Question

The polarizing angle for a piece of glass for green light is ${60}^{\circ }$. Find the angle of minimum deviation for green light for its passage through ${60}^{\circ }$ prism, made of same glass.

• A. ${30}^{\circ }$
• B. ${15}^{\circ }$
• C. ${60}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Given:
Polarizing angle, $i_p={60}^{\circ }$
Angle of prism, $A={60}^{\circ }$

According to Brewster's law

$\mu =\tan i_p=\tan {60}^{\circ }=\sqrt{3}$

Also, from prism formula

$\mu =\frac{\sin \left\{\frac{(A+{\delta }_m)}{2}\right\}}{\sin \left\{\frac{A}{2}\right\}}$

Where, ${\delta }_m$ is the angle of minimum deviation

$\Rightarrow \sqrt{3}=\frac{\sin \left\{\frac{{60}^{\circ }+\delta m}{2}\right\}}{\sin \left\{\frac{{60}^{\circ }}{2}\right\}}$

$\Rightarrow \sin \left\{\frac{{60}^{\circ }+{\delta }_m}{2}\right\}=\sqrt{3}\sin {30}^{\circ }$

$\Rightarrow \sin \left\{\frac{{60}^{\circ }+{\delta }_m}{2}\right\}=\frac{\sqrt{3}}{2}=\sin {60}^{\circ }$

$\Rightarrow \frac{{60}^{\circ }+\delta m}{2}={60}^{\circ }$

$\Rightarrow {\delta }_m={120}^{\circ }-{60}^{\circ }$

$\Rightarrow {\delta }_m={60}^{\circ }$

Hence, option $\left(\text{C}\right)$ is correct.