Calculate the desired incident polarization of a light beam

[Corwin_S]

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.

Cheers

 

[blue_leaf77]

Hi I want to calculate the necessary incident polarization of a light beam

What type of polarization is it, linear?

 

[Corwin_S]

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]

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?

 

[Corwin_S]

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.

 

[Corwin_S]

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