Question

The light ray is incident at angle of $60^{\circ}$ on a prism of angle $45^{\circ}$. When the light rays falls on the other surface at $90^{\circ}$, the refractive index of the material of prism mu and the angle of deviation $\delta$ are given by

• A. $\mu=\sqrt{2}, \delta=30^{\circ}$
• B. $\mu=1.5, \delta=15^{\circ}$
• C. $\mu=\frac{\sqrt{3}}{2}, \delta=30^{\circ}$
• D. $\mu=\sqrt{\frac{3}{2}}, \delta=15^{\circ}$

Answer 1

Answer: (D)

From the problem it is clear that $\angle e=\angle r_{2}=0$
From $A=r_{1}+r_{2}$
$\Rightarrow r_{1}= A =45^{\circ}$
$\therefore \mu=\frac{\sin i}{\sin r_{1}}=\frac{\sin 60^{\circ}}{\sin 45^{\circ}}=\sqrt{\frac{3}{2}}$
Also, from $i+e=A+\delta$
$\Rightarrow 60+0=45+\delta $
$\Rightarrow \delta=15^{\circ}$