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, ${\delta }^{\prime }=180^{\circ }-2r_2$
$\Rightarrow {\delta }^{\prime }=180^{\circ }-2\times 60^{\circ }=60^{\circ }$