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

 

[m2003]

You are correct in using Malus' law for this problem. However, the degree of polarization is defined as the ratio of the intensity of the polarized light to the total intensity of the incident light. In this case, the total intensity is given by I_0 and the polarized intensity is given by I = I_0 \cos^2 \theta. Therefore, the degree of polarization can be calculated as:

\Pi = \frac{I}{I_0} = \frac{I_0 \cos^2 \theta}{I_0} = \cos^2 \theta.

This is equivalent to the equation given in the homework statement:

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

So, the degree of polarization in the light scattered at an angle \theta is indeed \Pi as given in the homework statement.