Reflective Properties of a Semicircular Mirror

Date: 05/16/2000 at 10:44:33
From: Hunter DG
Subject: Reflective properties of a circle

I would like to know what happens to parallel rays that enter a 
semicircle at any angle. Does a semicircle have reflective properties 
similar to one half of the x created by the intersection of the 
equations y = x and y = -x?

         /\
        /  \
       /____\
      /|    |\
     / |    | \
       ^    V
       ^    V

Notice the ray (with the arrows) entering the two lines. Once it hits 
a point on the line y = x it is reflected as a horizontal line until 
it hits a point on the line y = -x and is then reflected back parallel 
to the incoming vertical line:

         /\
        /  \
       />>>>\  
      /^    V\
     / ^    V \
       ^    V
       ^    V
       ^    V

Will a circle have a similar property, like this? (Excuse the crude 
drawing)
         ____
       _/>>>>\_ 
      / ^    V \
     |  ^    V  |
        ^    V
        ^    V
        ^    V

Date: 05/16/2000 at 12:52:50
From: Doctor Rick
Subject: Re: Reflective properties of a circle (please answer)

Hi, Hunter.

No, the semicircle doesn't have this "retroreflective" property. You 
can use basic geometry to understand some things about reflection in a 
circle. My figure is a bit less crude than yours is, but still not 
very good -- I hope you can understand it:

             ********
          ***__  |   ***
        ** |\  \_|      **
       *   |a\a  |\__     *
      *    |  \  |   \__   *
     *     |   \a|      \__ *
     *     |    \|     __a_\*
     *     |     O____/  a /|
           |     |        / |
           |     |       / b|
           |     |      /   |
           |     |     /    |

I drew a line entering vertically (parallel to the centerline of the 
hemisphere.) It is reflected such that the angle between the reflected 
ray and the radius of the circle at the point of reflection equals the 
angle between the incoming ray and the radius. (Both angles are "a" .) 
The same happens at the second reflection, if there is one. It's easy 
to show that all the angles marked "a" are equal. The angle marked "b" 
is how far the outgoing ray is from being parallel to the incoming 
ray. You can show that:

     b = 180 degrees - 4a

For what angle a is angle b equal to 0? If you know a little 
trigonometry, you can figure out where the outgoing ray crosses the 
centerline, as a function of angle a; you will find that the rays do 
not converge at a single focal point, as they would if the reflector 
were a parabola.

You can do further explorations concerning the behavior of rays that 
aren't parallel to the centerline. Another interesting exploration is 
the 3-dimensional version of the angle reflector you drew. The corner 
of a cube has the retroreflective property; when I was your age, 
"corner reflectors" like this were placed on the moon by the Apollo 
astronauts for laser ranging experiments.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/