# Why does light get polarised when reflected and refracted?

May 31, 2018

Light is a transverse wave, which means that the electric field (as well as the magnetic field) is perpendicular to the direction of propagation of light (at least in isotropic media - but let's keep things simple here).

So when light is incident obliquely on the boundary of two media, the electric field can be thought of as having two components - one in the plane of incidence, and one perpendicular to it. For unpolarized light, the direction of the electric field fluctuates randomly (while staying perpendicular to the direction of propagation) and as a result, the two components are of equal order in size.

How much of each component is transmitted and reflected can be figured out by using the laws of electromagnetism. Without going into the mathematical details here let me quote a result - the $p$ polarization (the component of the electric field in the plane of incidence) does not get reflected at all if the angle of incidence $\theta$ satisfies

$\theta = {\theta}_{B} = {\tan}^{-} 1 \left({n}_{2} / {n}_{1}\right)$

where ${n}_{1}$ and ${n}_{2}$ are the two refractive indexes. This angle is called the Brewster angle. For light traveling from air to water this has a value of about ${53}^{\circ}$

The other, so called $s$ component, the one perpendicular to the plane of incidence, is both transmitted and reflected for all angles of incidence (though the fraction changes with $\theta$).

So, when light is incident at the Brewster angle - the reflected beam has no component in the plane of polarization. It is entirely $s$ polarized. This is why the Brewster angle is also called the polarizing angle.

For angles of incidence close to the Brewster angle, the reflected beam has a lot more $s$ polarization than $p$ polarization - and so it is partially polarized.

The transmitted beam has both components and hence is never completely polarized. However, at the Brewster angle, all of the $p$ component is transmitted (no reflection) and only a part of the $s$ component is transmitted (the rest is reflected) - so the transmitted light is partially $p$ polarized. This remains true for angles close to the Brewster angle as well.