# Polarization of Light

This is more of a conceptual question:

Where does the equation for the intensity of polarized light come from?

I know the standard form is:

I=(1/2)(I_0)

after the EM wave passes through the first polarizer. Why is that? Why is the intensity exactly half?

The polarizer ensures that the only polarizations of light that pass through are those oriented along the same axis as the polarizer. Since the polarization of the light is random, this means any number of the polarizations could be eliminated, right? I am having difficulty understanding how a polarizer automatically blocks out 50% of the light at all times, and how this is true regardless of the angle of the polarizer, since the polarizations are random to begin with.

## Answers and Replies

Doc Al
Mentor
Are you familiar with the law of Malus?

If you think of the light as an electromagnetic wave, the orientation of its E field represents its polarization angle. The nature of an ordinary dichroic polarizer is to strongly absorb the E field in one direction, allowing the field perpendicular to that direction to pass through. The light that makes it through the polarizing "filter" has its field realigned with the polarization of the filter. Ignoring loss, if polarized light ($E_0$) passes through a dichroic filter at an angle $\theta$, the transmitted beam is now polarized parallel to the orientation of the filter's axis and has a field strength of $E_0 \cos\theta$. The intensity of the beam is reduced to $I_0\cos^2\theta$--the law of Malus.

Now if the incoming light is unpolarized, you can think of its polarization as being randomly distributed over all angles. The average of $\cos^2\theta$ is 1/2. So the intensity of the transmitted light is $I = I_0/2$.

Make sense?

Yes! Thank you for the explanation, especially the last bit. We covered the Law of Malus in my class, but didn't refer to it by that title. I think the thing I was really missing was the fact that the average of cos(sq) is 0.5, which makes complete sense if I graph it and think about it.

Thanks again!