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Homework #6: Reflection and Refraction
1. The two mirrors in the figure below meet at a right angle. The beam of light in
the vertical plane P strikes mirror 1 as shown. (a) Determine the distance the
reflected light beam travels before striking mirror 2. (b) In what direction does
the light beam travel after being reflected from mirror 2?
2. How many times will the incident beam shown in the figure above right be
reflected by each of the parallel mirrors?
3. An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the
vertical. What is actual direction of the Sun?
4. 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.
5. 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.
6. 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?
7. A beam of light both reflects and refracts at the surface between air and glass,
as shown in the figure below. If the index of refraction of the glass is ng, find the
angle of incidence, θ1, in the air that would result in the reflected ray and the
refracted ray being perpendicular to each other. [Hint: Remember the identity
sin(90° − θ) = cos θ.]
8. For 589-nm light, calculate the critical angle for the following materials
surrounded by air: (a) diamond and (b) flint glass.
9. A beam of light is incident from air on the surface of a liquid. If the angle of
incidence is 30.0° and the angle of refraction is 22.0°, find the critical angle for the
liquid when surrounded by air.
Solutions:
1.
(a) From geometry, 1.25 m  dsin40.0 , so d  1.94 m
(b)
50.0 above horizontal , or parallel to the incident ray
2. The incident light reaches the left-hand mirror at
distance
1.00 m  tan5.00  0.087 5 m
above its bottom edge. The reflected light first
reaches the right-hand mirror at height
2 0.087 5 m   0.175 m
It bounces between the mirrors with this distance
between points of contact with either.
Since
1.00 m
 5.72 , the light reflects
0.175 m
five tim es from the right-hand m irrorand six tim es from the left
3. 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
4. (a) From Snell’s law, n2 
(b) 2 
(c)
f
(d) v2 
0
n2
c
0


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
5. 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
6. 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
7. As shown at the right, 1    2  180
When   90 , this gives 2  90  1
Then, from Snell’s law
sin1 
ng sin 2
nair
 ng sin  90  1   ng cos1
Thus, when   90 ,
sin1
 tan1  ng or 1  tan1 ng
cos1
 
8. When surrounded by air  n2  1.00 , the critical angle of a material is
 n2 
1 
1 

  sin 
 n1 
 nm aterial 
c  sin1 
 1 
(a) For diamond,  c  sin 1 
  24.4
 2.419 
 1 
(b) For flint glass,  c  sin 1 
  37.0
 1.66 
9. Using Snell’s law, the index of refraction of the liquid is found to be
nliquid 
nair sini  1.00 sin 30.0

 1.33
sin r
sin 22.0
 n 
 1.00 
Thus, c  sin1  air   sin1 
  48.5
 nliquid 
 1.33 

