How Does Scattering Angle Affect Light Polarization?

[Orbital]

Homework Statement

A beam of unpolarized radiation is incident upon an electron. Show that the degree of polarization in the light scattered at an angle \theta to the incident beam is \Pi where

\Pi = \frac{1- \cos^2\theta}{1+ \cos^2 \theta}.


2. The attempt at a solution
This is a Thomson scattering and the polarization is linear so I guess Malus' law must be used, i.e.

I = I_0 \cos^2 \theta.

I'm interpreting the degree of polarization as

\Pi = \frac{I_{max} - I_{min}}{I_{max} + I_{min}}

but I cannot get the correct result.

 

[Orbital]

Is it correct that the intensity is related to the amplitude as I \propto A^2 and the incoming amplitude is related to the scattered amplitude as A' = A \cos \theta so that the scattered intensity is I_s = A^2 \cos^2 \theta and the degree of polarization becomes
\Pi = \frac{I - I_s}{I + I_s} =\frac{A^2 (1 - \cos^2 \theta)}{A^2 (1 + \cos^2 \theta)} = \frac{1 - \cos^2 \theta}{1 + \cos^2 \theta}?