Download Optics HW #1: Refraction

Optics HW #1:
Refraction.
1. p. 487 of
textbook, #1
2. p. 487 of
textbook, #5
3. p. 510 of
textbook, #78
4. An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the
vertical. What is actual direction of the Sun?
5. 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)?
6. 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.
7. A ray of light is incident on the surface of a block of clear ice at an angle of 40.0° with
the normal. Part of the light is reflected and part is refracted. Find the angle between the
reflected and refracted light.
8. A flashlight on the bottom of a 4.00-m-deep swimming pool sends a ray upward and
at an angle so that the ray strikes the surface of the water 2.00 m from the point directly
above the flashlight. What angle (in air) does the emerging ray make with the water’s
surface?
9. 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:
#1 – #3: See textbook solutions online.
4. n1 sin 1  n2 sin  2
sin 1  1.333sin 45.0
sin 1  (1.333)(0.707) 0.943
1  70.5 
19.5 above the horizontal
5. 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
6. (a) From Snell’s law, n2 
(b) 2 
(c)
f
(d) v2 
0
n2
c
0


c
m
 1.333  1   0.500 10
2
3.00 108 m s
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
7. From Snell’s law,
 n1 sin1 
00 sin 40.0 
1   1.
  sin 
  29.4
1.309


 n2 
2  sin1 
and from the law of reflection,   1  40.0
Hence, the angle between the reflected and refracted rays is
  180  2    180  29.4  40.0  111
8. The angle of incidence is
 2.00 m 
  26.6
 4.00 m 
1  tan1 
Therefore, Snell’s law gives
 n1 sin1 

 n2 
 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
  90.0  2  90.0  36.6  53.4
9. 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