Brewster's Angle

At one particular angle of incidence, the light reflected from a dielectric surface is completely polarized parallel to that surface. The reason is geometric and rather satisfying. Reflected light is produced by electrons in the surface layer being driven into oscillation along the electric-field direction of the transmitted wave, and an oscillating charge radiates nothing at all along its own axis of oscillation. At the angle where the reflected and refracted rays would be exactly 90° apart, the direction the reflected ray would have to travel coincides with that axis for the component polarized in the plane of incidence — so that component cannot be radiated, and what leaves the surface is purely the perpendicular component. Impose $\theta_1 + \theta_2 = 90^\circ$ on Snell's law and the algebra collapses to $\tan\theta_B = n_2/n_1$.

For air onto water, $n_2/n_1 = 1.333$ and $\theta_B = \arctan(1.333) = 53.1^\circ$ from the normal — which is 36.9° above the horizontal, so the Sun about a third of the way up the sky. Air onto glass at 1.52 gives 56.7°. Those are the conditions under which reflected glare is at its most strongly polarized, and they are why a polarizing filter does its most dramatic work in the middle of the morning rather than at noon.

David Brewster established the relation in Scotland in 1815, generalising Étienne-Louis Malus's 1808 observation that light reflected from a window in Paris came off polarized. The applications are everywhere once you look. Polarizing sunglasses are cut with their transmission axis vertical so they reject the horizontally polarized glare bouncing off water, wet roads and car bonnets. Gas laser tubes are sealed with Brewster windows — a plate tilted to $\theta_B$ passes one polarization with literally zero reflection loss, which is how such a laser produces polarized output for free and why the windows sit at that peculiar angle. A photographer's circular polarizer is the same physics, and it works best on surfaces viewed near 53°.

Four things to get right. At Brewster's angle the reflection is fully polarized, not eliminated. The perpendicular component still reflects — about 15% of it, for water — so polarized sunglasses reduce glare substantially and never abolish it. Second, the effect is broad rather than knife-edged: reflected light is strongly polarized anywhere from roughly 30° to 70°, so the practical benefit does not vanish if you are off the exact angle. What does kill it is looking straight down into water at near-normal incidence, where no direction is preferred and there is nothing for the filter to reject — which is precisely why the glasses do so little at midday over a pool. Third, it is tangent, not sine: $\arctan(1.333) = 53.1^\circ$ while $\arcsin(1.333)$ is undefined and $\arcsin(1/1.333) = 48.6^\circ$ is the critical angle for the opposite direction. The two formulas look similar enough that swapping them is the commonest slip on this page, and the wrong answer is plausible rather than absurd. Fourth, this applies to dielectrics only. Metals have complex refractive indices and no angle of zero reflection exists; the nearest analogue is the pseudo-Brewster angle, which is a shallow minimum rather than a null.