I have to calculate the degree of s-polarization (perpendicular) of a transmitted unpolarised light ray when it goes through a glass plate ($n=1.5$) at Brewster's angle (Brewster's window).
The setup would be something like this:
Using Fresnel's equation I have that the transmission coefficientes are $t_s=0.86$ and $t_p=1$. This can be arranged into a matrix:
$$t=\left( \begin{array}{cc} 0.86 & 0 \\ 0 & 1 \end{array} \right)$$
In this type of problems, the unpolarized incident beam can be described by:
$$E_i=E_0\left( \begin{array}{c} \cos \theta \\ \sin \theta e^{i\delta} \end{array} \right)=\frac{E_0}{\sqrt 2}\left( \begin{array}{c} 1 \\ 1 \end{array} \right) \implies E_{t}=\frac{E_0}{\sqrt 2}\left( \begin{array}{c} 0.86 \\ 1 \end{array} \right) $$
The transmited intensity is: $I=\left( \frac{0.86^2}{2}+\frac{1}{2} \right)I_0=0.87I_0$.
If $E_i$ described polarized light the degree of s-polarization would be:
$$P_s=\frac{I_s}{I}=\frac{0.86^2/2}{0.87}=0.43$$
But this assumes that the incident ray is fully polarized, which is wrong. I think that the difference $(1-0.86^2)I_0=0.26I_0$ could be the polarized component. And then, the degree of s-polarization intensity could be:
$$I_s=0.26\frac{0.86^2}{1+0.86^2}=0.11$$
But this is just a guess . The only case I've seen unpolarized light treated is for a linear polarizer, in that case the transmitted light is fully polarized.
In general, how would one calculete the polarized and unpolarized components?
Now I think it more reasonable the the polarized compontent has pure s-polarization.