# Hyperpolarization

How many polarized lenses make a quarter turn of polarized light with the most intensity?

For the first polarizer offset at $\theta_1$, the resulting intensity would be $t\cos\theta_1$, where $t=0.99$ is the transmittance. So with $n$ polarizers we would have a final intensity of \begin{gather} I = \prod_i^n t \cos\theta_i \\ \text{with } \sum_i^n\theta_i = \frac\pi2 \end{gather}
Of course, if the polarizers had perfect transmittance for polarized light, i.e. $t=1$ then we would just want infinitely many lenses making small turns. Now, $\cos\theta$ is not a linear function, and even for small angles $$\cos\theta \approx 1 - \frac{\theta^2}2 + \frac{\theta^4}{24}$$ but for $t=0.99$, there definitely will not be a win for varying $\theta_i$. Taking $$\theta = \theta_i = \theta_j \Longrightarrow \theta = \frac\pi{2n}~.$$ Then we have simply $$I = t^n \cos^n\theta = t^n\cos^n\left(\frac\pi{2n}\right) ~.$$
For $t=0.99$, this is maximized for $n=11$, outputting 80.042% of the original intensity.