Suppose we have a variable laser beam attenuator consisting of a zero-order half-wave plate and a broadband polarizer. The half-wave plate has a perfect 0.5 wave retardation at 800 nm and 0.51 wave retardation at 850 nm. My question is as follows: how much the attenuation ratio will degrade at 850 nm.
You can easily compute the spurious transmission through the polarizer using Jones calculus: https://en.wikipedia.org/wiki/Jones_calculus
Using the arbitrary phase retarder matrix with \eta=1.02 \pi, I compute a transmission of 0.1%.
Guenter Steinmeyer, thank you very much for your quick response and for going forward and doing the calculation! I also found this paper: https://www.osapublishing.org/ao/abstract.cfm?uri=ao-56-4-1062
Eq. 3 in the paper seems a little bit different to me than the Wikipedia Equation for arbitrary birefringent material that you have used. It might as well be the same (I should have studied matrices and complex numbers better at school). The final result (Eq. 5) seems consistent with your 0.1% though.
Anyways, I wanted to ask a follow-up question: did you just take the square of exp(-0.5 * i * 1.02pi) in order to get the 0.1% value?
Thanks again!
You have to be a little bit careful with the Jones calculus as there are at least two different definitions around. There is a footnote on that Wikipedia page, and Hecht also comments on this in his textbook, strongly advising against mixing the two different definitions. Then there is a mysterious parameter \phi in the Wikipedia arbitrary matrix. I set this to zero because I then get the identical matrix for the half-wave plate.
What you actually have to compute is
P.M(\eta=1.02 \pi,\theta=45 Degree).|H>
The . is for matrix multiplication. P is one of the two linear polarizer matrices and |H> = (1,0). Note that you have to multiply in the reverse order of the components.
I computed (-0.0314,0) and (0,-0.9995 i) for the two polarizer orientations. And yes, you have to square this to compute the power transmission that we usually refer to.
One can further note that 0.0314 is just \pi/100. From that you can quickly develop a simple estimate for small deviations from half wave retardance.
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