# Brewster angle no reflection

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?

• You have a misconception. The rays that is refracted is not polarised, or slightly polarised, and not at all completely vertically polarised. The reflected ray is the one that is majorly plane polarized. I'm not posting an answer as I do not know the latter part, but this is for sure. Try this link: en.m.wikipedia.org/wiki/Brewster%27s_angle The picture itself will point your misconception. +1 for good observation. – Wrichik Basu May 14 '17 at 19:06
• What you can do is that buy a nicol's prism and try out the experiment if you have enough resources at hand. – Wrichik Basu May 14 '17 at 19:11

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