Question

A ray of monochromatic light suffers minimum deviation of $38^{\circ }$, while passing through a prism of refracting angle $60^{\circ }.$ Refractive index of the prism material is :

• A. 2.4
• B. 0.8
• C. 1.3
• D. 1.5

Answer 1

Answer: (D)

For a prism of angle $A$ and $\delta$ being angle of minimum deviation, refractive index of the material of prism is given by

$n=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
Given, $A=60^{\circ },$
${\delta }_m=30^{\circ }$
$n=\frac{\sin \left(\frac{60^{\circ }+38^{\circ }}{2}\right)}{\sin \frac{60^{\circ }}{2}}$
$=\frac{\sin 49^{\circ }}{\sin 30^{\circ }}$
$n=\frac{0.7547}{0.5}$
$=1.5$