Chapter 22
Reflection and
Refraction of Light
Problem Solutions
22.1
The total d istance the light travels is
d
2 Dcenter to
REarth
RMoon
center
2 3.84 108
Therefore,
22.2
d
t
v
6.38 106 1.76 106 m 7.52 108 m
7.52 108 m
2.51 s
(a) The energy of a photon is sin
sin
c
1.00
nprism
3.00 108 m s
nair nprism 1.00 nprism , w here Planck’ s constant is
c
sin 45 and the speed of light in vacuum is c 3.00 108 m s . If
1.00 10 10 m ,
6.63 10
E
(b) E
1.99 10
34
J s 3.00 108 m s
1.00 10
15
J
1 eV
1.602 10-19 J
-10
m
1.99 10
15
J
1.24 104 eV
(c) and (d )
For the X-rays to be m ore penetrating, the photons should be m ore
energetic. Since the energy of a photon is d irectly proportional to the frequency and
inversely proportional to the w avelength, the wavelength should decrease , w hich is
the sam e as saying the frequency should increase .
22.3
(a)
E
hf
6.63 10
34
J s 5.00 1017 Hz
1 eV
1.60 10 19 J
2.07 103 eV
355
356
CH APTER 22
(b) E
E
22.4
(a)
hf
c
f
34
J s 3.00 108 m s
1 nm
10 9 m
2
3.00 10 nm
6.63 10
0
6.63 10
hc
19
J
1 eV
1.60 10 19 J
3.00 108 m s
5.45 1014 Hz
6.63 10
19
J
4.14 eV
5.50 10
7
m
(b) From Table 22.1 the ind ex of refraction for benzene is n 1.501. Thus, the
w avelength in benzene is
0
n
(c)
E
hf
n
5.50 10 7 m
1.501
6.63 10
34
3.67 10
7
m
J s 5.45 1014 Hz
1 eV
1.60 10 19 J
2.26 eV
(d ) The energy of the photon is proportional to the frequency w hich d oes not change as
the light goes from one m ed ium to another. Thus, w hen the photon enters
benzene, the energy does not change.
22.5
The speed of light in a m ed ium w ith ind ex of refraction n is v c n , w here c is its
speed in vacuum .
(a) For w ater, n 1.333, and v
3.00 108 m s
1.333
(b) For crow n glass, n 1.52 , and v
(c) For d iam ond , n 2.419 , and v
22.6
2.25 108 m s
3.00 108 m s
1.52
3.00 108 m s
2.419
1.97 108 m s
1.24 108 m s
(a) From f c , the w avelength is given by
c f . The energy of a photon is E hf ,
so the frequency m ay be expressed as f E h , and the w avelength becom es
c
f
(b)
c
E h
hc
E
Higher energy photons have shorter wavelengths.
Reflection and Refraction of Light
22.7
From Snell’ s law , n2 sin
( n1 1.00 ), w e have sin
2
2
n1 sin 1 . Thus, w hen
1.00 sin 45 n2 .
(a) For quartz, n2 1.458 , and
2
(b) For carbon d isulfid e, n2 1.628 , and
(c) For w ater, n2 1.333 , and
2
sin
1
sin
45 and the first m ed ium is air
1
1.00 sin 45
1
1.458
2
sin
1
1.00 sin 45
1.00 sin 45
1.333
1.628
32
22.8
(a) From geom etry, 1.25 m d sin40.0 , so d
22.9
n1 sin
22.10
n2 sin
1
2
sin
1
1.333sin 45.0
sin
1
(1.333)(0.707) 0.943
1
(a)
1.94 m
50.0 above horizontal , or parallel to the incid ent ray
(b)
19.5 above the horizontal
70.5
n
c
v
3.00 108 m s
2.17 108 m s
1.38
29
26
357
358
CH APTER 22
(b) From Snell’ s law , n2 sin
sin
2
22.11
n1 sin
n2
1
sin
0
(c)
2
f
(d ) v2
n2
632.8 nm
1.52
1
1.00 sin 23.1
n1 sin 1
sin 2
(a) From Snell’ s law , n2
(b)
22.12
1
n1 sin 1 ,
2
1.38
sin
1.00 sin 30.0
sin19.24
1
0.284
1.52
416 nm
4.74 1014 Hz in air and in syrup
0
3.00 108 m s
632.8 10 9 m
c
n2
3.00 108 m s
1.52
1.97 108 m s
c
16.5°
(a) When light refracts from air n1 1.00 into the
Crow n glass, Snell’ s law gives the angle of
refraction as
2
sin
1
sin 25.0 nCrown glass
For first quad rant angles, the sine of the angle
increases as the angle increases. Thus, from the
above equation, note that 2 w ill increase w hen
the ind ex of refraction of the Crow n glass
d ecreases. From Figure 22.14, this m eans that
the longer w avelengths have the largest angles
of refraction, and deviate the least from the
original path.
Figure 22.14
Reflection and Refraction of Light
359
(b) From Figure 22.14, observe that the ind ex of refraction of Crow n glass for the given
w avelengths is:
and
400 nm:
nCrown glass 1.53 ;
650 nm:
nCrown glass 1.51
Thus, Snell’ s law gives:
22.13
500 nm:
nCrown glass 1.52 ;
400 nm:
sin
1
2
sin 25.0 1.53
16.0
500 nm:
sin
1
2
sin 25.0 1.52
16.1
650 nm:
sin
1
2
sin 25.0 1.51
16.3
From Snell’ s law ,
26.5
2
and from the law of reflection, sin
1
0.499
H ence, the angle betw een the reflected and refracted rays is
30.0
1
22.14
Using a protractor to m easure the angle of incid ence and the angle of refraction in
Active Figure 22.6b gives 1 55 and 2 33 . Then, from Snell’ s law , the ind ex of
refraction for the Lucite is
60.0
1
(a)
22.15
v2
3.00 108 m s
1.5
c
n2
(b)
90.0
(c)
0
2
n2
2
2.0 108 m s
30.0
6.328 10
1.5
7
m
4.2 10
The ind ex of refraction of zircon is
(a)
v
c
n
3.00 108 m s
1.923
7
m
3
30.0
1.56 108 m s
(b) The w avelength in the zircon is
nglass sin 3
1.66 sin30.0
sin 1
sin 1
4
n2
1.333
38.5
360
CH APTER 22
(c) The frequency is
22.16
c
1
The angle of incid ence is
1
tan
1
2.00 m
4.00 m
26.6
Therefore, Snell’ s law gives
n2
nglass sin
1.66 sin 60.0
1
1.44
and the angle the refracted ray m akes w ith the surface is
90.0
22.17
2
90.0
36.6
53.4
The incid ent light reaches the left-hand m irror at
d istance
1
2
2
above its bottom ed ge. The reflected light first
reaches the right-hand m irror at height
d
2 0.087 5 m
0.175 m
It bounces betw een the m irrors w ith d istance d
betw een points of contact a given m irror.
Since the full 1.00 length of the right-hand m irror is available for reflections, the num ber
of reflections from this m irror w ill be
1.00 m
0.175 m
N right
5.71
5 full reflections
Since the first reflection from the left-hand m irror occurs at a height of d 2 0.087 5 m ,
the total num ber of that can occur from this m irror is
N left
22.18
1
1.00 m 0.087 5 m
0.175 m
6.21
6 full reflections
(a) From Snell’ s law , the angle of refraction at the first surface is
90
1
Reflection and Refraction of Light
361
(b) Since the upper and low er surfaces are parallel, the norm al lines w here the ray
strikes these surfaces are parallel. H ence, the angle of incid ence at the low er surface
w ill be 2 19.5 . The angle of refraction at this surface is then
3
1
sin
nglass sin
glass
sin
nair
1
1.50 sin19.5
30.0
1.00
Thus, the light em erges traveling parallel to the incid ent beam .
(c) Consid er the sketch at the right and let h represent the d istance from point a to c
(that is, the hypotenuse of triangle abc). Then,
h
2.00 cm
cos 2
Also,
d
1
h sin
2.00 cm
cos19.5
2
30.0
2.12 cm
10.5 , so
19.5
2.12 cm sin10.5
0.386 cm
(d ) The speed of the light in the glass is
v
c
nglass
3.00 108 m s
1.50
2.00 108 m s
(e) The tim e required for the light to travel through the glass is
t
h
v
2.12 cm
1m
8
2.00 10 m s 102 cm
1.06 10
10
s
(f) Changing the angle of incid ence w ill change the angle of refraction, and therefore
the d istance h the light travels in the glass. Thus, the travel time will also change .
362
22.19
CH APTER 22
From Snell’ s law , the angle of incid ence at the air -oil interface is
sin
1
sin
1
noil sin
nair
oil
1.48 sin 20.0
30.4
1.00
and the angle of refraction as the light enters the w ater is
sin
22.20
1
noil sin
nwater
oil
sin
1.48 sin 20.0
1
22.3
1.333
Since the light ray strikes the first surface at norm al
incid ence, it passes into the prism w ithout d eviation.
Thus, the angle of incid ence at the second surface
(hypotenuse of the triangular prism ) is 1 45.0 as
show n in the sketch at the right. The angle of
refraction is
2
45.0
15.0
60.0
and Snell’ s law gives the ind ex of refraction of the prism m aterial as
n1
22.21
t
n2 sin 2
sin 1
1.00 sin 60.0
sin 45.0
time to travel 6.20 m in ice
t
6.20 m
vice
1.22
time to travel 6.20 m in air
6.20 m
c
c
n
Since the speed of light in a m ed ium of refractive ind ex n is v
t
6.20 m
1.309
c
1
c
6.20 m 0.309
8
3.00 10 m s
6.39 10
9
s
6.39 ns
Reflection and Refraction of Light
22.22
nmedium
sin 50.0
nliver
From Snell’ s law , sin
But,
nmedium
nliver
so,
c vmedium
c vliver
sin
1
363
vliver
vmedium
0.900
0.900 sin50.0
43.6
From the law of reflection,
12.0 cm
2
d
22.23
d
tan
h
6.00 cm , and
6.00 cm
tan 43.6
6.30 cm
(a) Before the container is filled , the ray’ s path is
as show n in Figure (a) at the right. From this
figure, observe that
sin
1
d
s1
d
h
1
2
d
2
hd
2
1
After the container is filled , the ray’ s path is
show n in Figure (b). From this figure, w e find
that
sin
d 2
s2
2
d 2
h
2
1
d 2
2
From Snell’ s law , 1.50sin
1.00
hd
2
1
4 hd
1.00 sin
90.0
2
2
1
1.00 sin60.0 , or
n
Sim plifying, this gives
(b) If 90.0
4 hd
and
4 hd
2
1 n2 h d
2
n2
1
90.0
90.0
1.50 sin 5.26
90.0
180 and
180
or
h
d
n nwater 1.333 , then
n2 1
4 n2
364
22.24
CH APTER 22
(a) A sketch illustrating the situation and the tw o triangles need ed in the solution is
given below :
(b) The angle of incid ence at the w ater surface is
sin
1
1.50sin5.26
7.91
(c) Snell’ s law gives the angle of refraction as
1
sin
2
nwater sin
nair
1
sin
1.333 sin 42.0
1
(d ) The refracted beam m akes angle
180
1
As show n at the right,
When
w ith the horizontal.
h (2.10 102 m) , the height of the target is
(e) Since tan
22.25
63.1
1.00
1
180
2
90 , this gives
2
90
1
Then, from Snell’ s law
sin
ng sin
1
2
nair
ng sin 90
Thus, w hen
90 ,
sin
cos
1
1
1
ng cos
tan
1
1
ng or
1
tan
1
ng
Reflection and Refraction of Light
22.26
From the d raw ing, observe that
1
tan
R
h1
1
1
tan
1.5 m
2.0 m
37
Applying Snell’ s law to the ray show n gives
2
sin
nliquid sin
1
1
sin
nair
1
1.5 sin 37
1.0
64
Thus, the d istance of the girl from the cistern is
d
22.27
h2 tan
1.2 m tan 64
2
2.5 m
When the Sun is 28.0° above the horizon, the angle of
incid ence for sunlight at the air-w ater bound ary is
1
90.0
28.0
62.0
Thus, the angle of refraction is
2
sin
1
sin
1
nair sin
nwater
1
1.00 sin 62.0
41.5
1.333
3.00 m
tan 2
The d epth of the tank is then h
22.28
3.00 m
tan 41.5
3.39 m
The angles of refraction for the tw o w avelengths are
red
and
blue
sin
sin
1
1
nair sin
nred
nair sin
nblue
1
1
sin
sin
1
1
1.00 0 sin 30.00
1.615
18.04
1.00 0 sin 30.00
1.650
17.64
Thus, the angle betw een the tw o refracted rays is
red
blue
18.04
17.64
0.40
365
366
22.29
CH APTER 22
Using Snell’ s law gives
red
and
22.30
blue
nair sin
nred
1
nair sin
nblue
1
sin
i
sin
i
(1.000)sin 83.00
1.331
1
(1.000)sin 83.00
1.340
1
sin
48.22
47.79
Using Snell’ s law gives
red
and
22.31
sin
violet
sin
nair sin
nred
1
sin
1
i
nair sin
nviolet
sin
i
(1.00)sin 60.0
1.512
1
sin
1
34.9
(1.00)sin 60.0
1.530
34.5
Using Snell’ s law gives
red
and
violet
sin
nair sin
nred
1
sin
1
i
nair sin
nviolet
Thus, the d ispersion is
red
sin
i
sin
violet
(1.000)sin 50.00
1.455
1
1
31.77
(1.000)sin 50.00
1.468
31.77
31.45
0.32
31.45
Reflection and Refraction of Light
22.32
367
For the violet light, nglass 1.66 , and
1r
sin
1
sin
1
90
and
2i
2r
nair sin
nglass
1.00sin 50.0
1.66
1r
62.5 ,
1
27.5
180.0
60.0
57.5 ,
32.5 . The final angle of refraction of the violet light is
90.0
sin
1i
nglass sin
2i
nair
sin
1
1.66sin 32.5
1.00
63.2
Follow ing the sam e steps for the red light nglass 1.62 gives
1r
28.2 ,
61.8 ,
58.2 ,
2i
31.8 , and
2r
58.6
Thus, the angular d ispersion of the em erging light is
Dispersion
22.33
2r violet
2r red
63.2
58.6
4.6
(a) The angle of incid ence at the first surface is
30 , and the angle of refraction is
1i
1r
sin
1
sin
1
Also,
90
nair sin
nglass
1i
1.0 sin 30
1.5
1r
19
71 and 1.40sin
1.60sin
1
Therefore, the angle of incid ence at the second surface is 1.20sin
angle of refraction at this surface is
1.00sin
2
1.20sin
1.40sin . The
368
CH APTER 22
(b) The angle of reflection at each surface equals the angle of incid ence at that surface.
Thus,
sin
22.34
1.60sin
2
1
, and
2 reflection
41
2i
As light goes from a m ed ium having a refractive ind ex n1 to a m ed ium w ith refractive
ind ex n2 n1 , the critical angle is given the relation sin c n2 n1 . Table 22.1 gives the
n1 sin 26.5
refractive ind ex for various substances at n2
.
sin 31.7
(a) For fu sed quartz surround ed by air, n3 sin 26.5
(b) In going from polystyrene ( sin
R
n1 sin 26.5
n3
n1 sin 26.5
0.392 ) to
sin
c
1
1.00 1.544
40.4 .
When light is com ing from a m ed ium of refractive ind ex n1 into w ater ( n2 1.333 ), the
critical angle is given by
sin 1 (1.333 n1 ) .
c
(a) For fused quartz, n1 1.458 , giving
c
sin
1
1.333 1.458
(b) In going from polystyrene ( n1 1.49 ) to w ater,
(c) From Sod ium Chlorid e ( n1 1.544 ) to w ater,
22.36
sin 31.7
n1 sin 36.7
23.1 .
R
(c) From Sod ium Chlorid e ( n1 1.544 ) to air,
22.35
sin 36.7 ,
n1 sin 36.7
.
sin 31.7
giving n3
air,
n1 sin 26.5
sin 31.7
n2 sin 36.7
c
c
sin
sin
1
1
66.1 .
1.333 1.49
1.333 1.544
Using Snell’ s law , the ind ex of refraction of the liquid is found to be
nliquid
Thus,
c
sin
nair sin
sin r
1
i
nair
nliquid
1.00 sin 30.0
sin 22.0
sin
1
1.00
1.33
1.33
48.5
63.5 .
59.7 .
Reflection and Refraction of Light
22.37
When light attem pts to cross a bound ary from one m ed ium of refractive ind ex n1 into a
new m ed ium of refractive ind ex n2 n1 , total internal reflection w ill occur if the angle of
incid ence exceed s the critical angle given by
22.38
(a) If n1 1.53 and n2
nair
(b) If n1 1.53 and n2
nwater
1.00, then
c
1.333, then
sin
c
sin 1
c
1
n2 n1 .
1.00
1.53
sin
1
1.333
1.53
40.8°
60.6°
The critical angle for this m aterial in air is
Thus,
22.39
369
1
c
sin
r
90.0
i
sin
1
nair
npipe
c
sin
1
1.00
1.36
47.3
42.7 and from Snell’ s law ,
npipe sin
r
nair
sin
1
1.36 sin 42.7
1.00
67.2
The angle of incid ence at each of the shorter faces of the prism is
45° as show n 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 surround ed by air, the
critical angle is given by sin c nair nprism 1.00 nprism , so it is
necessary that
sin
or
c
nprism
1.00
nprism
1.00
sin 45
sin 45
1.00
2 2
2
370
22.40
CH APTER 22
(a) The m inim um angle of incid ence for
w hich total internal reflection occurs
is the critical angle. At the critical
angle, the angle of refraction is 90° as
show n in the figure at the right. From
Snell’ s law , ng sin i na sin90 , the
critical angle for the glass-air interface is found to be
i
sin
c
1
na sin 90
ng
sin
1
1.00
1.78
34.2
(b) When the slab of glass has a layer
of w ater on top, w e w ant the angle
of incid ence at the w ater-air
interface to equal the critical angle
for that com bination of m ed ia. At
this angle, Snell’ s law gives
nw sin
na sin90
c
and
sin
c
1.00
1.00 nw
N ow , consid ering the refraction at the glass-w ater interface, Snell’ s law gives
ng sin i ng sin c . Com bining this w ith the result for sin c from above, w e find the
required angle of incid ence in the glass to be
i
sin
1
nw sin
ng
c
sin
1
nw 1.00 nw
ng
sin
1
1.00
ng
sin
1
1.00
1.78
34.2
(c) and (d )
Observe in the calculation of Part (b) that all the physical properties of the
intervening layer (w ater in this case) canceled , and the result of Part (b) is id entical
to that of Part (a). This w ill alw ays be true w hen the upper and low er surfaces of the
intervening layer are parallel to each other. N either the thickness nor the ind ex of
refraction of the intervening layer affects the result.
22.41
(a) Snell’ s law can be w ritten as
sin
sin
1
2
v1
. At the critical angle of incid ence
v2
the angle of refraction is 90° and Snell’ s law becom es sin
air bound ary,
c
sin
1
v1
v2
sin
1
343 m s
1850 m s
10.7
c
1
c
v1
. At the concretev2
,
Reflection and Refraction of Light
371
(b) Sound can be totally reflected only if it is initially traveling in the slow er m ed ium .
H ence, at the concrete-air bound ary, the sound m ust be traveling in air .
(c)
Sound in air falling on the wall from most directions is 100% reflected , so the w all is a
good m irror.
22.42
The sketch at the right show s 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 incid ence at this face
is the angle 1 in the triangle show n. N ote that one sid e of this
triangle is half the d iagonal of a face and is given by
2
d
2
2
2
2
Also, the hypotenuse of this triangle is
Thus, sin
1
d 2
L
2
3
L
2
d
2
2
2
3
2
2
2
1
3
2
For total internal reflection at this face, it is necessary that
sin
1
sin
c
nair
ncube
or
1
3
1.00
n
giving
n
3
372
22.43
CH APTER 22
If
42.0 at the bound ary betw een the prism glass
n2
and the surround ing m ed ium , then sin c
gives
n1
c
nm
nglass
sin 42.0
From the geom etry show n in the above figure,
90.0
and
r
22.44
48.0 ,
180
60.0
72.0
18.0 . Thus, applying Snell’ s law at the first surface gives
90.0
1
42.0
sin
1
nglass sin
r
sin
nm
1
sin r
nm nglass
sin
1
sin18.0
sin 42.0
The circular raft m ust cover the area of the surface
through w hich light from the d iam ond could em erge.
Thus, it m ust form the base of a cone (w ith apex at the
d iam ond ) w hose half angle is , w here is greater than
or equal to the critical angle.
The critical angle at the w ater-air bound ary is
c
sin
1
nair
nwater
sin
1
1.00
1.333
48.6
Thus, the m inim um d iam eter of the raft is
2 rmin
2 h tan
min
2 h tan
c
2 2.00 m tan 48.6
4.54 m
27.5
Reflection and Refraction of Light
22.45
373
At the air-ice bound ary, Snell’ s law gives the angle of refraction in the ice as
1
sin
1r
nair sin
nice
1i
sin
1.00 sin 30.0
1
22.5
1.309
Since the sid es of the ice layer are parallel, the angle of incid ence at the ice -w ater
bound ary is 2i
22.5 . Then, from Snell’ s law , the angle of refraction in the w ater
1r
is
sin
2r
22.46
1
nice sin
nwater
2i
sin
1.309 sin 22.5
1
22.0
1.333
When light, com ing from the surround ing
m ed ium is incid ent on the surface of the glass
slab, Snell’ s law gives ng sin r ns sin i , or
sin
ns ng sin
r
i
(a) If i 30.0 and the surround ing m ed ium is
w ater (ns 1.333) , the angle of refraction is
sin
r
1
1.333sin 30.0
23.7
1.66
(b) From Snell’ s law given above, w e see that as ns
ng w e have sin
the angle or refraction approaches the angle of incid ence,
(c) If ns
22.47
ng , then sin
(ns ng )sin
r
From Snell’ s law , ng sin
r
i
i
, or
r
i
i
sin i , or
30.0 .
.
ns sin i , w here n g is the refractive ind ex of the glass and ns
is that of the surround ing m ed ium . If ng
r
sin
r
r
1.52 (crow n glass), ns 1.333 (w ater), and
19.6 , the angle of incid ence m ust have been
i
sin
1
ng sin
ns
r
sin
1
1.52 sin19.6
1.333
22.5
From the law of reflection, the angle of reflection for any light reflecting from the glass
surface as the light is incid ent on the glass w ill be reflection
22.5 .
i
374
22.48
CH APTER 22
(a) For polystyrene surround ed by air, total internal
reflection at the left vertical face requires that
3
sin
c
1
nair
np
sin
1
1.00
1.49
42.2
From the geom etry show n in the figure at the right,
2
90.0
90.0
3
42.2
47.8
Thus, use of Snell’ s law at the upper surface gives
sin
n p sin
1
2
1.49 sin 47.8
nair
1.00
1.10
so it is seen that any angle of incidence 90
at the upper surface w ill yield total
internal reflection at the left vertical face.
(b) Repeating the steps of part (a) w ith the ind ex of refraction of air replaced by that of
w ater yield s 3 63.5 , 2 26.5 , sin 1 0.499 , and 1 30.0 .
(c) Total internal reflection is not possible since npolystyrene
22.49
(a) From the geom etry of the figure at
the right, observe that 1 60.0 .
Also, from the law of reflection,
60.0 . Therefore,
2
1
90.0
30.0 , and
2
90.0 180
30.0 or
3
30.0 .
3
Then, since the prism is im m ersed in
w ater n2 1.333 , Snell’ s law gives
4
sin
1
nglass sin
n2
3
sin
1
1.66 sin30.0
1.333
38.5
ncarbon disulfide
Reflection and Refraction of Light
(b) For refraction to occur at point P, it is necessary that
Thus,
n2
22.50
sin
c
nglass sin
n2
nglass
1
1
1
2
2
.
1.44
Applying Snell’ s law to this refraction gives
If
1
, w hich gives
1.66 sin 60.0
1
c
375
nglass sin
2
nair sin
1
, this becom es
nglass sin
sin 2
2
2sin
2
2
cos
2
or cos
nglass
2
2
Then, the angle of incid ence is
1
22.51
2
2
2cos
1
nglass
2
2cos
1
1.56
2
77.5
In the figure at the right, observe that
90
90
1 and
1 . Thus,
.
Sim ilarly, on the right sid e of the
prism ,
90
90
2 and
2,
giving
.
N ext, observe that the angle betw een
the reflected rays is B
so B 2
,
. Finally, observe that the
left sid e of the prism is sloped at angle from the vertical, and the right sid e is sloped at
angle . Thus, the angle betw een the tw o sid es is A
, and w e obtain the result
B 2
2A
376
22.52
CH APTER 22
(a) Observe in the sketch at the right that a ray originally
traveling along the inner ed ge w ill have the sm allest angle
of incid ence w hen it strikes the outer ed ge 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
nair
1
sin
sin c
or
c
npipe n
But,
sin
R d
,
R
w hich sim plifies to
(b) As d
0, R
(c)
22.53
R nd n 1
nd
d
d ecreases. This is reasonable behavior.
n 1 1 1n
1, Rmin increases. This is reasonable behavior.
nd
n 1
Rmin
1.40 100 m
1.40 1
350 m
Consid er light w hich leaves the low er end of the w ire and
travels parallel to the w ire w hile in the benzene. If the w ire
appears straight to an observer looking along the d ry portion
of the w ire, this ray from the low er end of the w ire m ust enter
the observers eye as he sights along the w ire. Thus, the ray
m ust refract and travel parallel to the w ire in air. The angle of
refraction is then 2 90.0 30.0 60.0 . From Snell’ s law ,
the angle of incid ence w as
1
sin
1
sin
1
1
n
0 . This is reasonable behavior.
As n increases, Rmin
As n
R d
R
so w e m ust have
nair sin 2
nbenzene
1.00 sin 60.0
1.50
and the w ire is bent by angle
35.3
60.0
1
60.0
35.3
24.7
377
Reflection and Refraction of Light
22.54
From the sketch at the right, observe that
the angle of incid ence at A is the sam e as
the prism angle at point O. Given that
60.0 , application of Snell’ s law at
point A gives
1.00 sin60.0 or
1.50sin
35.3
From triangle A OB, w e calculate the
angle of incid ence and reflection, , at
point B:
90.0
90.0
or
180
60.0
35.3
24.7
N ow , w e find the angle of incid ence at point C using triangle BCQ:
90.0
or
90.0
90.0
90.0
90.0
84.7
180
5.26
Finally, application of Snell’ s law at point C gives
sin
or
22.55
1
1.50sin5.26
1.00 sin
1.50 sin 5.26
7.91
The path of a light ray d uring a reflection
and / or refraction process is alw ays
reversible. Thus, if the em erging ray is
parallel to the incid ent ray, the path w hich the
light follow s through this cylind er m ust be
sym m etric about the center line as show n at
the right.
Thus,
1
sin
1
d 2
R
sin
1
1.00 m
2.00 m
30.0
Triangle A BC is isosceles, so
and
180
180 2 . Also,
w hich gives
1 2 15.0 . Then, from applying Snell’ s law at point A ,
ncylinder
nair sin
sin
1
1.00 sin 30.0
sin15.0
1.93
180
1
378
CH APTER 22
22.56
The angle of refraction as the light enters the left end of the slab is
2
sin
1
nair sin
nslab
1
sin
1
1.00 sin50.0
1.48
31.2
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 N th reflection occurs at
x 2 N 1 d w here
d
0.310 cm 2
tan
2
0.310 cm
2 tan 31.2
0.256 cm
Therefore, the num ber of reflections m ad e before reaching the other end of the slab at
x L 42 cm is found from L 2 N 1 d to be
N
1 L
1
2 d
1 42 cm
1
2 0.256 cm
82.5 or 82 complete reflections
Reflection and Refraction of Light
22.57
379
(a) If 1 45.0 , application of Snell’ s law at the
point w here the beam enters the plastic block
gives
1.00 sin 45.0
[1]
n sin
Application of Snell’ s law at the point w here
the beam em erges from the plastic, w ith 2 76.0
gives
n sin 90
or
1.00 sin 76
1.00 sin76
[2]
n cos
Divid ing Equation [1] by Equation [2], w e obtain
sin 45.0
sin 76
tan
0.729
and
n
Thus, from Equation [1],
36.1
sin 45.0
sin
sin 45.0
sin 36.1
1.20
(b) Observe from the figure above that sin
L d . Thus, the d istance the light travels
insid e the plastic is d L sin , and if L 50.0 cm 0.500 m , the tim e required is
t
22.58
d
v
L sin
cn
1.20 0.500 m
nL
c sin
8
3.00 10 m s sin36.1
Snell’ s law w ould pred ict that nair sin
sin
i
nwater sin
i
nwater sin
r
3.40 10
9
s
3.40 ns
, or since nair 1.00 ,
r
Com paring this equation to the equation of a straight line, y mx b , show s that if
Snell’ s law is valid , a graph of sin i versus sin r should yield a straight line that w ould
pass through the origin if extend ed and w ould have a slope equal to nwater .
i
deg
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
r
deg
7.50
15.1
22.3
28.7
35.2
40.3
45.3
47.7
sin
i
0.174
0.342
0.500
0.643
0.766
0.866
0.940
0.985
sin
r
0.131
0.261
0.379
0.480
0.576
0.647
0.711
0.740
380
CH APTER 22
The straightness of the graph line and the fact that its extension passes through the
origin d em onstrates the valid ity of Snell’ s law . Using the end points of the graph line to
calculate its slope gives the value of the ind ex of refraction of w ater as
nwater
22.59
slope
0.985 0.174
1.33
0.740 0.131
Applying Snell’ s law at points A, B, and C, gives
and
[1]
1.40sin
1.60sin
1.20sin
1.40sin
[2]
1.20sin
[3]
1.00sin
2
1
Com bining Equations [1], [2], and [3] yield s
sin
2
1.60sin
[4]
1
N ote that Equation [4] is exactly w hat Snell’ s law w ould yield if the second and third
layers of this “ sand w ich” w ere ignored . This w ill alw ays be true if the surfaces of all the
layers are parallel to each other.
(a) If
1
30.0 , then Equation [4] gives
2
sin
1
1.60sin30.0
(b) At the critical angle of incid ence on the low est surface,
[4] gives
1
sin
1
sin 2
1.60
sin
1
sin 90.0
1.60
38.7
2
53.1
90.0 . Then, Equation
Reflection and Refraction of Light
22.60
For the first placem ent, Snell’ s law gives,
n2
n1 sin 26.5
sin 31.7
In the second placem ent, application of Snell’ s law yield s
n3 sin 26.5
n2 sin 36.7
n1 sin 26.5
sin 31.7
sin 36.7 , or n3
Finally, using Snell’ s law in the third placem ent gives
sin
and
R
R
n1 sin 26.5
n3
23.1
n1 sin 26.5
sin 31.7
n1 sin 36.7
0.392
n1 sin 36.7
sin 31.7
381