Cross Polarizers with a wave plate retardation

jcbale1
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Homework Statement


"Consider a wave plate with a retardation τ=(η_0-n_e)*w*L/c. Show that when it is placed between crossed polarizers with its optical axis at an angle of θ with respect to the polarizer axis, the transmitted intensity is given by:

I_out/I_in= (sin^2(2θ)(sin^2(τ/2))

In practice, one often sets the optical axis at θ=45° to obtain maximum contrast.


Homework Equations




E=P_θ*P_0*E_0


I=E times E*


The Attempt at a Solution



I know that without any retardation I/I_0=cos^2(θ). I'm really not sure how to approach this problem
 
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Let's start by rewriting your given equations:

ψ = (n_o - n_e)Lω/c

since ψ is a radian angle so τ is not a good symbol since it usually has dimensions of time.

Then Io/Ii = sin2(2θ)sin2(ψ/2)

So you have one sinusoidally time-varying electric vector field Eo that passes thru the plate and then thru the polarizer and analyzer, and a second field Ee that does the same but is phase-shifted by ψ radians w/r/t Eo and is perpendicular to it.

Then at the output of the analyzer, add the two fields vectorially, then square and time-average to get the intensity, & off you go.
 
Thank you very much, that helps a lot!
 
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