Thursday, May 19, 2011

Fox 341

8 comments:

  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.

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

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

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

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  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?

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  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.

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  6. edit: natural numbers n >1.

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  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

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  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.

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