Download Refraction Practice Problems

Refraction Practice Problems
7.
An underwater scuba diver sees the Sun at an apparent angle of 45.0°
from the vertical. What is actual direction of the Sun?
8.
Light is incident normal to a 1.00-cm layer of water that lies on top of a flat
®
Lucite plate with a thickness of 0.500 cm. How much more time is required for
light to pass through this double layer than is required to traverse the same
distance in air (nLucite = 1.59)?
9. A laser beam is incident at an angle of 30.0° to the vertical onto a solution of
corn syrup in water. If the beam is refracted to 19.24° to the vertical, (a) what is
the index of refraction of the syrup solution? Suppose the light is red, with
wavelength 632.8 nm in a vacuum. Find its (b) wavelength, (c) frequency, and (d)
speed in the solution. Hint: The frequency stays constant when light travels
from one material to another.
21.
The light beam shown in Figure P22.21 makes an angle of 20.0° with the
normal line NN’ in the linseed oil. Determine the angles θ and θ’. (The refractive
index for linseed oil is 1.48.)
Figure P22.21
22.
A submarine is 300 m horizontally out from the shore and 100 m beneath
the surface of the water. A laser beam is sent from the sub so that it strikes the
surface of the water at a point 210 m from the shore. If the beam just strikes the
top of a building standing directly at the water’s edge, find the height of the
building.
Solutions:
22.7
n1 sin 1  n2 sin  2
sin 1  1.333sin 45.0
sin 1  (1.333)(0.707) 0.943
1  70.5 
22.8
19.5 above the horizontal
The speed of light in water is vw ater 
c
, and in Lucite® vLucite 
nw ater
total time required to transverse the double layer is
t1 
c
nLucite
. Thus, the
dwater dLucite dwaternwater  dLucitenLucite


vwater vLucite
c
The time to travel the same distance in air is t2 
dwater  dLucite
, so the additional
c
time required for the double layer is
t t1  t2 
1.00 10

dw ater  nw ater  1  dLucite  nLucite  1
2
22.9
c
m
 1.333  1   0.500 10
(a) From Snell’s law, n2 
(b) 2 
(c)
f
(d) v2 
0
n2
c
0


2
3.00 10 m s
8
m
 1.59  1 
n1 sin1  1.00 sin 30.0

 1.52
sin 2
sin19.24
632.8 nm
 417 nm
1.52
3.00 108 m s
 4.74 1014 H z in air and in syrup
632.8 109 m
c 3.00 108 m s

 1.98 108 m s
n2
1.52
2.09 1011 s
22.21
From Snell’s law, the angle of incidence at the air-oil interface is
 noil sin oil 

 nair 
  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
 noilsinoil 
48 sin20.0 
1   1.
  sin 
  22.3
1.333


 nwater 
   sin1 
22.22
The angle of incidence at the water
surface is
1  tan 1 
90.0 m 
  42.0
 100 m 
Then, Snell’s law gives the angle of
refraction as
 nwater sin1 
333 sin 42.0 
1   1.
  sin 
  63.1
nair
1.00




2  sin1 
so the height of the building is h 
210 m
210 m

 107 m
tan 2 tan63.1