This preview shows pages 1–5. Sign up to view the full content.
C
HAPTER
22
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.
207
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document CHAPTER 22
Problem Solutions
22.1
The total distance the light travels is
Therefore,
( )
866
2
23.84
10
6.38
1.76
m7
.
52
10 m
center to
Eart
hM
o
o
n
center
dD
R R
⎛⎞
∆=
−
−
⎜⎟
⎝⎠
=
×−×−×
=×
8
8
7.52
2.51 s
d
v
t
∆×
==
=
∆
8
3.00
s
×
22.2
If the wheel has 360 teeth, it turns through an angle of
1720 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
( )
5
1720 rev
5.05
s
27.5 rev s
t
θ
ω
−
=
×
Hence, the speed of light is
( )
5
2 7500 m
2
s
d
c
t
−
=
×
8
2.97
s
×
22.3
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,
()
12
rev
2 rad rev
rad
720
π
θπ
=
,
and
( )
( )
8
3
2.998
10 m s
2
22
7
2
0
2 11.45
c
td
c d
θθ
×
∆
∆∆
=
=
=
∆
×
114 rad s
208
CHAPTER 22
22.4
(a) The time for the light to travel to the stationary mirror and back is
( )
3
4
8
235.
01
0 m
2
2.33
10
s
3.00
10 m s
d
t
c
−
×
∆=
=
=
×
×
At the lowest angular speed, the octagonal mirror will have rotated
18 rev
in this
time, so
4
s
min
t
θ
ω
−
∆
==
=
∆×
536 rev s
(b) At the next higher angular speed, the mirror will have rotated
28 rev
in the
elapsed time, or
( )
2
22
5
3
6
r
e
v
s
min
ωω
=
3
1.07
10 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
. Therefore, if
c
is the speed of light,
d
t
c
, or
d
c
=
(b) If
2.
,
500 m
d
=
()
( )
11
rev
1 rev
degree
60
360 degree
60 360
⎛⎞
⎜⎟
⎝⎠
, and
5555 rev s
=
,
( )( )
360
2.500 m
5555
s1
c
8
3.000
10 m s
×
22.6
Mirror
1
Light
beam
Mirror
2
40.0°
P
1.25 m
Mirror 1
i
1
= 40°
40°
d
1.25 m
Mirror 2
40°
50°
50°
50°
i
2
= 50°
209
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentCHAPTER 22
(a) From geometry,
1.25
ms
i
n40.0
d
=
°
, so
d
=
1.94 m
(b)
50.
0 above horizontal
°
, or parallel to the incident ray.
22.7
11
2
si
ns
i
n
nn
2
θ
=
1
n1
.
333 sin 45.0
=°
1
n(
1.333)(0.707)
0.943
==
1
70.
5
→
19.
5 above the horizontal
°
θ
1
n
1
= 1.00
θ
2
n
2
= 1.333
22.8
The speed of light in water is
water
c
v
n
=
, and in Lucite
Lucite
. Thus, the total
time required to transverse the double layer is
1
Lucit
ewa
t
er
eLu
ite
dd
d
vv
+
∆=
The time to travel the same distance in air is
2
, so the additional time
required for the double layer is
(
)
12
i
te
dn
tt
−
+−
∆=∆
−∆
)(
22
8
1.0010
m1.
3331
0.50010
591
3.0010 ms
−−
This is the end of the preview. Sign up
to
access the rest of the document.
This homework help was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.
 Spring '08
 Chilton

Click to edit the document details