# Calculate the desired incident polarization of a light beam

## Main Question or Discussion Point

Hi I want to calculate the necessary incident polarization of a light beam at a given angle of incidence (theta_i) that reflects off BK7 glass (n = 1.5168) and is linearly polarized (i.e., 45 degrees). I know how to do similar calculations for incident natural unpolarized light, but not in the case of incoming polarized light.

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blue_leaf77
Homework Helper
Hi I want to calculate the necessary incident polarization of a light beam
What type of polarization is it, linear?

The reflected beam comes out linearly polarized (50/50 s/p). I'm trying to find the incoming beam's polarization (assume it is not circularly or ellipitically polarized).

blue_leaf77
Homework Helper
Forget my previous comment, if the reflected light is linearly polarized then so is the incoming one for the case of external reflection (##n_2 > n_1##).
Anyway, you have Fresnel equations for TE and TM components:
$$\frac{E_r^{TE}}{E_i^{TE}} = \frac{n_1\cos \theta_1 - n_2\cos \theta_2}{n_1\cos \theta_1 + n_2\cos \theta_2}$$
$$\frac{E_r^{TM}}{E_i^{TM}} = \frac{n_2\cos \theta_1 - n_1\cos \theta_2}{n_2\cos \theta_1 + n_1\cos \theta_2}$$
Now it's required that ##E_r^{TE}=E_r^{TM}## and that ##\theta_1## is given, from which ##\theta_2## will follow from Snell's law. So, isn't it straightforward to get the ratio of the components of the incoming light?

This is completely correct. Those can be called the reflection coefficients of the s and p polarized components. I believe the correct way of determining the polarization of the input is to find the degree of polarization:

V = Ip/(Ip+In)
And I believe, although am not sure, that Ip = Rs + Rp and In = 1/2(Rp + Rn).

Hence given the specs n = 1.5154 @ 650 nm, and the incidence/reflectance angle = 50 degrees, I compute the degree of polarization to be ~66.67%.

I don't know if this is right though.

This is completely correct. Those can be called the reflection coefficients of the s and p polarized components. I believe the correct way of determining the polarization of the input is to find the degree of polarization:

V = Ip/(Ip+In)
And I believe, although am not sure, that Ip = Rs + Rp and In = 1/2(Rp + Rn).

Hence given the specs n = 1.5154 @ 650 nm, and the incidence/reflectance angle = 50 degrees, I compute the degree of polarization to be ~66.67%.

I don't know if this is right though.
Correction, n =1.5145