If light strikes an interface so that there is a 90^o angle between the reflected and refracted rays, the reflected light will be linearly polarized. The direction of polarization (the way the electric field vectors point)is parallel to the plane of the interface.

The special angle of incidence that produces a 90^o angle between the reflected and refracted ray is called the Brewster angle, q_p.

A little geometry shows that tan(q_p) = n₂/n₁.

Why is the reflected light polarized? Let's say the incident light is unpolarized. When the incident light crosses the interface the light is absorbed temporarily by atoms in the seond medium. Electrons in these atoms oscillate back and forth in the direction of the electric field vectors in the refracted ray, perpendicular to the direction the refracted light is traveling.

The light is re-emitted by the atoms to form both the reflected and refracted rays. The electric field vectors in the light match the direction the electrons were oscillating, and they must be perpendicular to the direction of propagation of the wave. When light comes in at the Brewster angle the reflected wave has no electric field vectors parallel to the refracted ray, because the electrons do not oscillate along that direction. The reflected wave also has no electric field vectors parallel to the reflected ray, because that's the direction of propagation of the wave. The only direction possible is perpendicular to the plane of the picture, so the reflected ray is linearly polarized.

The refracted ray is partly polarized because it has more light with electric field vectors in the plane of the picture than perpendicular to it.

If the angle of incidence is something other than 0^o and the Brewster angle, the reflected ray is also partly polarized.