2
$\begingroup$

I am trying to understand the concept of depolarization of light by various particle shapes. For example, depolarization light scattered (re radiated) from a sphere illuminated with vertically polarized light.

Is my picture an accurate representation of why light is depolarized? Essentially are the electrons on the 'edge' of the sphere forced to oscillate on a curved path? If this is not correct could you please explain the physical mechanism behind why some scattered light is depolarized from particles.

For the sake of my picture we will say the incident wavelength is 623.8 nm, and the sphere is of diameter 1 micron.

Vertically polarized wave incident on a sphere

$\endgroup$

1 Answer 1

1
$\begingroup$

I think it really depends on what you mean by depolarized, but a spherical particle like this scatters light with a complicated, but wholly foretellable (by Mie theory) and reproducible polarization dependence on scattering direction. So you imagine a farfield radiation diagram for the scattered light. To get the full picture, a "radiation diagram" would be a complex Jones vector field on the surface of a unit sphere. Each point on the unit sphere represents the direction defined by the ray joining the sphere's centre and the surface point in question. At each point, two complex quantities define the relative intensity and polarisation of the scattered wave. This will be a complicated object: as shown in Born and Wolf "Principles of Optics" section 14.5 ("Diffraction by a conducting sphere: Theory of Mie), this polarization field can be exquisitely sensitive to position on our unit sphere. But there is one, consistent, reproducible and only one such polarization state for each point on the unit sphere sphere. You don't get the polarization state randomly varying with time if the Mie scatterer is lit by coherent, polarized light.

So some people call this complicated polarization scrambling "depolarization" and what you have drawn is a fairly accurate intuitive guide to how this polarization scrambling arises.

True depolarization arises from thes phenomenons:

  1. The scattering properties of your Mie object to fluctuate randomly with time: these properties can be (i) orientation, if the object is not spherically symmetric (e.g. tumbling polar molecule), (ii) length dimensions if a micro- as opposed to molecular sized object vibrates, (iii) relative positions of scatterers if more than one scatterer is involved.

  2. The mixing of many such scatterings as you have drawn above by objects whose relative positions and orientations fluctuate randomly with time.

  3. The relationship between polarizations of incoming and scattered light is also fundamentally bound up with the exchange of angular momentum between the light and the medium it interacts with. For further explanation of this statement, see Chapter 18 of the Third Volume of the "Feynman Lectures on Physics". This chapter is called "Angular Momentum".

There is a simple way to summarise all these complicated mechanisms: these are the interactions of light with a thermalized system of scatterers.

$\endgroup$
2
  • $\begingroup$ Thanks, that's a very thorough description. I will check out those references. There seems to be very little literature out there on the actual physical 'mechanics' of depolarization. Oh and by depolarization I mean the change in polarization via scattering when hit with a linearly polarized source. $\endgroup$ Sep 24, 2013 at 13:38
  • $\begingroup$ I checked out the references. That picture in Principles of optics is very nice. I guess my diagram sort of explains whats going on. One thing I don't quite understand is the 'first, second, third and fourth electric partial wave'. Haven't really heard that term before, but I will go through that chapter more thoroughly. Thanks again for the references. $\endgroup$ Sep 25, 2013 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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