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 $\sin i =\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}$
$\sin i =\sin 45^{\circ}$
$\therefore i =45^{\circ}$