# Polarization change upon reflection

by Chen
Tags: polarization, reflection
 P: 1,004 Hello, I'm trying to understand how exactly light changes its polarization when reflected from a mirror, for example. I'm quite familiar with Fresnel's equations and resulting coefficients, but I'm not sure how the phase of the TE and TM polarizations changes upon reflection. For example, let's say I have a mirror standing in the YZ plane, and I'm firing a laser towards the mirror with a k-vector that lies in the XY plane, and makes a small angle with the normal to the mirror. If the beam is originally linearly polarized thus / when looking towards the mirror (let's say 45 degrees with the Z axis), how will it be polarized after the reflection? Looking away from the mirror, would it be / still, or would it be transformed into \? (due to a phase change of the TM polarization) Looking at this page: http://scienceworld.wolfram.com/phys...Equations.html I see that the reflection coefficient for both polarizations differs only by a phase of pi for normal incident. How is this possible? I thought that for normal incidence, there is no difference between TE and TM, so how can there be a different reflection coefficient for them? Any help would be welcome. Thanks, Chen
P: 1,477
 Quote by Chen I thought that for normal incidence, there is no difference between TE and TM, so how can there be a different reflection coefficient for them?
I think you have the directions for the parallel and perpendicular components confused. The perpendicular component has the E-field pointing normal to the surface, whereas from your post you seem to think that it is parallel.

Claude.
 P: 1,004 None of the polarizations necessarily have the E component perpendicular to the surface. The classification is into two polarizations, one of which has E perpendicular to the plane of incidence (and therefore parallel to the reflection surface), and the other has E parallel to the plane of incidence. Chen
P: 1,477
Polarization change upon reflection

 Quote by Chen ...one of which has E perpendicular to the plane of incidence (and therefore parallel to the reflection surface), and the other has E parallel to the plane of incidence. Chen
You need to decompose the incoming polarisation state into a component parallel to the interface and one perpendicular to the interface. Trying to wobble atoms parallel to the interface and trying to wobble atoms perpendicular to the interface are two physically distinct cases, which is why each component needs to be analysed separately.

http://www.uta.edu/optics/research/e...lipsometry.htm

Claude.
 P: 2 It's because when people derived this formula they defined the coordinate system of the reflected light with the same handed as the input light. So, the direction of the Erp and the direction of the Eip are opposite, while the electrical field are actually pointing the same direction. I think that's why you would see a minus P reflectivity (p is the parallel polarized light with the incident plane) under the normal incidence. My answer on your question would be the relected light has a \ polarization. I'm not quite sure and also searching for the confirmation, if you've already got the correct answer, plz also tell me, thx. JS.Li
 Sci Advisor P: 1,477 I can confirm that the answer to Chen's original question is that you will get a \ polarisation back when viewed from the mirror, the shift in polarisation only comes because the direction you are viewing the wave from switched! The polarisation will remain / if you continue looking toward the mirror. Claude.
 P: 2 Thanks:)