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The qualitatively explanation of the Brewster angle (e.g., see https://en.wikipedia.org/wiki/Brewster%27s_angle) is that incident light produces small electric dipoles on the surface of the dieletric medium (glas), and these dipoles produce the reflected light. However, a dipole doesn't emit in the direction of the oscillation axis. If the incident light is $p$-polarized and hits the glass under the Brewster angle $$\alpha_{Brewster} = arctan\left(\frac{n_2 }{n_1}\right)$$ the dipole axis is in the direction of reflection and thus no light is reflected in this direction.

When testing it with glass it works perfectly. However, when testing it with a mirror, it doesn't work at all. My initial expectation was that $p$-polarized incident light under 45$^\circ$ angle should be reflected in $-45^\circ$, which is also the direction of the dipole axis. So there shouldn't be reflected light. But there is... (measured it).

This was also asked here Why do mirrors not follow brewster's angle? and the answer suggested that the refractive index of a metal is almost infinity ($n_2 \approx \infty$), meaning that the Brewster angle is $90^\circ$, which is of course no practical incidence angle. Unfortunately, this doesn't explain physically what is happening...

My explanation at the moment is this: Because the refractive index of metal is so large, the angle of refraction according to Snell's law os $0^\circ$ (so straight into the mirror...), and thus the axes of the induced dipoles on the mirror surface are all within the mirror plane (independent on the polarization or incidence angle of the incident light). And thus there is no polarization effect of the reflected (re-emitted) light. Is that explanation correct, resp. is that the mechanism how a mirror works at all?

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First, you confuse permittivity with refractive index. The refractive index of silver for the visible wavelengths is almost 0.

Secondly, all laws still apply (ie Maxwell eqs). However, I think its easier to think of it this way: metals have conducting electrons, and they oscillate mostly freely. They are not bound in a potential which affects their mobility and thus, also their optical properties. Reflection of metals is more akin to some plasma currents than to the thoughts of dipoles like in dielectrics, electrons respond immediately to "any" applied field (of course, metals still have an atomic structure which affects different frequencies differently) and generate electrical currents, while dielectrics do not.

If you go to this website, it immediately shows you Silver, for the green wavelength. If you scroll down, you can find the curves for p- and s- reflection for different incident angles. You can see the typical brewster's angle dip is around 75°, except the difference between s- and p- reflection percentages are 99% and 97% respectively.

Hope this can help.

(But of course, just like antennas, they generate small dipoles, except they are on the surface and along the E-field lines)

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  • $\begingroup$ Many thanks, that helps! To conclude: 1) Yes, I confused refractive index and permittivity... However, the answer referenced in my question <physics.stackexchange.com/questions/453820/…> is apparently incorrect, although the argumentation there is indeed with dielectric constant. Could you confirm? 2) So the dipole-explanation is in general not good for mirrors, only for dielectric materials? $\endgroup$
    – Charles Tucker 3
    Commented Jun 3, 2021 at 10:37
  • $\begingroup$ Ok, maybe I wasn't clear. Dipoles still apply, that's their definition, oscillating electrons. Just that in dielectrics they are bound in a potential, which in turn has a restitution force to the movement, effectively shaping their movement direction, range, etc. In a metal, the electrons are free, they oscillate directly with the field with almost no constraints. And all electromagnetic constants and the way they shape electromagnetic propagation are well defined and do not care if its a dielectric or metal, thats why refractive index of metals <<1 and in dielectrics >1, their formalism (1/2) $\endgroup$
    – José Andrade
    Commented Jun 3, 2021 at 11:08
  • $\begingroup$ is exactly the same in both cases. Just the reason and how these constants arise is different. In dielectrics $J$ will be 0, in metals, that is not the case. I might also be glossing over things a bit too generally, and not in a very precise manner. I would instead challenge you to take a book on classic electrodynamics and find how maxwell eqs are used to derived metal and dielectric properties. $\endgroup$
    – José Andrade
    Commented Jun 3, 2021 at 11:13
  • $\begingroup$ Yes, I will do so. It's a bit embarrassing but after all this years I realize that I don't know how a mirror actually (physically) works... And I don't want to be pushy with the dipoles, but is the equation for the Brewster angle above applicable to mirrors as well? Because with $n=0.05$ for silver (from the link you provided) the angle would be 2.8$^\circ$, so basically the frontal reflection should be polarized (hard to believe). Gold (same webpage) has $n=0.27$ (15$^\circ$) and Aluminium$n=1.19$. So this should be easy to observe (which I did not in my measurements...). $\endgroup$
    – Charles Tucker 3
    Commented Jun 3, 2021 at 15:46
  • $\begingroup$ No need to apologize. Well, although I am an optics expert, my fundamental theory is long long gone. My feeling is that the brewster formula applies because of how the refractive index arises in dielectrics, however (again, my feeling) it should not work for metals. Refractive index shows the phase shift of light, and its underlying physics are different in a conductor and in a dielectric, but of course both materials display measurable phase shifts. I am sorry, but my time is limited, but I would look into Fresnel eqs and Maxwell eqs and try to understand how EM fields and electrons interact. $\endgroup$
    – José Andrade
    Commented Jun 3, 2021 at 16:59

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