Degree of 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
$$\Pi = \frac{I - I_s}{I + I_s} =\frac{A^2 (1 - \cos^2 \theta)}{A^2 (1 + \cos^2 \theta)} = \frac{1 - \cos^2 \theta}{1 + \cos^2 \theta}$$?