4.3.2 In Class or Homework Exercise
1. Rays of the sun are seen to make a 31.0o angle to the vertical beneath the
water. At what angle above the horizon is the sun?
ni  1.00
nR  1.33
 R  31.0
h  ?
We must first find the angle of incidence:
ni sin i  nR sin  R
1.00sin i  1.33sin 31.0
i  43.2
 h  90.0  i
 90.0  43.2
 46.8
2. What is the index of refraction in a medium if the angle of incidence in air is
57o and the angle of refraction is 36o ?
i  57
 R  36
ni  1.00
nR  ?
ni sin i  nR sin  R
1.00sin 57  nR sin 36
nR  1.43
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3. How many more minutes would it take light from the sun to reach Earth if the
space between them were filled with water rather than a vacuum? The sun is
1.5 108 km from the Earth.
Vacuum
vv  3.00 108 m / s
d  1.5 108 km  1.5 1011 m
tv  ?
vv 
3.00 108 
d
tv
1.5 1011
tv
tv  500. s
Water
d  1.5 108 km  1.5 1011 m
nw  1.33
tw  ?
nw 
1.33 
c
vw
3.00  108
vw
vw  2.26  108 m / s
vw 
2.26 108 
d
tw
1.5 1011
tw
tw  664 s
t  t w  t v
 664  500
 164 s
 2.73min
4. Light travelling in air strikes a rectangular glass block (n=1.50) with an angle
of incidence of 60.0o .
a. What is the angle of refraction in the glass?
ni  1.00
nR  1.50
i  60.0
R  ?
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ni sin i  nR sin  R
1.00sin 60.0  1.50sin  R
 R  35.3
b. What is the angle of refraction in the air after the light emerges from the
glass block?
Using the Z pattern, we can see that the angle of incidence going from
the glass to the air is the same as the angle of refraction when going
from the air to the glass.
i  35.3
nR  1.50
ni  1.00
R  ?
ni sin i  nR sin  R
1.50sin 35.3  1.00sin  R
 R  60.0
c. What can you say about the directions of the final refracted ray and the
initial incidence ray?
Since the light came out of the block at t he same angle as it initially
went in, it must be parallel to the original incident ray.
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5. In searching the bottom of a pool at night, a watchman shines a narrow beam
of light from his flashlight (1.3 m above the water) onto the surface of the
water at a point 2.7 m from the edge of the pool. Where does the light hit the
bottom of the pool if the pool is 2.1 m deep?
First, we must find the angle of incidence using trigonometry:
2.7
1.3
i  64
tan i 
Now we can use Snell’s Law:
ni sin i  nR sin  R
1.00sin 64  1.33sin  R
 R  43
x
2.1
x
tan 43 
2.1
x  2.0m
tan  R 
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The light hits the pool 4.7 m from the edge of the pool.
6. A thick sheet of plastic (n=1.500) is used as the side of an aquarium tank.
Light reflected from a fish in the water has an angle of incidence of 35.0o . At
what angle does the light enter the air?
There are two interfaces involved here – water to plastic and plastic to air.
Water to plastic
ni  1.33
nR  1.50
i  35.0
R  ?
ni sin i  nR sin  R
1.33sin 35.0  1.50sin  R
 R  30.6
The angle of incidence for the second interface will be the same as the angle
of refraction for the first interface (Z pattern)
Plastic to air
ni  1.50
nR  1.00
i  30.6
R  ?
ni sin i  nR sin  R
1.50sin 30.6  1.00sin  R
 R  49.8
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7. What is the critical angle for the interface between
a. water and plexiglass (n=1.51)?
ni  1.51
nR  1.33
 R  90
c  ?
ni sin i  nR sin  R
1.51sin  c  1.33sin 90
c  61.7
b. water and diamond (n=2.42)?
ni  2.42
nR  1.33
 R  90
c  ?
ni sin i  nR sin  R
2.42sin c  1.33sin 90
c  33.3
8. The critical angle for a certain material in air is 41o . What is the critical angle if
the material is immersed in water?
First, we must find the index of material for the unknown material. Since air
has an index of refraction of 1.00, the index of refraction for the unknown
material must be higher than this.
nR  1.00
 R  90
 c  41
ni  ?
ni sin i  nR sin  R
ni sin 41  1.00sin 90
ni  1.52
For the second part, the unknown material has an index of refraction greater
than water so light must go into the water second for there to be a critical
angle.
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ni  1.52
nR  1.33
 R  90
c  ?
ni sin i  nR sin  R
1.52sin c  1.33sin 90
c  61.0
9. You notice that when a light ray enters a certain liquid from water, it is bent
toward the normal, but when it enters the same liquid from crown glass, it is
bent away from the normal. What can you conclude about the index of
refraction?
Since it is bent toward the normal going from water to the liquid, it must be
slowing down; therefore n  nwater . Since it is bent away from the normal going
from the crown glass to the liquid, n  ncrown glass .
nwater  nliquid  ncrown glass
1.33  nliquid  1.52
10. Could a material ever have an index of refraction less than one? Why or why
not?
No. Since light cannot go any faster in any material than it does in a vacuum,
the ratio of the speed in a vacuum to the speed in the material cannot be less
than one.
11. How might you determine the speed of light in a solid, rectangular, transparent
object?
You could use a laser to direct light into the material at an angle. The index of
refraction for the material can then be calculated using Snell’s Law. Knowing
the index of refraction, the speed of light in the solid can then be calculated
c
using n  .
v
12. You are looking diagonally down at a fish in a pond. From the fish's point of
view, does your head appear higher or lower than it really is? Why?
From the fish’s point of view, your head appears to be higher than it really is.
Since the light coming from you bends toward the normal when it hits the
water, it appears to the fish that it came from higher than it really did.
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13. When you look down into a swimming pool, are you likely to underestimate or
overestimate the depth? Explain.
You are likely to underestimate the depth (see the coin demonstration at the
beginning of this section).
14. If you crack the windshield in your car, the two pieces of glass separate at the
crack and there will be air in between them. You will see a silvery line along
the crack. Explain why.
Since there is now air between the two pieces of glass, light must now travel
from glass to air, and back to glass. Since it is going from a high index (glass)
to a low index (air), there will be a critical angle. Any light that hits this
interface at an angle greater than the critical angle will be reflected and the
interface will look shiny since it is acting as a reflecting surface.
15. If you were to use quartz and crown glass to make an optical fibre, which
would you use for the coating layer? Why?
The index of refraction for quartz is 1.46 and for crown glass is 1.52. Since the
light must be in the higher index material for total internal reflection to occur,
the crown glass must form the core with the quartz as the coating layer.
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