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# What Are The S and P Polarization States?

Summary:

*Question: What are the S and P polarization states, and how do I enter a beam that is s-polarized?*

Authored By:

Mark Nicholson

Published On:

08/01/2007

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**What Are The S and P Polarization States?**

*Question: What are the S and P polarization states, and how do I enter a beam that is s-polarized?*

It's an error to consider a ray, or a beam of light, as being s- or p- polarized. This is because s and p are defined relative to the plane of incidence of the ray on the surface, and are not characteristics of the beam itself.

The amplitude and polarization state of the electric field is described by a vector E which has components {Ex, Ey, Ez} which are all complex-valued. The ray propagation vector k has components {l, m, n} where l, m, and n are the direction cosines of the ray in the x, y and z directions. The electric field vector E must be orthogonal to the propagation vector k so that

k.E = 0

and therefore

Ex.l + Ey.m +Ez.n = 0

Now of course what we choose to call the x, y and z axes is arbitrary. For example, I may choose z to go from left to right across the screen, y from bottom to top, and x into the screen. Or, I might choose z to point out of the screen, y to go left to right and x to go bottom to top. As long as I am everywhere consistent in my definition of coordinate axes, there is no problem.

But, when a ray intercepts the surface of an optical component, we define a plane, called the plane of incidence, which is not arbitrary. The plane of incidence contains both the k vector and the surface normal vector n at the intercept point. The s component of the field is the projection of E that lies along the axis orthogonal to the plane of incidence, while the p projection lies within the plane of incidence. The electric field E is then divided into Es and Ep components, both of which are complex valued.

So, s and p are defined relative to the plane of incidence of the ray on the surface, and are not characteristics of the beam itself.