4
$\begingroup$

This is a followup question of (Dependence of the angle of rotation on the wavelength of plane polarized light).

Ron's answer tells about the difference between left and right circularly polarized light. But the explanation isn't in the answer.

So, what's the reason behind different index of refraction?

$\endgroup$
4
$\begingroup$

The answer to this is really a combination of the two answers given in the question you linked.

Namely, chiral molecules are unique in that they make a distinction between left and right circularly polarized light.

As porphyrin pointed out, the index of refraction is dependent on the ability of a dipole to be induced in a molecule by electromagnetic radiation. Linearly polarized light is itself achiral, thus any given wavelength of linearly polarized light has only one magnitude of dipole it will induce in a molecule, all else being the same. Circularly polarized light, on the other hand, is itself chiral. Thus, when we bring two chiral objects together, the molecule and the circularly polarized light, there is no reason at all to suspect the dipole induced in the molecule should be the same in both cases. The symmetry necessary to make such a claim simply is not present.

Thus, if you imagined tabulating the index of refraction for some molecule as a function of wavelength using circularly polarized light, you would need to have two columns. One would be for right-polarization and the other for left-polarization.


Also, because it is relevant to the answer, I will reiterate what Ron was saying in the comments. The photon is an inherently quantum object. It is a boson and thus has two possible values of spin angular momentum, $\pm\hbar$. These two values correspond to two allowable states the photon can be in, one which is polarized to the right and the other which is polarized to the left. That is, the electric field vector spins in a circle one way or the other. There also allowable solutions to the state of a photon which are some linear combination of these two polarization states. A 50/50 linear combination gives linearly polarized light. Thus, it seemed like you might have been picturing two photons whose polarizations were cancelling each other out. This is incorrect. Rather, you have to think of a single photon in a superposition of two polarization states.

$\endgroup$
  • $\begingroup$ "Rather, you have to think of a single photon in a superposition of two polarization states." I have thought in this way. Not as photon, but the plane polarized light to be superposition of left and right circularly polarized light. $\endgroup$ – Mockingbird May 5 '17 at 5:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.