Question

The polarizing angle for a piece of glass for green light is ${54.74}^{\circ }$. The angle of minimum deviation for an equilateral prism made of same glass is,
(Given : $\tan (54.74{)}^{\circ }=1.414$)

• A. ${45}^{\circ }$
• B. ${54.74}^{\circ }$
• C. ${60}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is D ${30}^{\circ }$

As we know,
$\mu =\tan i_p=\tan (54.74{)}^{\circ }=1.414=\sqrt{2}$

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

Here, $A={60}^{\circ }\sin \left(\frac{A}{2}\right)=\frac{1}{2}$

And $\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}$

$\therefore \sin \left(\frac{{60}^{\circ }+{\delta }_m}{2}\right)=\frac{1}{\sqrt{2}}$


$\sin \left(\frac{{60}^{\circ }+{\delta }_m}{2}\right)=\sin {45}^{\circ }$

$\frac{{60}^{\circ }+{\delta }_m}{2}={45}^{\circ }$

${\delta }_m={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

Hence, $\left(D\right)$ is the correct answer.