Question

A ray of light traveling in water hits a glass surface. The index of refraction of the water is 1.33, and that of the glass is 1.50. At what angle with the plane of the surface must the incident ray strike the glass in order that the polarization of the reflected ray is the greatest

Answer 1

Answer:

$\mu=41.5\textdegree$

Explanation:

From the question we are told that:

Water index of refraction $i_w=1.33$

Glass index of refraction $i_g=1.50$

Generally the equation for Brewster's law is mathematically given by

 $\theta=tan^{-1}(\frac{i_g}{i_w})$

 $\theta=tan^{-1}(\frac{1.50}{1.33})$

 $\theta=48.44 \textdegree$

Therefore Angle of incident to plane  \mu (normal at 90 degree to the surface)

 $\mu=90\textdegree-\theta$

 $\mu=90\textdegree-48.44\textdegree$

 $\mu=41.5\textdegree$