S vs P Polarization: Explaining Ex Component

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SUMMARY

The discussion focuses on the differences between S-polarized and P-polarized light, specifically in relation to their electric field components when propagating along the x-direction. S-polarized light exhibits electric field components only in the y-direction (Ei=(0, Ey, 0)), while P-polarized light includes components in both the x and z directions (Ei=(Ex, 0, Ez)). The essential distinction lies in the orientation of the electric field relative to the plane of incidence, with S-polarization being perpendicular and P-polarization being within the plane.

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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, parallel 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, parallel 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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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.
 
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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