Question

A prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ }$ . At what angle a ray must be incident on it so that it suffers a minimum deviation?

• A. $45^{\circ }$
• B. $60^{\circ }$
• C. $80^{\circ }$
• D. $180^{\circ }$

Answer 1

Answer: (A)

The relation for refractive index of prism is
$\mu =\frac{\sin i}{\sin r}$ ...(i)
The condition for minimum deviation is
$r=\frac{A}{2}=\frac{60^{\circ }}{2}=30^{\circ }$
Putting the given values of $\mu =\sqrt{2}$ and $r=30^{\circ }$ in Eq. (i), we get
$\sqrt{2}=\frac{\sin i}{\sin 30^{\circ }}$
or $\sin i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}$
$\sin i=\sin 45^{\circ }$
$\therefore i=45^{\circ }$