Question

An equilateral prism produces an angle of deviation $60^{\circ} $ when the angle of incidence is $60^{\circ} $ . The angle of deviation produced when the ray is incident normal to the surface is

• A. $30^{o}$
• B. $45^{o}$
• C. $60^{o}$
• D. can't be determined

Answer 1

Answer: (C)

From angle of deviation formula, $\delta=i+e$ $-A$
$\Rightarrow 60^{\circ}=60^{\circ}+e-60^{\circ}$
$\Rightarrow e=60^{\circ}$, Here $i=e$, so it angle of minimum deviation.
So from the formula, we have $\mu=\frac{\sin \frac{A+\delta}{2}}{\sin \frac{A}{2}}$
$\Rightarrow \mu=\frac{\sin \frac{60^{\circ}+60^{\circ}}{2}}{\sin \frac{60^{\circ}}{2}} $
$\Rightarrow \mu=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
So critical angle for the case is, $C=\sin ^{-1} \frac{1}{\mu}$ $=\sin ^{-1} \frac{1}{\sqrt{3}}=35.26^{\circ}$
So there must be total internal reflection takes place.
So for the given case,
$i^{\prime}=90^{\circ}, r_1=0^{\circ}$
We know, $r_1+r_2=A=60^{\circ}$
$\Rightarrow 0^{\circ}+r_2=60^{\circ}$
$\Rightarrow r_2=60^{\circ}$
So angle of deviation is, $\delta^{\prime}=180^{\circ}-2 r_2$
$\Rightarrow \delta^{\prime}=180^{\circ}-2 \times 60^{\circ}=60^{\circ}$