Question

Question: Light that is traveling in water (with an index of refraction of 1.33) is incident on a plane of glass (with index of refraction of 1.53). At what angle of incidence does the reflected light end up fully polarized?

Short answer

The angle of incidence for which the reflected ray is fully polarized is${49}^o$.

Step-by-step solution

Given

1. The refractive index of water is$n_1=1.33$
2. The refractive index of glass is $n_2=1.53$

Understanding the concept

Brewster’s angle is the angle of incidence, of a light travelling from the rarer to the denser medium, for which the reflected ray is polarized.

Formula:

Brewster’s angle

${\theta }_B={\tan }^{-1}\left(\frac{n_2}{n_1}\right)$

Calculate the value of n3.

The angle of incidence, at which the reflected light is fully polarized, is-

$\begin{array}{rcl}{\mathrm{\theta }}_{\mathrm{B}}& =& {\tan }^{-1}\left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right)\\ & =& {\tan }^{-1}\left(\frac{1.53}{1.33}\right)\\ & =& {49}^{\circ }\end{array}$

The reflected light is fully polarized for the angle of incidence ${49}^{\circ }$.