# Polarization with successive polarizers for minimal loss of energy

So, I am currenly studying polarization at a relatively basic level, and there is this qualitative question in a list of suggested exercises I am pretty sure I know the answer, but not exactly how to arrive on it. Furthermore, I think that if I am right about the answer, it would actually be quite tough to prove it mathematically.

So, the question is that given a certain number of polarizers and a polarized EM-wave source, what would be the best way to minimize energy loss to rotate the wave by a given degree.

Ok, so, I fell, intuitively, that the answer would be to turn every polarizer the least in relation to the one before it so that in the end the EM wave is rotated by the final angle.

I think that it is not obvious why the TOTAL energy loss should be minimal using that approach. What guarantees that there isn't another function of polarizer rotation in relation to how many degrees the light has already been turned that will minimize loss even further?

The proof might have something to do with showing that the average loss of the function that turns every polarizer $\theta/n$ in relation to the previous one, where $\theta$=Final Angle and $n$=total number of polarizers, is equal or smaller than the average loss of any function in the set of functions of polarizer turn angle.

The easiest way to prove this is to ask yourself what minimizes the energy loss from A to B with just one additional polarizer. That is, starting at 0° and ending at $\beta$, what is the intermediate angle $\alpha$ you need to place a polarizer to get the least total power loss.
Amplitude after the first polarizer: $$A_{\alpha} = A_0 \cos\alpha$$
Amplitude after the second: \begin{align}A_{\beta} &= A_{\alpha} \cos(\beta - \alpha) \\ &= A_0 \cos\alpha \cos(\beta-\alpha)\end{align}
Differentiating this expression with respect to $\alpha$ and setting the result to zero:
$$-\sin\alpha \cos(\beta-\alpha) - \cos\alpha \sin(\beta-\alpha) = 0\\ \tan\alpha = \tan(\beta-\alpha)\\ \alpha = \frac12 \beta$$
Just as your hunch predicted. So assuming you have $n-1$ polarizers set to the optimal angle, the last one needs to divide the remaining interval by two. But that is true no matter what the last polarizer is that you insert... so if you keep removing one polarizer and putting it back "in the middle" in the slot where you had removed it, you will eventually have all the polarizer angles evenly spaced.