Degree of polarization

  • Thread starter Orbital
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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 [itex]\theta[/itex] to the incident beam is [itex]\Pi[/itex] where

[tex]\Pi = \frac{1- \cos^2\theta}{1+ \cos^2 \theta}[/tex].


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.

[tex]I = I_0 \cos^2 \theta[/tex].

I'm interpreting the degree of polarization as

[tex]\Pi = \frac{I_{max} - I_{min}}{I_{max} + I_{min}}[/tex]

but I cannot get the correct result.
 

Answers and Replies

  • #2
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Is it correct that the intensity is related to the amplitude as [itex]I \propto A^2[/itex] and the incoming amplitude is related to the scattered amplitude as [itex] A' = A \cos \theta[/itex] so that the scattered intensity is [itex]I_s = A^2 \cos^2 \theta[/itex] and the degree of polarization becomes
[tex]\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}[/tex]?
 

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