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S vs P polarization

by geo_alchemist
Tags: polarization
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geo_alchemist
#1
Apr29-10, 03:20 PM
P: 35
Let us imagine a surface, with X and Y axes on it and Z axis normal to it.
well, the theory, (in my particular case, some article) says that S polarized light, propagating along the x-directioin possesses only electric field components, Ei, paralel to the surface (||y-direction), i.e. transversal electric waves have Ei=(0, Ey, 0), while P-polarized light has Ei=(Ex, 0, Ez).

The question:

I understand that S-polarized light, propagating along x-direction has only electric field components, Ei, paralel to the surface. Ei=(0, Ey, 0), since in case of s-polarization E is perpendicular to plane of incidence.

Also I understand that P-polarized light, propagating in the same x-direction has Ez component, since in case of P-polarization E is in the plane of incidence.


what I do not understand is, why p-polarized light has Ex component.
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Andy Resnick
#2
Apr29-10, 03:48 PM
Sci Advisor
P: 5,523
A picture is worth 1000 words:

http://jolisfukyu.tokai-sc.jaea.go.j...5IMG/05_11.jpg

Does this help? Otherwise, please refer to the diagram and tell me which directions you are calling x,y, and z.
geo_alchemist
#3
Apr29-10, 04:10 PM
P: 35
It seems that I understand bt it still is a bit foggy for me. Any further clarification would be greatfully appreciated.
Here is modified image with x, y and z.


Uploaded with ImageShack.us

Andy Resnick
#4
Apr30-10, 08:10 AM
Sci Advisor
P: 5,523
S vs P polarization

Do you see the essential difference between s- and p- polarization? It's not the (x,y,z) components per se, it's whether or not the electric field is in the plane of incidence.
geo_alchemist
#5
Apr30-10, 11:19 AM
P: 35
Essential difference I see perfectly. what I need is to describe this difference in terms of Ex, Ey and Ez.

Anyway, now I understand it, thanks for picture ;)


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