## Thursday, May 19, 2011

1. If and only if the laser beam forms a regular polygon, right? I'm not sure if this is the answer you were looking for though. If it is, I can explain further.

2. I'm sorry, regular polygons are not the only case. There are more.

3. tan^-1(x/y) = pi

Where (x,y) is the hole. I'll explain later.

4. Alright, WLOG assume the hole (x,y) is in the top right quadrant. Then the arc length formed from the first beam (before reflection) is
S = 2r*arctan(x/y). In order for the light to escape; we must have 2*n*r*arctan(x/y) = 2*k*pi*r, where n is the number of arc lengths it takes to to sum up to k revolutions of the circle.

or, arctan(x/y) = (k/n)*pi.

where k is the number of revolutions of the circle, and n is the number of arcs S it takes to equal that revolution. So we have nS = k.

thus, arctan(x/y) = S*pi.

And finally, we get x/y = 2*pi*r.

I feel like I have made a mistake somewhere though. Can anyone see my error?

5. To trap the laser beam, the angle between the beam and the tangent plane at the hole cannot be 180deg/n for all natural numbers n.

6. edit: natural numbers n >1.

7. Let @ be the angle between the beam and the tangent plane at the hole.
The angles that allow the beam to escape are:
@=90 deg and
@=180m/n where m and n are natural numbers, n>3 and m<(n/2)

jyu

8. I believe if the angle with the tangent at the first reflection is any rational number, then the central angles, which will always all be equal, are rational numbers, say a/b. Then after 360b reflections the sum of the central angles is 360a, which is a multiple of 360, so the light escapes. I guess this means that the original angle must be irrational to trap the light.