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