Question

If refractive index of a material of equilateral prism is $\sqrt{3}$, then angle of minimum deviation of the prism is

• A. $30^{\circ }$
• B. $45^{\circ }$
• C. $60^{\circ }$
• D. $75^{\circ }$

Answer 1

Answer: (C)

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