Answer 1—The front-surface reflection—I’ll call this the in-phase reference reflection—has intensity K, and the light inside the mirror for the first internal reflection has intensity 1 – K.

The light that comes out after one pass through the mirror is out-of-phase and has an intensity of (1 – K)². Light with intensity (1 – K) × K makes another bounce inside the mirror.

The light that comes out after the second internal bounce is in-phase with the reference reflection, so it adds (1 – K)² × K to its intensity. The light after the third bounce [(1 – K)² × K²)] is out-of-phase again, and so on. You end up with two series, one describing the total in-phase reflection and another describing the total out-of-phase reflection.

In order to get total cancellation, we need to make these two sums equal to each other and solve for K.

The two summations can be combined and a closed-form equivalent can be found, which works out to –1/(1 + K). The equation reduces to: 

This is readily solved for K = 1/3.

Bonus question: Assuming that nothing in the system actually absorbs light, and given that there’s no net reflection, what happens to the incident light energy? I don’t know the answer to this, so I’d like to hear your thoughts on the matter.

Contributor: David Tweed