# Intensity of light transmitted by a polarizer when the incident light is unpolarized

my question is similar to this older one, but I have not enough privileges to answer it or comment on it.

I do realize that the question has received detailed answers by Selene Routley, and I sort of get them.

My concern is for high school students who barely know integrals, let alone distributions or quantum mechanics.

Our textbook does not mention photons in this context. Light is simply described as a linearly polarized transverse wave, and Malus' Law is given. Unpolarized light is described as a (uniform) superposition of polarized waves with random uncorrelated polarization planes.

From time to time our textbook tries to make contact between physics and maths. In this instance it goes like "In order to prove the 1/2 factor in the transmitted intensity, let us calculate the average over the polarization angles" and then proceeds with the integration over the angles.

Personally, I find this a bit confusing, perhaps due to excessive conciseness. Why an average? I have a bunch of electromagnetic waves hitting a surface. I should be summing their intensities, not averaging them.

I thought we could reason as follows. The light hitting the polarizer is a superposition of linearly polarized waves, with random polarization angles. It is reasonable to assume that the overall intensity of light at a given polarization angle $$S_\theta$$ does not depend on such angle. Therefore the intensity $$S_i$$ of the overall incident light is obtained as the sum (integral) $$S_i = \int_0^{2\pi} S_\theta d\theta = 2\pi S_\theta$$ Since each polarized wave is subject to Malus Law, it is transmitted with intensity $$S_{t\theta} = S_{\theta} \cos^2(\theta)$$. Therefore the overall transmitted intensity is $$S_t = \int_0^{2\pi} S_\theta \cos^2(\theta) d\theta = 2\pi S_\theta \frac{1}{2}$$

Upon eliminating $$S_\theta$$, these results entail $$S_t = \frac{1}{2} S_i$$

I do realize that, mathematically, this is equivalent to calculating an average. One can say: let us assume that the incident light contains $$N$$ waves with the same intensity $$S_{i1}$$. Then, $$S_i=N S_{i1}$$. By definition of average, the overall transmitted intensity can be written as $$S_t = N \bar S_{t1}$$, where $$\bar S_{t1} = \frac{1}{2\pi}\int_0^{2\pi} S_{i1} \cos^2 \theta d\theta = \frac{S_{i1}}{2}$$ is the average transmitted intensity for an individual wave. Multiplying both hands of this equation by $$N$$ gives once again $$S_t = S_i/2$$.

In my opinion this less concise argument is better than the hasty one by our textbook. However I still think the first one I made is better, especially when this is an "excuse" to put the recently learned tool of integration to use. Of course, one can subsequently point out the relation with averaging.

But what is the reason for choosing an approach based on the averaging procedure?

Thanks for any insight on this Francesco

## 1 Answer

The integration steps you have done to get $$1/2$$ is perfectly alright since the math is right and the physics is right.

A Simple Intuitive Picture -

Before I answer your question I would like to show how $$1/2$$ appears with a very simple argument that has nothing to do with averaging or integration, which is perfectly valid.

Unpolarized light by definition as same intensity at every polarization angle and this also means that decomposing the unpolarized beam into two perpendicular components will each have the intensity.

Any vector including the polarization vector can be decomposed to two perpendicular components. Hence for a polarized light incident on a Polarizer, simply decomposed the polarization of the unpolarized light in two components one parallel to the axis of polarization of the polarizer and other perpendicular to it. Now since the incident light is unpolarized both these components will be equal and each will contain half the intensity so that the total intensity adds to the original intensity. Only the one parallel to the polarization axis will pass through hence the output intensity will be just one component which has a value of half the original intensity.

Back to your question - Why averages?

Averaging is very common because the standard way of thinking about light and polarization is the photon picture (quantum mechanics approach). In a very simple language whenever we have a system with large number of particles/ensembles (here large number of polarized beams in random direction) any observed quantity is average of the effect due to each individual particle/ensemble. Here the property is 'intensity consistent with Malus Law' and the actually observed intensity is average of this property on all individual particles (beams polarized in random direction). Whatever I have described is a statistical technique called Ensemble Average which in Quantum theory is called Quantum Ensemble Average. And this is the reason why you see averages everywhere when dealing with a large number of constituents.

Just to add a bit more complication, to bring the Quantum picture in, we require quantum operator that correspond to the observable we are interested in (here its the Malus law), hence here the operator is simply $$c.{cos}^2\theta$$.

To simplify the entire discussion, I will present how we think of this in Quantum Mechanical point of view. Please take this with a pinch of salt, as I am only trying to convey the idea and have absolutely no intention in conveying the exact detailed calculation:

In Quantum Mechanics, observing anything that is physically observable using the appropriate Operator, the wavefunction (think of it as the actual particle but expressed in mathematical language) collapses to some predefined states called Eigen States of the respective operator (A Postulate of Quantum Mechanics). For our discussion, the particles are photons(they are the fundamental building blocks of light), the operator is the polarizer/Malus Law and think of the eigen states as polarization direction, for simplicity just consider two perpendicular directions. And because its an unpolarized light, the wavefunction will be a superimposition of both polarization direction with equal probability(defining property of unpolarized).

Now what happens is, there photons incident on the polarizer is 'observed' and hence collapses to either one of the eigen states i.e. each photon can be either along the axis of polarizer, in which case it passes through, or perpendicular to axis in which case they are blocked out.

Having the background explained now consider 10 unpolarized photons hitting the Polarizer. Because it is unpolarized, about 5 of these will be polarized in one direction and other 5 in other direction (this is what should be observed using the Malus Law). Hence 5 out of 10 will appear in the other side. What is the average number of Photons that appear on the other side? its $$5/10=1/2$$.

Hence given total number of unpolarized photons, say $$N_0$$ on average only $$N_0/2$$ photons will pass through the Polarizer.

• Uhm, I'm not sure your argument has nothing to do with integration or averaging. I do agree that the polarization vector of a wave can be decomposed along the transmission axis and perpendicular to it. But there is no reason for these component to be equal, for an individual wave. You get what you get, randomly.
– PFB
Commented Mar 5, 2021 at 19:26
• (sorry a bit of my comment was missing). I expect the average components to be equal. I am aware of ensemble average, and they're ok for averages. But my point is a different one. I would still say that overall intensity is the sum of the individual contributions, not its average. Of course the overall intensity can be calculated as the average times the number of particles, if you like.
– PFB
Commented Mar 5, 2021 at 19:38
• I understand your concern. What you might be thinking is the fact that here we dont have one wave but a large number in random polarization and so one has to average it to get 1/2. That is perfectly valid and that is what is commonly done. But, here in my 1st part I directly used one of the defining property of unpolarized light as in here. Since its definition I dont have to prove it, thats the idea in the simple intutive 1st part. Commented Mar 5, 2021 at 19:46
• No no you missed the whole point of ensemble average. If I have a system of particles, any physically observed property is NOT the sum of that property of individual particles, but rather an AVERAGE of the individual particle property. Commented Mar 5, 2021 at 19:50
• Yes, I think you're right about the ensemble average. I was still thinking of the average over the individual "waves". Apologies. I'll look into the defining property of unpolarized light mentioned in the wikipedia page you linked. I'm unfamiliar with that, and the sources mentioned in the page. Anyway both points of view seem a little too advanced for high school students. My goal is having the students understand the point, in the context of the textbook, which never mentions photons or ensemble averages.
– PFB
Commented Mar 5, 2021 at 21:12