Brewster angle no reflection

Question

I'm a bit confused about the Brewster angle given by $$\theta_{p} = \tan ^{-1} {\left(\frac{n_2}{n_1}\right)}\;.$$ I think I am correct in saying that if unpolarized light is incident at the Brewster angle, then all light reflecting from the surface is horizontally polarized (so the light gets split up? The horizontal polarization of the incident ray is reflected while the vertical polarization component is refracted?).

But what happens if the incident light is 100% vertically polarized? Will there be no reflection? And will there be no refraction if the incident light is 100% horizontally polarized? If this is wrong, what is the criteria to have no reflection at the Brewster angle?

Answer 1

It is true that at the Brewster angle, the reflected light must be only horizontally polarized.

From this you can conclude that

1. vertically polarized light incident at the BA will be totally refracted
2. when unpolarized light is incident at the BA, the reflected ray is horizontally polarized

Neither of these things tell you that there is "no" refracted component of horizontal light. To understand this we look at the Fresnel equations: these tell us that there is a refracted component of horizontally polarized light at all angles of incidence (until the reflection is 100% which only happens for glancing incidence).

Note - all the above assumes that $n_1>n_2$ so we are talking about cases where the is no total (interal) reflection.

So the answer to your final question: there is no reflection if the light is incident at the Brewster angle and is 100% vertically polarized. Refraction occurs at all polarizations and at all angles of incidence below the critical angle.