Question

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

• A. $ 45^{\circ} $
• B. $ 60^{\circ} $
• C. $ 90^{\circ} $
• D. $ 180^{\circ} $

Answer 1

Answer: (A)

In the position of minimum deviation angle of incidence is equal to angle of emergence.
Let a ray of monochromatic light $P Q$ be incident on face $A B$. $P Q R S$ is path of light ray,

where $i$ is angle of incidence $r$ angle of refraction, $r$ angle of incidence and $i$ angle of emergence. In position of minimum deviation
$i^{'}=i, r^{'}=r, \delta=\delta m$
$2 r=A $ or $ r=\frac{A}{2}$
Given $A=60^{\circ}$
$r=\frac{60^{\circ}}{2}=30^{\circ}$
Also from Snell's law
$\mu=\frac{\sin i}{\sin r}=\frac{\sin i}{\sin 30^{\circ}}$
$\sqrt{2}=\frac{\sin i}{\sin 30^{\circ}}$
$\Rightarrow \sin i=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow i=45^{\circ}$