286
22.4
Chapter 22
c 3.00 × 10 8 m s
= 5.50 × 10 −7 m
=
f
5.45 × 1014 Hz
(a)
λ0 =
(b)
From Table 22.1 the index of refraction for benzene is n = 1.501. Thus, the wavelength in
benzene is
λn =
(c)
(d)
22.5
22.6
1 eV


E = hf = ( 6.63 × 10 −34 J ⋅ s ( 5.45 × 1014 Hz 
= 2.26 eV
 1.60 × 10 −19 J 
)
The speed of light in a medium with index of refraction n is v = c n, where c is its speed in
vacuum.
3.00 × 10 8 m s
= 2.25 × 10 8 m s
1.333
(a)
For water, n = 1.333, and v =
(b)
For crown glass, n = 1.52, and v =
(c)
For diamond, n = 2.419, and v =
(a)
From λ f = c, the wavelength is given by λ = c f . The energy of a photon is E = hf , so the
frequency may be expressed as f = E h, and the wavelength becomes
(b)
56157_22_ch22_p281-304.indd 286
)
The energy of the photon is proportional to the frequency, which does not change as the
light goes from one medium to another. Thus, when the photon enters benzene,
the energy does not change .
λ=
22.7
λ0 5.50 × 10 −7 m
=
= 3.67 × 10 −7 m
n
1.501
3.00 × 10 8 m s
= 1.97 × 10 8 m s
1.52
3.00 × 10 8 m s
= 1.24 × 10 8 m s
2.419
c
c
hc
=
=
f E h
E
Higher energy photons have shorter wavelengtths.
From Snell’s law, n2 sin θ 2 = n1 sin θ1. Thus, when θ1 = 45° and the first medium is air
(n1 = 1.00 ), we have sin θ 2 = (1.00 ) sin 45° n 2.
 (1.00 ) sin 45° 
θ 2 = sin −1 
 = 29°

1.458
(a)
For quartz, n2 = 1.458,
(b)
For carbon disulfide, n2 = 1.628,
(c)
For water, n2 = 1.333,
and
and
and
 (1.00 ) sin 45° 
θ 2 = sin −1 
 = 26°

1.628
 (1.00 ) sin 45° 
θ 2 = sin −1 
 = 32°

1.333
3/19/08 7:54:46 PM
Reflection and Refraction of Light
287
22.8
(a)
(b)
22.9
From geometry, 1.25 m = d sin 40.0°, so d = 1.94 m
50.0° above horizontal , or parallel to the incident ray
n1 sin θ1 = n2 sin θ 2
sin θ1 = 1.333 sin 45.0°
sin θ1 = (1.333)(0.707 ) = 0.943
θ1 = 70.5° →
22.10
19.5° above the horizontal
c 3.00 × 10 8 m s
=
= 1.38
v 2.17 × 10 8 m s
(a)
n=
(b)
From Snell’s law, n2 sin θ 2 = n1 sin θ1,
 n sin θ1 
 (1.00 ) sin 23.1° 
−1
θ 2 = sin −1  1
= sin −1 
 = sin ( 0.284 ) = 16.5°

1.38
 n2 


22.11
56157_22_ch22_p281-304.indd 287
(a)
From Snell’s law, n2 =
(b)
λ2 =
(c)
f =
(d)
v2 =
n1 sin θ1 (1.00 ) sin 30.0°
=
= 1.52
sin θ 2
sin 19.24°
λ0 632.8 nm
=
= 416 nm
n2
1.52
c
3.00 × 10 8 m s
= 4.74 × 1014 Hz in air and in syrup
=
λ0 632.8 × 10 −9 m
c 3.00 × 10 8 m s
=
= 1.97 × 10 8 m s
n2
1.52
3/19/08 7:54:47 PM
288
22.12
Chapter 22
(a)
When light refracts from air ( n1 = 1.00 ) into the
Crown glass, Snell’s law gives the angle of
refraction as
(
θ 2 = sin −1 sin 25.0° nCrown glass
)
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 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:
nCrown glass = 1.53 ; λ = 500 nm:
λ = 650 nm:
nCrown glass = 1.51
Thus, Snell’s law gives:
22.13
Figure 22.14
nCrown glass = 1.52 ;
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°
From Snell’s law,
 n sin θ1 
 (1.00 )sin 40.0° 
= sin −1 
θ 2 = sin −1  1

 = 29.4°
1.309

 n2 
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
56157_22_ch22_p281-304.indd 288
Using a protractor to measure the angle of incidence and the angle of refraction in Active
Figure 22.6b gives θ1 = 55° and θ 2 = 33°. Then, from Snell’s law, the index of refraction for the
Lucite is
n2 =
n1 sin θ1 (1.00 ) sin 55°
=
= 1.5
sin θ 2
sin 33°
(a)
v2 =
c 3.00 × 10 8 m s
=
= 2.0 × 10 8 m s
n2
1.5
(b)
f=
(c)
λ2 =
c 3.00 × 10 8 m s
=
= 4.74 × 1014 Hz
λ0 6.328 × 10 −7 m
λ0 6.328 × 10 −7 m
=
= 4.2 × 10 −7 m
n2
1.5
3/19/08 7:54:48 PM
Reflection and Refraction of Light
22.15
22.16
289
The index of refraction of zircon is n = 1.923.
c 3.00 × 10 8 m s
=
= 1.56 × 10 8 m s
n
1.923
(a)
v=
(b)
The wavelength in the zircon is λn =
(c)
The frequency is f =
λ0 632.8 nm
=
= 329.1 nm
n
1.923
c 3.00 × 108 m s
v
=
=
= 4.74 × 1014 Hz
λn λ0 632.8 × 10 −9 m
The angle of incidence is
 2 .00 m 
= 26.6°
θ1 = tan −1 
 4.00 m 
Therefore, Snell’s law gives
 n sin θ1 
 (1.333) sin 26.6° 
θ 2 = sin −1  1
= sin −1 

 = 36.6°
1.00


 n2 
and the angle the refracted ray makes with the surface is
φ = 90.0° − θ 2 = 90.0° − 36.6° = 53.4°
22.17
The incident light reaches the left-hand mirror at
distance
d 2 = (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
d = 2 ( 0.087 5 m ) = 0.175 m
It bounces between the mirrors with distance d
between points of contact with a given mirror.
Since the full 1.00 m length of the right-hand mirror is
available for reflections, the number of reflections from this mirror will be
N right =
1.00 m
= 5.71 →
0.175 m
5 full reflections
Since the first reflection from the left-hand mirror occurs at a height of d 2 = 0.087 5 m, the total
number of reflections that can occur from this mirror is
N left = 1 +
56157_22_ch22_p281-304.indd 289
1.00 m − 0.087 5 m
= 6.21 →
0.175 m
6 full reflections
3/19/08 7:54:49 PM
290
22.18
Chapter 22
(a)
From Snell’s law, the angle of refraction at the first surface is
 n sin θ1 
−1  (1.00 ) sin 30.0° 
θ 2 = sin −1  air
 = sin 
 = 19.5°
1.50
 nglass 


(b)
Since the upper and lower surfaces are parallel, the normal lines where the ray strikes these
surfaces are parallel. Hence, the angle of incidence at the lower surface will be θ 2 = 19.5° .
The angle of refraction at this surface is then
 nglass sin θ glass 
−1  (1.50 ) sin 19.5° 
θ 3 = sin −1 
 = sin 
 = 30.0°
n
1.00




air
Thus, the light emerges traveling parallel to the incident beam.
(c)
Consider the sketch above and let h represent the distance from point a to c (that is, the
hypotenuse of triangle abc). Then,
h=
2 .00 cm 2 .00 cm
=
= 2 .12 cm
cos θ 2
cos 19.5°
Also, α = θ1 − θ 2 = 30.0° − 19.5° = 10.5°, so
d = h sin α = ( 2 .12 cm ) sin 10.5° = 0.386 cm
(d)
The speed of the light in the glass is
v=
(e)
22.19
nglass
=
3.00 × 10 8 m s
= 2.00 × 10 8 m s
1.50
The time required for the light to travel through the glass is
t=
(f)
c
h
2.12 cm  1 m 
−10
=

 = 1.06 × 10 s
v 2.00 × 10 8 m s  10 2 cm 
Changing the angle of incidence will change the angle of refraction, and therefore the distance h the light travels in the glass. Thus, the travel time will also change .
From Snell’s law, the angle of incidence at the air-oil interface is
 n sin θoil 
θ = sin −1  oil

 nair

 (1.48 ) sin 20.0° 
= sin −1 
 = 30.4°
1.00

and the angle of refraction as the light enters the water is
 n sin θ oil 
 (1.48 ) sin 20.0° 
θ ′ = sin −1  oil
= sin −1 

 = 22.3°
1.333


 nwater 
56157_22_ch22_p281-304.indd 290
3/19/08 7:54:49 PM
Reflection and Refraction of Light
22.20
291
Since the light ray strikes the first surface at normal incidence, it passes into the prism without deviation. Thus,
the angle of incidence at the second surface (hypotenuse
of the triangular prism) is θ1 = 45.0°, as shown in the
sketch at the right. The angle of refraction is
θ 2 = 45.0° + 15.0° = 60.0°
and Snell’s law gives the index of refraction of the prism material as
n sin θ 2 (1.00 ) sin ( 60.0°)
n1 = 2
= 1.22
=
sin θ1
sin ( 45.0°)
22.21
∆t = ( time to travel 6.20 m in ice ) − ( time to trravel 6.20 m in air )
∆t =
6.20 m 6.20 m
−
c
vice
Since the speed of light in a medium of refractive index n is v =
c
n
 1.309 1  ( 6.20 m )( 0.309 )
= 6.39 × 10 −9 s = 6.39 ns
∆t = ( 6.20 m ) 
− =
 c
c
3.00 × 10 8 m s
22.22
n

From Snell’s law, sin θ =  medium  sin 50.0°
 nliver 
nmedium c vmedium
v
But,
=
= liver = 0.900
nliver
c vliver
vmedium
so,
θ = sin −1 ( 0.900 ) sin 50.0°  = 43.6°
From the law of reflection,
d=
22.23
(a)
6.00 cm
12 .0 cm
d
= 6.00 cm, and h =
=
= 6.30 cm
2
tanθ tan ( 43.6°)
q1
Before the container is filled, the ray’s path
is as shown in Figure (a) at the right. From
this figure, observe that
d
sin θ1 = =
s1
d
h2 + d 2
h
1
=
( h d )2 + 1
d
(a)
After the container is filled, the ray’s path is shown
in Figure (b). From this figure, we find that
sin θ 2 =
d 2
=
s2
d 2
h2 + (d 2)
2
q1
1
=
4 (h d ) + 1
2
h q2
From Snell’s law, nair sin θ1 = n sin θ 2, or
1.00
(h d )
2
+1
=
s1
q1
n
d/2
and 4 ( h d ) + 1 = n 2 ( h d ) + n 2
2
4 (h d ) + 1
2
)
2
2
Simplifying, this gives ( 4 − n ( h d ) = n − 1
2
s2
2
or
h
=
d
(b)
n2 − 1
4 − n2
continued on next page
56157_22_ch22_p281-304.indd 291
3/19/08 7:54:50 PM
292
Chapter 22
(b)
If d = 8.0 cm and n = nwater = 1.333, then
h = ( 8.0 cm )
22.24
(1.333)2 − 1
2
4 − (1.333)
= 4.7 cm
(a)
A sketch illustrating the situation and the two triangles needed in the solution is given below:
(b)
The angle of incidence at the water surface is
 90.0 m 
θ1 = tan −1 
= 42.0°
 1.00 × 10 2 m 
(c)
Snell’s law gives the angle of refraction as
n
 (1.333) sin 42 .0° 
sin θ1 
= sin −1 
θ 2 = sin −1  water
 = 63.1°


nair
1.00


(d)
The refracted beam makes angle φ = 90.0° − θ 2 = 26.9° with the horizontal.
(e)
Since tan φ = h (2.10 × 10 2 m), the height of the target is
)
h = ( 2.10 × 10 2 m tan ( 26.9°) = 107 m
22.25
As shown at the right, θ1 + β + θ 2 = 180°.
When β = 90°, this gives θ 2 = 90° − θ1
Then, from Snell’s law
sin θ1 =
ng sin θ 2
nair
= ng sin ( 90° − θ1 ) = ng cos θ1
Thus, when β = 90°,
56157_22_ch22_p281-304.indd 292
sin θ1
= tan θ1 = ng or θ1 = tan −1 ng
cos θ1
( )
3/19/08 7:54:51 PM
Reflection and Refraction of Light
22.26
293
From the drawing, observe that
 R
 1.5 m 
θ1 = tan −1   = tan −1 
= 37°
 h1 
 2 .0 m 
Applying Snell’s law to the ray shown gives
 nliquid sin θ1 
−1  1.5 sin 37° 
θ 2 = sin −1 
 = 64°
 = sin 
1.0
nair

Thus, the distance of the girl from the cistern is
d = h2 tan θ 2 = (1.2 m ) tan 64° = 2 .5 m
22.27
When the Sun is 28.0° above the horizon, the angle of
incidence for sunlight at the air-water boundary is
θ1 = 90.0° − 28.0° = 62.0°
Thus, the angle of refraction is
 n sin θ1 
θ 2 = sin −1  air

 nwater 
 (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°)
The angles of refraction for the two wavelengths are
 n sin θ1 
 1.00 0 sin 30.00° 
θ red = sin −1  air
= sin −1 
 = 18.04°


1.615
 nred 
and
 n sin θ1 
 1.00 0 sin 30.00° 
θ blue = sin −1  air
= sin −1 
 = 17.64°

1.650
 nblue 
Thus, the angle between the two refracted rays is
∆θ = θ red − θ blue = 18.04° − 17.64° = 0.40°
22.29
Using Snell’s law gives
 n sin θi 
 (1.000 )sin 83.00° 
= sin −1 
θ red = sin −1  air
 = 48.22°


1.331
 nred 
and
22.30
 n sin θi 
 (1.000 ) sin 83.00° 
θ blue = sin −1  air
= sin −1 
 = 47.79°


1.340
 nblue 
Using Snell’s law gives
 n sin θi 
 (1.00 )sin 60.0° 
θ red = sin −1  air
= sin −1 
 = 34.9°

1.512
 nred 
and
56157_22_ch22_p281-304.indd 293
 n sin θi 
 (1.00 )sin 60.0° 
θ violet = sin −1  air
= sin −1 
 = 34.5°

1.530
 nviolet 
3/19/08 7:54:52 PM
294
22.31
Chapter 22
Using Snell’s law gives
 n sin θi 
 (1.000 )sin 50.00° 
θ red = sin −1  air
= sin −1 
 = 31.77°


1.455
 nred 
and
 n sin θi 
 (1.000 )sin 50.00° 
θ violet = sin −1  air
= sin −1 
 = 31.45°

1.468
 nviolet 
Thus, the dispersion is θ red − θ violet = 31.77° − 31.45° = 0.32°
22.32
For the violet light, nglass = 1.66, and
 n sin θ1i 
θ1r = sin −1  air

 nglass 
 1.00 sin 50.0° 
= sin −1 
 = 27.5°

1.66
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 
 1.66 sin 32.5° 
= sin −1 
θ 2r = sin −1 
 = 63.2°


nair
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 2 r
22.33
(a)
violet
− θ2 r
red
= 63.2° − 58.6° = 4.6°
The angle of incidence at the first surface is
θ1i = 30°, and the angle of refraction is
 n sin θ1i 
θ1r = sin −1  air

 nglass 
 1.0 sin 30° 
= sin −1 
 = 19°

1.5
Also, α = 90° − θ1r = 71° and
β = 180° − 60° − α = 49°
Therefore, the angle of incidence at the second surface is θ 2 i = 90° − β = 41° . The angle of
refraction at this surface is
 nglass sin θ 2i 
1.5 sin 41° 
θ 2r = sin −1 
= sin −1 
= 77°


1.0 
nair


(b)
The angle of reflection at each surface equals the angle of incidence at that surface. Thus,
(θ1 )reflection = θ1i =
56157_22_ch22_p281-304.indd 294
30° , and (θ 2 )reflection = θ 2i = 41°
3/19/08 7:54:53 PM
Reflection and Refraction of Light
22.34
295
As light goes from a medium having a refractive index n1 to a medium with refractive index
n2 < n1, the critical angle is given the relation sin θ c = n2 n1. Table 22.1 gives the refractive index
for various substances at λ0 = 589 nm .
(a)
For fused quartz surrounded by air, n1 = 1.458 and n2 = 1.00, giving
θ c = sin −1 (1.00 1.458 ) = 43.3° .
22.35
22.36
(b)
In going from polystyrene (n1 = 1.49 ) to air, θ c = sin −1 (1.00 1.49 ) = 42.2° .
(c)
From sodium chloride (n1 = 1.544) to air, θ c = sin −1 (1.00 1.544 ) = 40.4° .
When light is coming from a medium of refractive index n1 into water (n2 = 1.333), the critical
angle is given by θ c = sin −1 (1.333 n1 ).
(a)
For fused quartz, n1 = 1.458 , giving θ c = sin −1 (1.333 1.458 ) = 66.1° .
(b)
In going from polystyrene (n1 = 1.49) to water, θ c = sin −1 (1.333 1.49 ) = 63.5° .
(c)
From sodium chloride ( n1 = 1.544 ) to water, θ c = sin −1 (1.333 1.544 ) = 59.7° .
Using Snell’s law, the index of refraction of the liquid is found to be
nliquid =
nair sin θi (1.00 ) sin 30.0°
= 1.33
=
sin θ r
sin 22.0°
 n 
 1.00 
= 48.5°
Thus, θ c = sin −1  air  = sin −1 
 1.33 
n
 liquid 
22.37
22.38
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
 1.00 
θ c = sin −1 
= 40.8°
 1.53 
 1.333 
= 60.6°
θ c = sin −1 
 1.53 
The critical angle for this material in air is
 n 
 1.00 
θ c = sin −1  air  = sin −1 
= 47.3°
 1.36 
 npipe 
Thus, θ r = 90.0° − θ c = 42 .7° and from Snell’s law,
 npipe sin θ r 
−1  (1.36 ) sin 42 .7° 
θi = sin −1 
 = 67.2°
 = sin 
1.00
nair

56157_22_ch22_p281-304.indd 295
3/19/08 8:32:08 PM
296
22.39
22.40
Chapter 22
The angle of incidence at each of the shorter faces of the prism is
45°, as shown in the figure at the right. For total internal
reflection to occur at these faces, it is necessary that the critical
angle be less than 45°. With the prism surrounded by air, the
critical angle is given by sin θ c = nair nprism = 1.00 nprism , so it is
necessary that
sin θ c =
1.00
< sin 45°
nprism
or
nprism >
1.00
1.00
=
=
sin 45°
2 2
(a)
The minimum angle of incidence for
which total internal reflection occurs
is the critical angle. At the critical
angle, the angle of refraction is 90°,
as shown in the figure at the right.
From Snell’s law, ng sin θi = na sin 90°,
the critical angle for the glass-air
interface is found to be
2
 n sin 90° 
−1  1.00 
θi = θ c = sin −1  a
 = sin  1.78  = 34.2°
ng


(b)
When the slab of glass has a layer
of water on top, we want the angle
of incidence at the water-air
interface to equal the critical angle
for that combination of media. At
this angle, Snell’s law gives
nw sin θ c = na sin 90° = 1.00
and
sin θ c = 1.00 nw
Now, considering the refraction at the glass-water interface, Snell’s law gives
ng sin θi = ng sin θ c. Combining this with the result for sin θ c from above, we find the
required angle of incidence in the glass to be
 n (1.00 nw ) 
 1.00 
 n sin θ c 
−1  1.00 
θi = sin −1  w
= sin −1 
= sin −1  w

 = sin  1.78  = 34.2°

ng


 ng 
 ng 
(c) and (d) Observe in the calculation of part (b) that all the physical properties of the intervening
layer (water in this case) canceled, and the result of part (b) is identical to that of part (a).
This will always be true when the upper and lower surfaces of the intervening layer are
parallel to each other. Neither the thickness nor the index of refraction of the intervening
layer affects the result.
56157_22_ch22_p281-304.indd 296
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Reflection and Refraction of Light
22.41
(a)
297
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,
 343 m s 
v 
θ c = sin −1  1  = sin −1 
= 10.7°
 1 850 m s 
 v2 
(b)
(c)
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 .
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
!2 + !2
!
=
2
2
2
Also, the hypotenuse of this triangle is
Thus,
sin θ1 =
3
!2
 d
L = !2 +   = !2 +
=!
 2
2
2
1
d 2
! 2
=
=
L
3
! 3 2
)
(
For total internal reflection at this face, it is necessary that
sin θ1 ≥ sin θ c =
22.43
nair
ncube
or
1
1.00
≥
n
3
giving
n≥ 3
If θ c = 42 .0° at the boundary between the prism glass
n
and the surrounding medium, then sin θ c = 2 gives
n1
nm
= sin 42.0°
nglass
From the geometry shown in the figure at the right,
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
 sin θ r 
 nglass sin θ r 
 sin 18.0° 
θ1 = sin −1 
= sin −1 
= sin −1 
= 27.5°


 sin 42.0° 
nm


 nm nglaass 
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298
22.44
Chapter 22
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 q, where q is greater than
or equal to the critical angle.
The critical angle at the water-air boundary is
 n 
 1.00 
θ c = sin −1  air  = sin −1 
= 48.6°
 1.333 
 nwater 
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
22.45
At the air-ice boundary, Snell’s law gives the angle of refraction in the ice as
 n sin θ1i 
 (1.00 ) sin 30.0° 
= sin −1 
θ1r = sin −1  air
 = 22.5°

1.309
 nice 
Since the sides of the ice layer are parallel, the angle of incidence at the ice-water boundary is
θ 2 i = θ1r = 22.5°. Then, from Snell’s law, the angle of refraction in the water is
 n sin θ 2 i 
 (1.309 ) sin 22.5° 
= sin −1 
θ 2 r = sin −1  ice
 = 22 .0°

1.333
 nwater 
22.46
When light coming from the surrounding medium
is incident on the surface of the glass slab, Snell’s
law gives ng sin θ r = ns sin θi , or
(
)
sin θ r = ns ng sin θi
(a)
If θi = 30.0° and the surrounding medium is
water (ns = 1.333), the angle of refraction is
 1.333 sin ( 30.0°) 
θ r = sin −1 
 = 23.7°
1.66


(b)
From Snell’s law given above, we see that as ns → ng we have sin θ r → sin θi, or
the angle or refraction approaches the angle of incidence, θ r → θi = 30.0° .
(c)
22.47
If ns > ng , then sin θ r = (ns ng )sin θi > sin θi, or θ r > θi .
From Snell’s law, ng sin θ r = ns sin θi , where ng is the refractive index of the glass and ns
is that of the surrounding medium. If ng = 1.52 (crown glass), ns = 1.333 (water), and θ r = 19.6°,
the angle of incidence must have been
 ng sin θ r 
−1  (1.52 ) sin 19.6° 
θi = sin −1 
 = sin 
 = 22.5°
1.333
n




s
From the law of reflection, the angle of reflection for any light reflecting from the glass surface as
the light is incident on the glass will be θ reflection = θ i = 22.5° .
56157_22_ch22_p281-304.indd 298
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Reflection and Refraction of Light
22.48
(a)
299
For polystyrene surrounded by air, total internal
reflection at the left vertical face requires that
n 
 1.00 
θ 3 ≥ θ c = sin −1  air  = sin −1 
= 42 .2°
 1.49 
 np 
From the geometry shown in the figure at the right,
θ 2 = 90.0° − θ 3 ≤ 90.0° − 42 .2° = 47.8°
Thus, use of Snell’s law at the upper surface gives
sin θ1 =
n p 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.
(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° .
22.49
(c)
Total internal reflection is not possible since npolystyrene < ncarbon disulfide
(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°.
Then, since the prism is immersed in
water ( n2 = 1.333), Snell’s law gives
 nglass sin θ 3 
−1  (1.66 ) sin 30.0° 
θ 4 = sin −1 
 = 38.5°
 = sin 
1.333
n2

(b)
For refraction to occur at point P, it is necessary that θ c > θ1 .
 n 
Thus, θ c = sin −1  2  > θ1 , which gives
 nglass 
n2 > nglass sin θ1 = (1.66 ) sin 60.0° = 1.44
22.50
Applying Snell’s law to this refraction gives nglass sin θ 2 = nair sin θ1
If θ1 = 2 θ 2, this becomes
nglass sin θ 2 = sin ( 2 θ 2 ) = 2 sin θ 2 cos θ 2 or cos θ 2 =
nglass
2
Then, the angle of incidence is
 nglass 
 1.56 
θ1 = 2 θ 2 = 2 cos −1 
= 2 cos −1 
= 77.5°

 2 
 2 
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3/19/08 7:54:59 PM
300
22.51
Chapter 22
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 .
Next, observe that the angle between
the reflected rays is B = (α + β ) + (γ + δ ) ,
so B = 2 (α + γ ) . Finally, observe that the
left side of the prism is sloped at angle a from the vertical, and the right side is sloped at
angle g . 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
θ ≥ θc
But, sin θ =
or
sin θ ≥ sin θ c =
R−d
,
R
nair
1
=
npipe n
so we must have
R−d 1
≥
R
n
which simplifies to R ≥ nd ( n − 1)
(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)
56157_22_ch22_p281-304.indd 300
Rmin =
(1.40 ) (100 µm )
nd
= 350 µ m
=
n −1
1.40 − 1
3/19/08 7:55:00 PM
Reflection and Refraction of Light
22.53
301
Consider light which leaves the lower end of the wire and
travels parallel to the wire while in the benzene. If the wire
appears straight to an observer looking along the dry portion
of the wire, this ray from the lower end of the wire must enter
the observers eye as he sights along the wire. Thus, the ray
must refract and travel parallel to the wire in air. The angle of
refraction is then θ 2 = 90.0° − 30.0° = 60.0°. From Snell’s law,
the angle of incidence was
 n sin θ 2 
θ1 = sin −1  air
 nbenzene 
 (1.00 ) sin 60.0° 
= sin −1 
 = 35.3°

1.50
and the wire is bent by angle θ = 60.0° − θ1 = 60.0° − 35.3° = 24.7° .
22.54
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
1.50 sin β = (1.00 ) sin 60.0°
or
β = 35.3°
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 (1.00 ) sin φ = (1.50 ) sin ( 5.26°)
or
22.55
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.
 1.00 m 
 d 2
Thus, θ1 = sin −1 
= 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 =
56157_22_ch22_p281-304.indd 301
nair sin θ1 (1.00 ) sin 30.0°
= 1.93
=
sin α
sin 15.0°
3/19/08 7:55:01 PM
302
Chapter 22
22.56
The angle of refraction as the light enters the left end of the slab is
 n sin θ1 
 (1.00 ) sin 50.0° 
= sin −1 
θ 2 = sin −1  air
 = 31.2°

1.48
 nslab 
Observe from the figure that the first reflection occurs at x = d, the second reflection is at x = 3d,
the third is at x = 5d, and so forth. In general, the Nth reflection occurs at x = ( 2 N − 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 = ( 2 N − 1) d to be
N=
22.57
(a)
1  L  1  42 cm

+ 1 = 82.5
 + 1 = 
2  d  2  0.256 cm 
or
82 complete reflections
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]
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
[2]
Dividing Equation [1] by Equation [2], we obtain
tan φ =
sin 45.0°
= 0.729 and
sin 76°
Thus, from Equation [1],
(b)
sin 45.0° sin 45.0°
=
= 1.20
sin φ
sin 36.1°
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 =
56157_22_ch22_p281-304.indd 302
n=
φ = 36.1°
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 × 10 8 m s sin 36.1°
(
)
3/19/08 7:55:02 PM
Reflection and Refraction of Light
22.58
303
Snell’s law would predict that nair sin θi = nwater sin θ r , or since nair = 1.00 ,
sin θi = 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 θi 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 .
θi ( deg ) θ r ( deg )
sin θi
sin θ r
10.0
7.50
0.174
0.131
20.0
15.1
0.342
0.261
30.0
22.3
0.500
0.379
40.0
28.7
0.643
0.480
50.0
35.2
0.766
0.576
60.0
40.3
0.866
0.647
70.0
45.3
0.940
0.711
80.0
47.7
0.985
0.740
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 =
22.59
0.985 − 0.174
= 1.33
0.740 − 0.131
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]
Combining Equations [1], [2], and [3] yields
sin θ 2 = 1.60 sin θ1
[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 
 sin 90.0° 
θ1 = sin −1 
= 38.7°
= sin −1 
 1.60 
 1.60 
56157_22_ch22_p281-304.indd 303
3/19/08 7:55:03 PM
304
Chapter 22
22.60
For the first placement, Snell’s law gives
n2 =
n1 sin 26.5°
sin 31.7°
In the second placement, application of Snell’s law yields
 n sin 26.5° 
sin 36.7°,
n3 sin 26.5° = n2 sin 36.7° =  1
 sin 31.7° 
or
n3 =
n1 sin 36.7°
sin 31.7°
Finally, using Snell’s law in the third placement gives
sin θ R =
and
56157_22_ch22_p281-304.indd 304
 sin 31.7° 
n1 sin 26.5°
= ( n1 sin 26.5°) 
= 0.392
n3
 n1 sin 36.7° 
θ R = 23.1°
3/19/08 7:55:04 PM