Question

Glass plate of refractive index $1.732$ is to be used as a polariser, its polarising angle is _________.

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${60}^o$

The diagram shows a glass plate to which a light ray is incident at angle of incidence equal to the polarising angle$I_p.$

Most part of this ray passes through the glass after refraction as refracted ray at an angle of refraction $r.$

Some part of the incident ray gets reflected and comes out on the air. Angle of reflection is obviously equal to angle of incidence $I.$

When $I=I_p,$ the angle formed by refracted ray and reflected ray is ${90}^o$

From the figure we can understand that $I_p+90+r=180$ $\therefore r=90-I_p$

By snell's law, refractive index $\mu =\frac{sin\left(I\right)}{sin\left(r\right)}$

In this case $\mu =\frac{sin\left(I_p\right)}{sin(90-I_p)}=\frac{sin\left(I_p\right)}{cos\left(I_p\right)}=tan\left(I_p\right)$

$\therefore I_p=ta{n}^{-1}\mu$

In the given problem, $\mu =1.732$ $\therefore I_p=ta{n}^{-1}\left(1.732\right)={60}^o$