Unpolarized light incidence

Question

A beam of unpolarized light carries 2000W/m2 down onto an air–plastic interface. It is found that of the light reflected at the interface 300W/m2 is polarized with its $E$-field perpendicular to the plane of incidence and 200W/m2 parallel to the plane of incidence. Determine the net transmittance across the interface.

So let's call the initial intensity $I_0$. Being unpolarized, we know that the field can be writen as $E^2 = E_{0\parallel}^2 + E_{0\perp}^2$, and that $I_0 = kE^2$.

Let's call $E_{0\parallel} = E\times x$ and $E_{0\perp} = E \times y$

So we want $T = 1 - R = 1 - \frac{R_{\perp} + R_{\parallel}}{2}$

And we know that $E_{\parallel} = \sqrt{R_{\parallel}} E_{0\parallel} = \sqrt{R_{\parallel}}xE$ ,and $E_{\perp} = \sqrt{R_{\perp}} E_{0\perp} = \sqrt{R_{\perp}}yE$

And also notice that $x^2+y^2=1$ and $I_{\parallel} = k E_{\parallel} ^2$ and $I_{\perp} = k E_{\perp} ^2$

Now, i am feeling that i am walking around a circle from more than one hour! That is, i am not being able to express the answer only in terms of the values given. So i guess i am missing something.

At first i had the suspicious that, being unpolarized, $x=y=1\sqrt{2}$, but this gives $T \approx 0.99$, obviously wrong. What am i missing here?

Answer 1

The reflectance $R$ is the ratio of the reflected intensity to the incident intensity. In your case, $R=(200+300)/2000=0.25$, which leads to $T = 1-R = 0.75$. Notice we must sum the intensity from both reflected polarizations to have the correct ratio, as the incident beam contains both of them (i.e., it is unpolarized).