Discussion Question 14A
P212, Week 14
Refraction
In this question, we will take the point of view of a fish swimming d = 4m below the surface of a
pond. A man of height h = 180 cm is standing by the side of the pond, and a shark is swimming
at the same depth as the fish. If the fish looks up at a viewing angle  = 42˚ (measured with
respect to the vertical), it sees the man's hat. The refractive index of water is 1.33.
h = 180 cm
x2
x1
d=4m

( = 42)
d=4m
(a) What is the horizontal distance x1 + x2 from the fish to the
man?
Note  is angle of incidence and  is angle of refraction
1.33sin 420  1.0sin     
x1  d tan  from small . x 2  h tan  from big 
dist = x1  x2  d tan   h tan   4 tan 420  1.8 tan   7.11 m
(b) At what viewing angle  must the fish look to see the man's
feet?
To see the feet, the light from the feet must be parallel to the
shore. This means the angle of refraction must be 900 .
1
1
n sin    sin 900  1  sin   
   
n 1.33
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h
d
h = 180 cm
x2
x1
d=4m

d=4m
( = 42)
(c) Where does the man appear to be? In fact, what does the whole sky look like to the
fish?
The man appears to be floating above the water.
The whole sky is compressed into a cone directly above the fish.
(d) The fish is astonished to see the shark reflected in the surface of the water, and apparently
floating above the pond! How far away must the shark be?
(i.e. what is the minimum horizontal distance between the fish and the shark for total internal
reflection to occur?)
d
/
The path for the fish to see the shark inverted is a reflection off of the
water surfact. The distance is L  2d sin  . The minimal distance is
the smallest angle  for total internal reflection or sin 

1
=
 48.750
n 1.33
Hence L=2(4) tan48.750  9.12 m
2
Discussion Question 14B
P212, Week 14
A Simple Movie Projector
A simple movie projector consists of a single converging lens placed 50 cm in front of the film.
The focal length of the lens is 49 cm.
film
lens (f = 49 cm)
screen
50 cm
(a) How far away from the projector lens should the screen be placed?
1 1
1
 
 s'  2450 cm
50 s' 49
24.5 m away from the lens.
(b) Should the film be oriented right-side-up or upside-down in the projector?
Did you make a ray-trace diagram in part (a)? If so, this question is a snap. 
m-
s'
2450

 0 hence image is inverted  film should be upside down
s
40
(c) The movie screen has actually been placed 30 m away from the projector lens, which
causes the picture to be out of focus. To correct this, you decide to add a second lens right
after the first one. (The lenses are so small that you can assume they are at the same
position). What focal length must your second lens have? Is it diverging or converging?
This may be a bit tricky … you know where you want the image to be, but where's the
object? Well, the object for your second lens is simply the image created by the first lens.
Now think carefully about sign conventions … (you might want to read the summary sheet)
The object for the second lens is the image due to the first lens. This image is in back
of the second lens a distance of 24.5 m. Hence it is a virtual object with s  -24.5
We want the image to be at s '  30, hence
1
1
1


 f  133.64 m
f 24.5 30
This is a diverging lens.
(d) That's a very long focal length you found in part (c) ... does that mean that your
correcting lens is strong or weak? What would an infinite focal length mean?
weak. infinite focal length indicates a plane mirror or a piece of glass (check radius of
curvature)