A perfect mirror covered the inside surface of a sphere (assumption: there is no any loss during reflection and reflections continue endless) and there is a very small laser on point $A$ in the surface of the sphere and the direction of the laser light goes to inside and reflects from other point $B$ in the surface of the sphere.
What are the whole possible mathematical conditions to get periodical reflections. (Passing again from $A$ to $B$)

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Thanks a lot for answers


Without loss of generality, we can rotate such that $A$ is at the top and the original laser beam lies in the $x$-$z$ plane. Then all the reflected beams will also lie in that plane, and the reflections will all take place on a great circle. Thus the problem is in fact two-dimensional, and we're just asking how a laser beam must be pointed in a circle such that it hits the same spot again. Since all the reflections are at the same angle, the laser beam traverses the same angle $\alpha$ between any two reflections, so the answer is that it must traverse a rational multiple of $\pi$ between any two reflections. Since the angle of incidence it forms with the surface normal on the sphere is $(\pi-\alpha)/2$, an equivalent condition is that the angle of incidence must be a rational multiple of $\pi$.

  • $\begingroup$ @Mathlover: Any three points lie on a plane. $\endgroup$ – joriki May 7 '12 at 14:19
  • $\begingroup$ Yes you are right I got it. Thanks $\endgroup$ – Mathlover May 7 '12 at 14:26
  • $\begingroup$ _@Joriki:How many reflection happens before passing from A again? How can I define the number of reflection that depends $\alpha$ Because Also can be Star shapes as well regular polygons $\endgroup$ – Mathlover May 7 '12 at 14:36
  • $\begingroup$ @Mathlover: If $\alpha=2\pi(p/q)$, with $p$ and $q$ coprime integers (i.e. $p/q$ a reduced fraction), it takes $q$ reflections to return to $A$. The special case of a polygon is attained for $p=1$. $\endgroup$ – joriki May 7 '12 at 14:44
  • $\begingroup$ Thanks a lot for your answer. Really very quick and helpful $\endgroup$ – Mathlover May 7 '12 at 15:01

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