Not Exactly A Fluorescence Polarizer

$tan (\theta)$ has a Cauchy distribution,

$f(x)=\cfrac{1}{\pi(1+x^2)}$

for $-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}$

then $\sqrt{|tan(\theta)|}$ has a distribution

$f(x)=\cfrac{1}{\pi(1+x^4)}|2x|$

for $-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}$  in blue below.

These set of peaks for the probability density of $\sqrt{tan(\theta)}$ are not at $\cfrac{\pi}{2}$ apart but $\cfrac{2}{\sqrt[4]{3}}$ apart.

$\cfrac{2}{\sqrt[4]{3}}=1.5197\,\, \small{rad}=87.071^o\ne90^o$

From,

$\cfrac{E_{p\,v}}{\hbar}=\left\{1-\sqrt{tan(\theta_v)}\right\}\cfrac{c}{x_v}$

The probability density of $1-\sqrt{tan(\theta_v)}$ has been right shifted by $1$.  The peak intensity of $E_p$ occurs not at $\theta=0\,\,\small{rad}$ but at $\theta=1\,\, \small{rad}=57.296^o$.

$a=\cfrac{c}{x_v}$

is assumed constant.

Note:  Also,

$\cfrac{E_{p\,v}}{\hbar}=\left\{\sqrt{tan(\theta_v)}-1\right\}\cfrac{c}{x_v}$

in which case two peaks occurs centered about $\theta=-1\,\, \small{rad}$,  $\theta=1.5197\,\, \small{rad}$ apart.  A notch occurs at $\theta=-1\,\, \small{rad}$.