Answers to Even Numbered Conceptual Questions
2.
14.
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).
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.
Problem Solutions
22.1
The total distance the light travels is
⎛
⎞
∆d = 2⎜ D center to − R Earth − R M oon ⎟
⎝ center
⎠
= 2( 3.84× 108 − 6.38× 106 − 1.76× 106 ) m = 7.52× 108 m
v=
Therefore,
22.7
∆d 7.52× 108 m
=
= 3.00× 108 m s
∆t
2.51 s
n1 sin θ1 = n2 sin θ 2
q1
n1 = 1.00
sin θ1 = 1.333sin 45.0°
q2
sin θ1 = (1.333)(0.707)= 0.943
n2 = 1.333
19.5° above the horizontal
θ 1 = 70.5° →
22.11
(a)
λwater =
λ0
nw ater
, so λ0 = nwaterλwater = ( 1.333)( 438 nm ) = 584 nm
(b) λ0 = nw aterλw ater = nbenzeneλbenzene
and
nbenzene λwater 438 nm
=
=
= 1.12
nwater λbenzene 390 nm
233
CHAPTER 22
22.16
The angle of incidence is
⎡ 2.00 m ⎤
θ1 = tan −1 ⎢
⎥ = 26.6°
⎣ 4.00 m ⎦
Therefore, Snell’s law gives
q2
2.00
m
air
water
4.00 m
234
q1 q1
⎡ n1 sinθ1 ⎤
⎥
⎣ 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
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,
h=
2.00 cm
2.00 cm
=
= 2.12 cm
cosθ 2
cos19.5°
Also, α = θ 1 − θ 2 = 30.0° − 19.5° = 10.5° , so
d = hsinα = ( 2.12 cm ) sin10.5° = 0.388 cm
22.29
Using Snell’s law gives
⎛ nair sin θ i ⎞
000) sin 83.00° ⎞
−1 ⎛ ( 1.
⎟ = sin ⎜
⎟ = 48.22°
1.331
⎝
⎠
⎝ nred ⎠
θ red = sin −1 ⎜
and
⎛ nair sin θ i ⎞
000) sin 83.00° ⎞
−1 ⎛ ( 1.
⎟ = sin ⎜
⎟ = 47.79°
1.340
⎝
⎠
⎝ nblue ⎠
θ blue = sin −1 ⎜
q1
a
2.00 cm
22.19
b
q1
q2
a
c d
f
Reflection and Refraction of Light
22.32
For the violet light, nglass = 1.66 , and
235
nair = 1.00
60.0°
⎛ n sinθ1i ⎞
θ1r = sin −1 ⎜ air
⎟
⎜ nglass ⎟
⎝
⎠
q1i = 50.0°
a
⎛ 1.00sin 50.0° ⎞
= sin −1 ⎜
⎟ = 27.5°
1.66
⎝
⎠
b
q1r q2i
q2r
nglass
α = 90° − θ 1r = 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 ⎞
66sin 32.5° ⎞
−1 ⎛ 1.
⎟ = sin ⎜
⎟ = 63.2°
nair
1.00
⎝
⎠
⎝
⎠
θ 2r = sin −1 ⎜
(
)
Following the same steps for the red light nglass = 1.62 gives
θ 1r = 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
D ispersion = θ 2r violet − θ 2r red = 63.2° − 58.6° = 4.6°
22.37
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
θc = sin−1 ⎛⎜
1.00 ⎞
⎟ = 40.8°
⎝ 1.53 ⎠
θc = sin−1 ⎛⎜
1.333 ⎞
⎟ = 60.6°
⎝ 1.53 ⎠
236
CHAPTER 22
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 ⎞
00 ⎞
−1 ⎛ 1.
⎟ = sin ⎜
⎟ = 48.6°
⎝ 1.333 ⎠
⎝ nw ater ⎠
θ c = sin −1 ⎜
Thus, the minimum diameter of the raft is
2rm in = 2htanθm in = 2htanθc = 2( 2.00 m ) tan 48.6° = 4.54 m
q
h
rmin
q
q
Reflection and Refraction of Light
237