# Polarization upon reflection?

The physical mechanism for this can be qualitatively understood from the manner in which electric dipoles in the media respond to p-polarized light. One can imagine that light incident on the surface is absorbed, and then reradiated by oscillating electric dipoles at the interface between the two media. The polarization of freely propagating light is always perpendicular to the direction in which the light is travelling. The dipoles that produce the transmitted (refracted) light oscillate in the polarization direction of that light. These same oscillating dipoles also generate the reflected light. However, dipoles do not radiate any energy in the direction of the dipole moment. If the refracted light is p-polarized and propagates exactly perpendicular to the direction in which the light is predicted to be specularly reflected, the dipoles point along the specular reflection direction and therefore no light can be reflected.
http://en.wikipedia.org/wiki/Brewster's_angle

The dipoles that produce the transmitted (refracted) light oscillate in the polarization direction of that light. Why?

Does that last sentence actually say no light is reflected back if the angle of impact is orthogonal to the surface?

Simon Bridge
Homework Helper
http://en.wikipedia.org/wiki/Brewster's_angle

The dipoles that produce the transmitted (refracted) light oscillate in the polarization direction of that light. Why?
How else would it oscillate? Short answer: that is how EM waves are made.

Does that last sentence actually say no light is reflected back if the angle of impact is orthogonal to the surface?
No. The last sentence says:

[IF] the refracted light is
(1) p-polarized and
(2) propagates exactly perpendicular to the direction in which the light is predicted to be specularly reflected,
[THEN] no light can be reflected.​

i.e. if light only has a component parallel to the plane of incidence, and the light is incident at Brewster's angle, then you get no reflection.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polref.html

How else would it oscillate? Short answer: that is how EM waves are made.

Dipoles making EM waves is the 2nd phase, but before EM waves are reflected they first have to make those dipoles oscillate.

EM waves are reflected so their electric fields always end up polarized parallel to the surface. They are polarized upon reflection in this particular way because the dipoles in the surface which re-emitted them are oscillating parallel to do surface. However, the original photons which impacted with the surface were unpolarized, so the question is why would they make those dipoles in the surface always oscillate parallel to the surface regardless of their random polarization?

Simon Bridge
Homework Helper
OK back up a bit.

"photons" belong to the particle theory of light.
The theory you are reading is all about the wave theory of light.
It is very important not to mix them up.

In the wave theory, light is a self-propagating EM wave with electric and magnetic field vectors that oscilate.
When there is a charge in the same place as there is an electric field vector, it experiences a force ##\vec F = q\vec E## which accelerates the charge. If the E field is oscilating then the charge wiggles back and forth - accelerating charges radiate.

If that charge is part of a material, then it also experiences the forces that keep it in the material.

If that material has a particular shape, then the charges are constrained by that shape. eg. Electrons in a thin wire can only move along the wire, if they are on the surface of a sheet of metal, then they can only move in the surface. It follows that E fields perpendicular to the surface cannot make the charges move.

If that material has a particular shape, then the charges are constrained by that shape. eg. Electrons in a thin wire can only move along the wire, if they are on the surface of a sheet of metal, then they can only move in the surface. It follows that E fields perpendicular to the surface cannot make the charges move.

Ok. I can see the surface would provide paths of less resistance for the dipole oscillation plane, but I still wouldn't expect it to be 100% uniform. It seems final polarization and uniformity of it could heavily depend on molecular structure of the material. When unpolarized sunlight reflects of horizontal glass and wooden tables, is reflected light in both cases 100% horizontally polarized?

Simon Bridge
Homework Helper
You are right - IRL, it is not 100% uniform and there is a slight possibility to oscillate out of the surface... just not very much and they tend to average out. Not all reflected light is 100% horizontally polarized - unless it is reflected at Brewser's angle.

Classical theories are usually constructed out of idealized circumstances, and the real world fuzzyness gets added on later. For a rough surface, not all the incident light can will have the same angle of incidence everywhere on the surface.

1 person
It makes sense now. Thanks. Would you also know what happens with polarization when light is passing through glass or crystal? For example, does ordinary window glass polarize, unpolarize, or preserves original light polarization?

ehild
Homework Helper
In case of oblique incidence, the materials behave differently when the light is polarized parallel with the plane of incidence (the plane defined by the normal of the interface and the incident ray) than in case of perpendicularly polarized light. The incident light can be decomposed into parallel and normal components and both components keep their direction upon reflection and transmission, if the material is transparent, isotropic, but the transmittance and reflectance are different for the different polarizations. That will change the polarization state of the reflected or transmitted light.
In case the material is anizotropic or has optical activity, the situation is even more complicated.

The direction of polarization does not change at normal incidence when reflected or transmitted from a sheet of glass, but changes at oblique incidence.

ehild