
The hiker sees a greater percentage of the violet
inner edge, so we consider the red outer edge. The
radius R of the circle of droplets is
42.0°
V
R
R = ( 8.00 km ) sin 42.0° = 5.35 km
Then the angle φ, between the vertical and the radius
where the bow touches the ground, is given by
2.00 km 2.00 km
cos φ =
=
= 0.374
R
5.35 km
or
8
0
.0
km
42°
R
R
f
φ = 68.1°
The angle filled by the visible bow is 360° − ( 2 × 68.1° ) = 224° , so the visible bow is
224°
= 62.2% of a circle
360°
2.00
km
Reflection and Refraction of Light
22.59
263
Applying Snell’s law at the first surface in the figure
at the right gives the angle of incidence as
 n sin θ 2
 nair
θ1 = sin −1 

−1
 = sin ( n sin θ 2 )

f
q1
(1)
q2
q3
Since the sum of the interior angles of a triangle equals
180°, observe that φ + ( 90° − θ 2 ) + ( 90° − θ 3 ) = 180° , which
reduces to θ 2 = φ − θ 3 . Thus, equation (1) becomes
θ1 = sin −1  n sin (φ − θ 3 )  = sin −1  n ( sin φ cosθ 3 − cos φ sin θ 3 ) 
At the smallest allowed value for θ1 , θ 3 is equal to the critical angle at the second
n
1
surface, or sin θ 3 = sin θ c = air = . Then,
n
n
cosθ 3 = 1 − sin 2 θ 3 = 1 −
and
 
 
θ1 = sin −1  n  sin φ
Note that when tan φ < 1
1
=
n2
n2 − 1
,
n

n2 − 1 1
− cos φ   = sin −1

n
n

(
n2 − 1 sin φ − cos φ
)
n2 − 1 , the value of θ1 given by this result is negative. This
simply means that, when tan φ < 1 n2 − 1 and the refracted beam is striking the second
surface of the prism at the critical angle, the incident beam at the first surface will be on
the opposite side of the normal line from what is shown in the figure.
264
CHAPTER 22
22.60
Snell’s law would predict that nair sin θ i = nwater sin θ r , or since nair = 1.00 ,
sin θ i = nwater sin θ r
Comparing this equation to the equation of a straight line, y = mx + b , shows that if
Snell’s law is valid, a graph of sin θ i versus sin θ r should yield a straight line that would
pass through the origin if extended and would have a slope equal to nwater .
θ r ( deg )
sin θ i
sin θ r
1.00
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
7.50
15.1
22.3
28.7
35.2
40.3
45.3
47.7
0.174
0.342
0.500
0.643
0.766
0.866
0.940
0.985
0.131
0.261
0.379
0.480
0.576
0.647
0.711
0.740
0.80
Sin qi
θ i ( deg )
0.60
0.40
0.20
0.00
0.00
0.20
0.40
Sin qr
0.60
0.80
The straightness of the graph line and the fact that its extension passes through the
origin demonstrates the validity of Snell’s law. Using the end points of the graph line to
calculate its slope gives the value of the index of refraction of water as
nwater = slope =
22.61
0.985 − 0.174
= 1.33
0.740 − 0.131
(a) If θ1 = 45.0° , application of Snell’s law at the
point where the beam enters the plastic block
gives
( 1.00 ) sin 45.0° = n sin φ
[1]
q1
90° – f
L
d
f
Application of Snell’s law at the point where
the beam emerges from the plastic, with θ 2 = 76.0°
gives
n sin ( 90° − φ ) = ( 1.00 ) sin 76°
or
( 1.00 ) sin 76° = n cos φ
Dividing Equation [1] by Equation [2], we obtain
tan φ =
sin 45.0°
= 0.729
sin 76°
Thus, from Equation [1],
and
n=
n
f
φ = 36.1°
sin 45.0° sin 45.0°
=
= 1.20
sin φ
sin 36.1°
q2
[2]
Reflection and Refraction of Light
265
(b) Observe from the figure above that sin φ = L d . Thus, the distance the light travels
inside the plastic is d = L sin φ , and if L = 50.0 cm = 0.500 m , the time required is
∆t =
22.62
1.20 ( 0.500 m )
d L sin φ
nL
=
=
=
= 3.40 × 10 −9 s = 3.40 ns
v
cn
c sin φ ( 3.00 × 108 m s ) sin 36.1°
B
(a) At the upper surface of the glass, the critical angle
is given by
n
B¢
Consider the critical ray PBB′ : tan θ c =
But, also, tan 2 θ c =
2
Thus,
t
qc
n
1
sin θ c = air =
nglass n
P
d¤4
d
d4 d
=
t
4t
1 n2
sin 2 θ c
sin 2 θ c
1
=
=
= 2
2
2
2
cos θ c 1 − sin θ c 1 − 1 n
n −1
1
 d
or
  = 2
n −1
 4t 
 4t 
n2 − 1 =  
 d
2
 4t 
giving n = 1 +  
 d
(b) Using the result from Part (a) and solving for d gives
Thus, if n = 1.52 and t = 0.600 cm , then
d=
d=
4 ( 0.600 cm )
2
( 1.52 ) − 1
2
4t
n2 − 1
= 2.10 cm
(c) The color at the inner edge of the white halo is associated with the wavelength
forming the smallest diameter dark spot. Considering the result from above,
d = 4t n2 − 1 , we see that this is the wavelength for which the glass has the highest
refractive index. From a dispersion graph, such as Figure 22.14 in the textbook, this
is seen to be the shorter wavelength. Thus, the inner edge will appear violet .
266
CHAPTER 22