Polarized Express ... er, Classic

I have a light source that’s polarized in the vertical direction, but I want it to be polarized in the horizontal direction. To make that happen I need … polarizers! When light passes through a polarizer, only the light whose polarization aligns with the polarizer passes through. When they aren’t perfectly aligned, only the component of the light that’s in the direction of the polarizer passes through.

Unfortunately, that means I can’t turn vertically polarized light into horizontally polarized light with a single polarizer. But I can do it with two polarizers, as shown below.

A vector is vertical. Its component along a 45 degree angle is shown. That second vector's component along another 45 degree angle is also shown. The final vector has half the length of the original.

If the first polarizer is positioned at an acute angle with respect to the light, and then the second polarizer is positioned at a complementary angle with respect to the first polarizer, some light will be horizontally polarized in the end.

Now, I have tons of polarizers, and each one also reflects 1 percent of any light that hits it — no matter its polarization or orientation — while polarizing the remaining 99 percent of the light.

I’m interested in horizontally polarizing as much of the incoming light as possible. How many polarizers should I use?

Let's say that we have $N$ polarizers which are placed with polarizing angle $0 \lt \theta_1 \lt \theta_2 \leq \cdots \leq \theta_N = \frac{\pi}{2}$ (with respect to the vertical initial angle). Given the gain of $g = 0.01$ that is reflected at each polarizer, the ultimate amount of horizontally polarized light will be $$I_N(\theta_1, \dots, \theta_{N-1}, \theta_N \mid g) = (1-g)^N \cos \theta_1 \prod_{i=2}^N \cos (\theta_i - \theta_{i-1})$$

Either by happenstance or by base case, we will first show that if you only use two polarizers that the optimal positioning is to place one with angle $\theta_1^* = \frac{\pi}{4}.$ In this case, $$I_2(\theta_1, \theta_2 \mid g) = (1-g)^2 \cos \theta_1 \cos (\theta_2 - \theta_1),$$ which has derivative $$\frac{\partial}{\partial\theta_1} I_2(\theta_1, \theta_2 \mid g) = (1-g)^2 \left( -\sin \theta_1 \cos (\theta_2 - \theta_1) + \cos \theta_1 \sin (\theta_2 - \theta_1) \right),$$ or equivalently, $\tan \theta_1 = \tan (\theta_2 - \theta_1).$ Since $y = \tan x$ is a monotonically increasing function on $(-\pi/2, \pi/2)$ and has a period of $\pi,$ then we have $\tan \theta = \tan \phi$ if and only if $\theta - \phi \in \pi \mathbb{Z}.$ Thus, since we have $\theta_1, \theta_2 - \theta_1 \in [0, \pi / 2)$ if $\tan \theta_1 = \tan (\theta_2 - \theta_1)$ then we must have $\theta_1 = \theta_2 - \theta_1,$ or equivalently $\theta_1 = \frac{\theta_2}{2}.$ Note that when $\theta_2 = \pi/2,$ then $I_2^*(g) = \frac{(1-g)^2}{2}.$

Assume that for some $N-1$ and any $\theta_{N-1} \in (0, \pi/2),$ the optimally spaced polarizers have polarizing angles $\theta_i = \frac{i \theta_{N-1}}{(N-1)},$ for $i = 1, \dots, N-1.$ In this case, we would then have $$I_{N-1}^*(g) = \left( (1-g) \cos \frac{\pi}{2(N-1)} \right)^{N-1}.$$

Let's choose any arbitrary $\theta_1 \in (0,\pi/2).$ From our induction step, we would then place the remaining $N-1$ polarizers with angles equidistantly spaced out between $\theta_1$ and $\pi/2,$ that is $$\theta_i = \theta_1 + (i-1) \frac{\frac{\pi}{2} - \theta_1}{N-1},$$ for $i = 2, \dots, N.$ In this case, we have $$I_N(\theta_1 \mid g) = I_N(\theta_1, \theta_2, \dots, \theta_N \mid g) = (1-g)^N \cos \theta_1 \cos^{(N-1)} \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right).$$ In this case, we see that the derivative is \begin{align*}\frac{d}{d\theta_1} I_N &= (1-g)^N \left( -\sin \theta_1 \cos^{(N-1)} \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right) + \cos \theta_1 \sin \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right) \cos^{(N-2)} \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right) \right) \\ &= (1-g)^N \cos^{(N-2)} \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right) \left( -\sin \theta_1 \cos \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right) + \cos \theta_1 \sin \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right) \right).\end{align*} Thus the derivative will be equal to zero if either $\theta_1 = \frac{\pi}{2}$ or $$\tan \theta_1 = \tan \left( \frac{\frac{\pi}{2} - \theta_1}{N-1} \right)$$ or equivalently $$\theta_1 = \frac{\frac{\pi}{2} - \theta_1}{N-1} \Rightarrow \theta_1 = \frac{\pi}{2N}.$$ In this case, we have again $\theta_i = \frac{i \pi}{2N}$ for $i = 1, \dots, N$ and $$I_N^*(g) = \left( (1-g) \cos \frac{\pi}{2N} \right)^N.$$

Therefore, by induction, for any number of polarizers $N,$ the optimal amount of horizontally polarized light is $$I_N^*(g) = \left( (1-g) \cos \frac{\pi}{2N} \right)$$ for any $N = 2, 3, \dots.$ The partial derivative with respect to $N$ is given by $$\frac{\partial}{\partial N} I_N^*(g) = I_N^*(g) \left( \ln(1-g) + \ln \cos \frac{\pi}{2N} + \frac{\pi}{2N} \tan \frac{\pi}{2N} \right).$$ While solving this $\frac{\partial}{\partial N} I_N^*(g) = 0$ analytically would be a pain, we see that $$\frac{\partial^2}{\partial N^2} \ln I_N^*(g) = - \frac{\pi^2}{4N^3} \sec^2 \frac{\pi}{2N} \leq 0,$$ so the logarithmic derivative is monotonically decreasing, and there is a unique global maximum in $N$ of $I_N^*(g).$ If we then start calculating $I_N^*(g)$ for $N = 2, 3, \dots$ until we see the sequence start to decrease, we can then stop with our search. Luckily, it does not take that long and see from the table below that the optimal solution occurs with $N = 11$ polarizers with polarization angles $\theta_i = \frac{i \pi}{22}$ for $i = 1, \dots, 11,$ as measured from the positive $y$-axis.

\[\begin{array}{|c|c|}\hline N & I_N^*(0.01) \\\hline 2 & 0.4901 \\\hline 3 & 0.6302 \\\hline 4 & 0.6998 \\\hline 5 & 0.7400 \\\hline 6 & 0.7647 \\\hline 7 & 0.7803 \\\hline 8 & 0.7901 \\\hline 9 & 0.7959 \\\hline 10 & 0.7990 \\\hline 11 & 0.8000 \\\hline 12 & 0.7995 \\\hline\end{array}\]