Polarization by Reflection — Question 1

Question 1

Perpendicular reflected and refracted rays determine Brewsters angle

Light meets a flat interface between lossless nonmagnetic dielectrics of unequal indices n1 and n2. At Brewster incidence the reflected p-polarized amplitude vanishes and the reflected and refracted rays are perpendicular. Here p means electric field in the incidence plane.

1. Combine this geometry with Snells law to derive the Brewster angle.

2. For incidence from air into index 1.50, find the incident and refracted angles and identify the polarization of the reflected part of an unpolarized beam.

Question 1 - Solution

Perpendicular reflected and refracted rays determine Brewsters angle

(a) Method and calculation

$$\theta_B+\theta_t=90^\circ,\quad n_1\sin\theta_B=n_2\sin\theta_t=n_2\cos\theta_B.$$$$\boxed{\tan\theta_B=n_2/n_1}.$$

(b) Interpretation and further calculation

$$\boxed{\theta_B=\tan^{-1}(1.50)=56.3099^\circ,\quad\theta_t=33.6901^\circ}.$$$$R_p=0\Rightarrow\boxed{\text{reflected light is }s\text{-polarized: }\mathbf E\perp\text{incidence plane}}.$$

Physical check

The incident unpolarized beam still has an s component that reflects; zero p reflection does not imply zero total reflection.