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S vs P polarization |
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| Apr29-10, 03:20 PM | #1 |
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S vs P polarization
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. |
| Apr29-10, 03:48 PM | #2 |
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Recognitions:
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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. |
| Apr29-10, 04:10 PM | #3 |
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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 |
| Apr30-10, 08:10 AM | #4 |
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Recognitions:
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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.
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| Apr30-10, 11:19 AM | #5 |
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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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