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Rotation and Polarisation of Light using Jones Matrices
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[QUOTE="Steve4Physics, post: 6463992, member: 681522"] Edits made. Assuming your final equation is correct, you've done the hard part. You just hit a 'blind spot' for the final simple steps: $$\sin2\theta(B^2-A^2) = -2AB\cos2\theta\cos\delta$$$$\tan2\theta=\frac {2AB\cos\delta}{(A^2-B^2) }$$$$\theta=\frac 1 2 \left[ \tan^{-1}\frac {2AB\cos\delta}{(A^2-B^2) } \right]$$Being a bit picky, I'll also note that the question describes ##\vec J## as a 'Jones matrix, but this should be 'Jones vector'. [/QUOTE]
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Rotation and Polarisation of Light using Jones Matrices
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