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Assume polarised light travels through an optically active medium (e.g. water with lots of sugar) with the specific angle $\alpha_s$ of depth $d$ at which end it is reflected (so in total it travels $2d$). Will the light at the end where it initially entered the material be polarised with a relative angle of $\alpha_sd + (-\alpha_s)d = 0$ or with $\alpha_sd + \alpha_sd = 2\alpha_sd$? I believe it is the first, because the rotation is reversed by the medium, but is that actually correct? |
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Correct, the net polarization rotation is zero. There's a neat way to see this. When you calculate the effects of a mirror, you can always do the calculation by imagining that the mirror is not a mirror, but a portal into an (imaginary) mirror-image copy of everything on the other side of the mirror. An atom-by-atom mirror image of a solution of D-glucose is a solution of L-glucose. Therefore, you imagine shining the light through a distance d of D-glucose, then through the mirror into the imaginary mirror-world, then through a distance d of L-glucose, then out the other end. Obviously the rotation of D-glucose cancels the rotation of L-glucose, so the net rotation is zero. |
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