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How does a Fresnel rhomb work (half and quarter wave plate)?

I am aware of birefringence, which creates a phase shift of $\Delta\phi=\dfrac{2\pi\Delta nL}{\lambda_0}$. But this doesn't explain how a plane polarised light shifts it's polarization angle after a half-waveplate or how a linearly polarised light becomes elliptical after a quarter-waveplate.

For the half one let's assume that wikipedia has an answer but why this electric field

$E(\hat{f}+i\hat{s})e^{i(kz-\omega t})$ describes an elliptical polarised light?

What I need is a mathematical description of how the retarders work...

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Have you taken a look at this in the literature? This is explained for instance in E.Hecht: Optics. Are you asking how a Fresnel rhomb works or how wave plates work? Both can be described with the equation you posted, so I don't get what you are asking for. – seb Mar 20 '13 at 12:54
There are several formalisms for describing the polarization of light. one of them is the Stokes formalism. Another one is Jones calculus. There is still more, for instance the Mueller Matrix formalism. You can transform all of them into each other. Depends on what you want to achieve. – seb Mar 20 '13 at 12:57
sorry, I could have written an answer by now. check out this wolfram demonstration. It's for a polarizer, but just exchange the phase shift and the you get the math for a wave plate. – seb Mar 20 '13 at 13:01
@sebastian: Thank you very much for your comments! What I actually need is to take an linearly electric field, make it pass through a quarter/half wave plate and see that the output is either shifted linear or circularly polarised. – Thanos Mar 20 '13 at 13:56
Ok, so I don't know how much time you want to put into this. I wrote a bachelor thesis about this topic. It includes some calculations and references. this is the link. I'd love to help you out with actual calculations, but I'm just on my mobile right now. But the general way to proceed is $E_{out} = Q^{-1} E_{in} Q$, where Q is the matrix for the retarder and E is your wave (in either formalism, Jones or Stokes, I favour Stokes), – seb Mar 20 '13 at 14:10

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