# How to treat partially polarized light with Jones vectors?

I have to calculate the degree of s-polarization (perpendicular) of a transmitted unpolarised light ray when it goes through a glass plate ($n=1.5$) at Brewster's angle (Brewster's window).

The setup would be something like this:

Using Fresnel's equation I have that the transmission coefficientes are $t_s=0.86$ and $t_p=1$. This can be arranged into a matrix:

$$t=\left( \begin{array}{cc} 0.86 & 0 \\ 0 & 1 \end{array} \right)$$

In this type of problems, the unpolarized incident beam can be described by:

$$E_i=E_0\left( \begin{array}{c} \cos \theta \\ \sin \theta e^{i\delta} \end{array} \right)=\frac{E_0}{\sqrt 2}\left( \begin{array}{c} 1 \\ 1 \end{array} \right) \implies E_{t}=\frac{E_0}{\sqrt 2}\left( \begin{array}{c} 0.86 \\ 1 \end{array} \right)$$

The transmited intensity is: $I=\left( \frac{0.86^2}{2}+\frac{1}{2} \right)I_0=0.87I_0$.

If $E_i$ described polarized light the degree of s-polarization would be:

$$P_s=\frac{I_s}{I}=\frac{0.86^2/2}{0.87}=0.43$$

But this assumes that the incident ray is fully polarized, which is wrong. I think that the difference $(1-0.86^2)I_0=0.26I_0$ could be the polarized component. And then, the degree of s-polarization intensity could be:

$$I_s=0.26\frac{0.86^2}{1+0.86^2}=0.11$$

But this is just a guess . The only case I've seen unpolarized light treated is for a linear polarizer, in that case the transmitted light is fully polarized.

In general, how would one calculete the polarized and unpolarized components?

Now I think it more reasonable the the polarized compontent has pure s-polarization.

The Fresnel transmission coefficients at the Brewster angle between two media of $n=1$ and $n=1.5$ are not 0.86 and 1.

I haven't checked the Maths, but looking at this: https://www.geogebratube.org/student/m325541 I would say the transmission coefficients are $t_s=0.6$ and $t_p=0.66$. The reflection coefficients $r_s=-0.4$ and $r_p=0$.

The transmission coefficients expressed in terms of power are about 0.86 and 1. Recall that the Transmittance, is $$T_p = \frac{n_2}{n_1}\frac{\cos\theta_2}{\cos\theta_1} t_p^{2}$$

It's hard to follow what you are asking in the rest of the question. Using these transmission coefficients and the fact that unpolarised light can be treated as 50% p-polarized and 50% s-polarised, then the transmitted light will be p-polarised as $$\frac{E_s}{E_{total}} = \frac{0.5 \times t_s}{0.5\times t_p + 0.5\times t_s} \simeq 47.6\%$$.

Or, in terms of power $$\frac{I_{s}}{I_{total}}= \frac{0.5 \times t_s^{2}}{0.5 \times t_s^{2} + 0.5 \times t_{p}^{2}} \simeq 45.2\%,$$ because the magnification factors are the same for both polarizations.

This of course only deals with the transmission through the first interface. You must then tackle the glass/air interface in a similar way if you wish to see what happens to light transmitted all the way through a block.

EDIT: As an aside; completely unpolarised light can be treated as equal amounts of light (i.e. equal E-field amplitudes) in two perpendicular directions in the plane perpendicular to the wave motion. The two components must not have any fixed phase relationship (otherwise you get elliptically polarised light). You can choose the two perpendicular directions to suit the problem. In this case, choosing the directions parallel (p-) and perpendicular (s-) to the plane of incidence is sensible!

• Oh, sorry I should have added an image. The goes in and gets out (so I get the square of $t$). Anyway, are you saying that 47.6% is s-polarized and 52.4% is p-polarized? Wouldn't an unpolarized compontent have to pass the plate? I think I don't fully understand unpolarized light. Dec 25, 2014 at 13:08
• @jinawee Depends what you define as polarisation: intensity or E-field? Either way, it is certainly "partially p-polarised" (that is, the majority of the intensity is p-polarised). Dec 25, 2014 at 13:20
• So that 47.6% and 52.4% are the light components without taking into account whether they are in phase or not? How would you calculate the polarized component and the unpolarized component? I think that is my main doubt. Dec 25, 2014 at 13:33
• @jinawee I can't say without you defining what you mean by polarised and unpolarised component. Note also that the calculation I show only applies to the air/glass interface. If you want to deal with the whole block then you have the glass/air interface to consider too. Dec 25, 2014 at 13:36
• @jinawee If I were you I'd have a good read through something like this teknik.uu.se/ftf/education/ftf2/Optics_FresnelsEqns.pdf Dec 25, 2014 at 13:40

Ok, I think I've figured it out.

As I mentionend, the resulting electric field is:

$$E_{t}=\frac{E_0}{\sqrt 2}\left( \begin{array}{c} 0.86 \\ 1 \end{array} \right)$$

Although it's not necessary, I found things clearer using the associated density matrix. The outgoing light is given by:

$$\rho=\frac{I_0}{2}\left( \begin{array}{cc} 0.74 && 0 \\ 0 && 1 \end{array} \right)$$

A way to calculate the unpolarized component would be to separate $\rho$ into $I_\mathrm{unpol}\rho_\mathrm{unpol}+I_\mathrm{pol}\rho_\mathrm{pol}$.This means:

$$\rho=0.74 I_0\left( \begin{array}{cc} 1/2 && 0 \\ 0 && 1/2 \end{array} \right)+\frac{0.26 I_0}{2}\left( \begin{array}{cc} 0 && 0 \\ 0 && 1 \end{array} \right)$$

Here one clearly sees the unpolarized part and the p-polarized part.

If we define the degree of polarization considering that unpolarized light has 50% polarization in any component, as @RobJeffries did:

$$\frac{I_s}{I}=\frac{0.74}{0.74+1}=43\%$$

But one could consider that unpolarized light has no polarization components (more or less what I was thinking when I asked the question). After all, the transmitted intensity will be constant for every orientation of the polarizer (linear or circular). In that case, there is no s-polarization at all and the degree of p-polarization is:

$$\frac{I_p}{I}=\frac{0.26/2}{0.74+0.26/2}=15\%$$

This is real p-polarization, if you place linear polarizer in the same direction, all of it will pass. In a similar fashion, if the polarizer is placed orthogonal, none of that light will pass.

• The numbers in your first matrix appear incorrect for a air-glass interface. Or, is this the transmitted E-field all the way through the block? Dec 29, 2014 at 11:16
• @RobJeffries Yep, "all the way", $r_s=0.86=0.62\cdot 1.38$ and $r_p$ is always one for a plano parallel plates at Brewster's angle. Dec 29, 2014 at 13:13