# Simple, nonmathematical argument for partial polarization by reflection

I'm teaching calculus-based electricity and magnetism with a sequence of topics in which students learn the basics of electromagnetic waves before the semester in which they get a more general introduction to topics such as wave kinematics, reflection, inverting and uninverting reflections, partial transmission, the optical density, and refraction. They also learn this topic before we do the electrical properties of materials, so they know the distinction between a conductor and an insulator, but they don't know about dielectric constants and so on.

I'm having them do simple experiments with polarizing films and calcite crystals, which works fine as a hands-on way of making sense of the geometry of an electromagnetic plane wave. I also have them look at their cell phones through the polaroids and also at glancing reflections from tabletops in order to see that the reflections are partially polarized.

For students at this stage, is there some very simple hand-waving argument I can present as to why reflections should be at least partially polarizing when the direction of incidence is not normal? Obviously it's easy to show by symmetry that for normal incidence, there is no polarization. I think it would be way too much for students at this stage to present a full treatment of the incident, reflected, and refracted waves with superposition and matching of boundary conditions. I'm thinking that there may be some conceptual simplification possible if one considers the case of an extreme grazing angle, and if we don't care about a detailed quantitative result for the amount of polarization, Brewster's angle, etc. Is there perhaps some simplification that can be made in the case where the surface is highly absorptive? My students do know about dipoles. Is there some simple argument that gives a qualitatively correct result if you treat the surface as a sheet of dipoles?

• This is a good question. Do you think that seeking a qualitative explantion for the boundary conditions of the fields is equivalent? Or are you trying to avoid even that? Oct 8, 2019 at 0:56
• @dmckee: I'm open to anything that's simple and doesn't require physics that my students don't yet know. However, there's a lot that they don't yet know at this stage :-) Ideally I'd like to have some kind of very simple visual thing that could be explained in 1-2 minutes.
– user4552
Oct 8, 2019 at 1:42
• This question physics.stackexchange.com/questions/431829/… quotes (without attributing the source) something that seems to be claiming to be a simple qualitative explanation. However, I don't understand their argument (they seem to be leaving out a lot), and they also seem to be claiming that polarization by reflection is 100%, which would be false.
– user4552
Oct 8, 2019 at 1:58

Quite simple approach is the scattering model, which considers reflected and transmitted wave as a scattering pattern by the dipoles induced in the second medium. This model is originally due to Sagnac, but it has been included in a number of textbooks. A good resource for this model is a paper

• Doyle, W. T. (1985). Scattering approach to Fresnel’s equations and Brewster’s law. American Journal of Physics, 53(5), 463-468, doi:10.1119/1.14201.

From the paper:

In the usual scattering model of Brewster's law the reflected beam is created by oscillations of the induced dipoles in the second medium, driven by the electric field of the transmitted wave. When the incident wave is $$p$$-polarized, the reflected beam has zero intensity at Brewster's angle where the reflected and transmitted beams are mutually perpendicular. Since the transmitted beam is transverse, the axes of the induced dipoles then point in the direction of the reflected beam, and a dipole cannot radiate along its own axis. In this model no Brewster angle can occur with $$s$$-polarized incident waves, because the induced dipoles are always perpendicular to the direction of the reflected beam and each dipole radiates isotropically in its own equatorial plane.

The paper further extends the original model to also include induced magnetic dipoles, and offers elementary derivation of Fresnel's equations and Brewster's law.

One thing the paper emphasizes is that all of the dipoles in the media are contributing to the reflected beam. So, “a sheet of dipoles” would be a wrong model for the situation.

• One thing the paper emphasizes is that all of the dipoles in the media are contributing to the reflected beam. So, “a sheet of dipoles” would be a wrong model for the situation. I was thinking that this might be a useful approximation in the case where there is strong absorption.
– user4552
Oct 12, 2019 at 17:42
• The argument quoted from Doyle is an argument that applies only at Brewster's angle, so it doesn't immediately tell us that we also get a similar type of partial s polarization at other angles.
– user4552
Oct 12, 2019 at 18:30
• Indeed. But we could imagine dipole radiation intensity diagram oriented so that the axis of dipole coincides with E-field of p-wave in second medium, while for s-wave we always have radiation in the direction orthogonal to dipole. So the reflected p-wave would have less intensity than s-wave, Oct 12, 2019 at 18:39
• You might want to edit that in as part of your answer, since it seems like a logically necessary piece.
– user4552
Oct 12, 2019 at 19:20

There is a nice treatment of this topic in the Feynman lectures, I-33-6, which can be reduced to a fairly simple argument that reflection produces partial polarization perpendicular to the plane of incidence. This is probably not the optimal answer for students with the precise background knowledge I describe in my question, but it is pretty simple and requires almost no math, so I'll post it as a self-answer.

We have a wave I in vacuum, incident on some matter in which light propagates losslessly. Wave I would hypothetically have gone straight and made a fictitious wave S if there had been no matter. In reality, it produces a transmitted wave T and a reflected wave R. The charges in the matter are driven by I to oscillate, and their oscillations not only have to produce waves R and T, they also have to produce a wave -S that cancels wave S, the minus sign indicating a flipped amplitude.

In the case where the incident wave has its electric field perpendicular to the plane of incidence (the $$\perp$$ case), the oscillations of the charges are perpendicular to the plane, and are therefore in the right direction to produce -S as efficiently as possible. But in the case of polarization parallel to the plane of incidence ($$\parallel$$), only one component of these oscillations is in the right direction to produce -S. Therefore the oscillations have to be stronger in the $$\parallel$$ case in order to (less efficiently) give -S. Since these oscillations are what produce T, transmission is stronger in the $$\parallel$$ case. (The ratio of the transmitted amplitudes is $$|A_\parallel/A_\perp|=1/\cos(\theta_i-\theta_t)$$.) By conservation of energy, this means that reflection is weaker in the $$\parallel$$ case.