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 $PQ$ be incident on face $AB$. $PQRS$ 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
$i^{{}^{\prime }}=i,r^{{}^{\prime }}=r,\delta =\delta m$
$2r=A$ or $r=\frac{A}{2}$
Given $A=60^{\circ }$
$r=\frac{60^{\circ }}{2}=30^{\circ }$
Also from Snell's law
$\mu =\frac{\sin i}{\sin r}=\frac{\sin i}{\sin 30^{\circ }}$
$\sqrt{2}=\frac{\sin i}{\sin 30^{\circ }}$
$\Rightarrow \sin i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow i=45^{\circ }$