
#1
May612, 02:33 PM

P: 261

This is from a paper that can be found at http://www.opticsinfobase.org/ao/abs...=ao26245221 (needs a university proxy/payment to view)
The setup is essentially light passing through a polarizer, a thin film sample and then an analyzer followed by a detector. The analyzer is a second polarizer that can be rotated. The paper starts with an expression for the intensity of the light from the analyzer. [itex]I=k_{0}+k_{1}cos2A+k_{2}sin2A[/itex] where [itex]k_{0}=n(cos^{2}P+tan^{2}\psi sin^{2}P)[/itex] [itex]k_{1}=n(cos^{2}Ptan^{2}\psi sin^{2}P)[/itex] [itex]k_{2}=n*tan\psi*sin2P*cos\Delta[/itex] where n is an arbitrary factor that relates to the intensity of the light, P and A are the angle of the polarizer and analyzer respectively from the with respect to the plane of incidence and the reflection coefficients of the s and p components of the light from the sample are related by [itex]\frac{r_{p}}{r_{s}}=tan\psi * e^{i\Delta}[/itex] Can anyone help me see how this intensity relation is obtained? It is just polarizers and reflections but I'm having trouble seeing how it is derived. Thank you. 


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