Reflection inside spherical mirror Suppose you are inside a perfectly spherical mirror. You shoot one beam of light and it reflects on the walls of the mirror. Considering the intensity is constant will the beam of light hit you again? 

I can shoot it from anywhere inside the sphere and I am a point.

One answer mentions that the beam will always come arbitrarily close. How to prove this?

I thought of this question while sleeping. But even after a day of thinking about this problem I couldn't solve it.

Note

The only answer is brief and does not explain or prove what is says. Looking for detailed answers preferably using illustration and valid mathematical proofs for the statements they make. Would like to read about possible generalizations and variations of this question. Please don;t restrict yourself to this specific situation and do not hesitate to write long answers which give a wide overview of these type problems. Bounty-awarded answer is likely to be long, detailed and self-sufficient mathematically in terms of proving the facts it states.
By the laws of optics, the reflections will occur in the plane formed by the initial ray and the center of the sphere, and this is a planar problem.

Now consider the circle where you are standing and the successive intersection points of the reflected rays with it.



By symmetry, these points will be located at angles that follow an arithmetic progression, and the angular delta from the initial point is $n\alpha\bmod1=\{n\alpha\}$, where the angles are expressed in turns. Now it is clear that the ray will come back iff $\alpha$ is a rational number. If it is irrational it will never come back exactly but you can always find an $n&gt;0$ such that $\{n\alpha\}&lt;\epsilon$ for any $\epsilon$.

The situation is similar for the second intersection points with the ray, at angles $n\alpha+\beta$, where $\beta$ is independent of $\alpha$*. But then, it may turn out that $n\alpha+\beta$ is a integer by coincidence so that even with irrational $\alpha$ there can be an exact return (only one because $n'\alpha+\beta$ cannot be another integer).

To summarize:

The return angles are $\{n\alpha\}$ and $\{n\alpha+\beta\}$, corresponding to


irrational $\alpha\to$ zero or one exact return plus infinitely many close ones, or
rational $\alpha\to$ infinitely many exact periodic returns but no close ones.




*As said somewhere else, the figure has two degrees of freedom: the (relative) distance to the center and the (relative) starting ray direction.
As has been mentioned, one should consider the plane formed by the source of the light, the first contact point, and the center of the sphere. The light beam will be confined to this plane, because the vector orthogonal to this plane will be tangent to the sphere at every point of contact. Now the problem is reduced to a similar argument, but considering a circle intead of a sphere.

Using Snell's Law (as Oscar Lanzi has pointed out), the length of each chord formed by the light beam will be the same. Now consider the angle 'a' around the circle formed by one of the chords. We will define the initial chord along which the beam was fired to be the 0'th chord, the next one to be the 1st chord, etc.

There are now two possibilities: 

1) If 'a' is some rational multiple of 2*pi, say (p/q)*2pi (assume gcd(p,q) = 1), then we are guaranteed that the (n+q)'th chord will simply be the n'th chord rotated around the center by an angle of p*2pi. Since p is an integer, the (n+q)'th chord will therefore be identical to the n'th chord, meaning the 0'th chord will be identical to the q'th chord. Therefore, intersection will occur every q*b reflections for every non-negative integer b.

2) If 'a' is an irrational multiple of 2pi, let m[0] be the least positive integer such that:

m[0]*a mod 2pi &lt; a.

We are guaranteed that such an m[0] exists, indeed m[0] will simply be the least positive integer such that m[0]*a > 2pi. Notice this implies that the angle between the m[0]'th chord and the 0'th chord is less than 'a'. 

We can simply repeat this process by defining m[n+1] to be the least positive integer multiple of m[n] such that:

m[n+1]*a mod 2pi &lt; m[n]*a mod 2pi

The geometrical interpretation of this is that the angle between the m[n+1]'th chord and the 0'th chord is less than the angle between the m[n]'th chord and the 0'th chord (for every n > 0). We can prove the existence of such an m[n+1] by letting j be the least positive integer such that:

j*(m[n]*a mod 2pi) > 2pi;

m[n+1] is simply j*m[n].

Define k[n] to be the angle between the m[n]'th chord and the 0'th chord, where k[0] is let to be 'a'. It is simple to see that the sequence k[n] is positive and decreasing. This means the sequence is convergent; in other words:

As n -> infinity , k[n+1]-k[n] -> 0.

So, for every real number e > 0 , there exists an n such that k[n+1]-k[n] &lt; e.

Since the m[n+1]'th chord is at an angle k[n+1] from the 0'th chord, and the m[n]'th chord is at an angle k[n] from the 0'th chord, this implies that as n goes to infinity, the angle between chord m[n+1] and chord m[n] approaches 0.
Notice that for any positive integers b,c, and an integer d (not greater than min(b,c)) the angle between chord b and chord c is identical to the angle between chord (b-d) and chord (c-d). This implies that the angle between chord m[n+1] and chord m[n] is identical to the angle between chord (m[n+1]-m[n]) and chord 0.

As stated, for any real e>0, there exists an n such that the angle between the m[n+1]'th chord and the m[n]'th chord is less than e, implying the angle betweeen chord (m[n+1]-m[n]) and chord 0 is less than e. This guarantees arbitrary precision.

It could also be the case that one chooses wisely the initial firing point in a way that 'a' is an irrational multiple of 2pi, but the initial firing point is met anyway; consider, for instance, the case where one fires the beam of light from the intersection of chord 0 and chord m[0]. In this case, the beam of light will clearly pass through the firing point, but will only do so once, since a given point on a given chord is shared only with exactly one other chord (of the same length), and for 'a' being an irrational multiple of 2pi, we are guaranteed that no chord is ever repeated.

Sorry for the lack of fancy formatting, I'm new here and haven't had a chance to learn how to do that properly.

I realize this is a bit of a tome, so feel free to ask me about any uncertainties that might arise!