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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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@Georg: My understanding (apparently as well as Steve B's) is that it rotates the light in both cases in the same direction, say right. Thus, for a still standing observer, the direction appears to change. I just wasn't sure if this was accurate. –  bitmask Nov 10 '11 at 14:29
You are right, and I need a coffee! –  Georg Nov 10 '11 at 14:33

1 Answer 1

up vote 4 down vote accepted

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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Thanks, this confirms my intuition :) –  bitmask Nov 10 '11 at 14:40

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