Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

When a light beam reaches a dielectric surface, the incident and reflected beams have different intensities depending on polarization. For the so-called Brewster's angle, the reflected light is lineary polarized.

My question is: how does this law work in case of mirror-like surface, when (ideally) all the light is reflected?

share|improve this question
add comment

1 Answer 1

up vote 6 down vote accepted

The easy answer is to say that Brewster's law only applies to reflection from the interface with a transparent medium, and a mirror isn't transparent. Indeed for an ideal perfect mirror, all light of both polarizations is reflected perfectly, so there is nothing to say.

For an actual real-world mirror, the metal mirror surface will have a finite skin depth, and can be considered a dielectric medium with a very large, complex index of refraction. This does lead to a small polarization dependence of the reflection coefficient for near grazing incidence angles. The analogue of Brewster's angle occurs at an angle given by 2*pi*(skin depth) / wavelength above grazing, where the parallel polarized reflection coefficient will reach a minimum ( but not zero, still close to 1). For more details, see for example Landau and Lifshitz, volume 8, section 87.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.