Question

The monochromatic beam of light is incident at $60^{\circ }$ on one face of an equilateral prism of refractive index $n$ and emerges from the opposite face making an angle $\theta (n)$ with the normal (see the figure). For $n=\sqrt{3}$ the value of $\theta$ is $60^{\circ }$ and $\frac{d\theta }{dn}=m$. The value of $m$ is

Answer 1

Answer: 2

Here $\angle MPQ+\angle MQP=60^{\circ }$. If $\angle MPQ=r$ then $\angle MQP$
$=60-r$
Applying Snell’s law at $P$
$sin60^{\circ }=nsinr...(i)$
Differentiating w.r.t ‘$n$’ we get
$O=sinr+ncosr\times \frac{dr}{dn}...(ii)$

Applying Snell’s law at $Q$
$sin\theta =nsin(60^{\circ }-r)...(iii)$
Differentiating the above equation w.r.t ‘$n$’ we get
$cos\theta \frac{d\theta }{dn}=sin(60^{\circ }-r)+ncos(60^{\circ }-r)[-\frac{dr}{dn}]$
$\therefore cos\theta \frac{d\theta }{dn}=sin(60^{\circ }-r)-ncos(60^{\circ }-r)$
$[-\frac{tanr}{n}]$[from (ii)]
$\therefore \frac{d\theta }{dn}=\frac{1}{cos\theta }[sin(60^{\circ }-r)+cos(60^{\circ }-r)tanr]...(iv)$
From eq. (i), substituting $n=\sqrt{3}$ we get $r=30^{\circ }$
From eq (iii), substituting $n=\sqrt{3},r=30^{\circ }$
we get $\theta =60^{\circ }$
On substituting the values of $r$ and $\theta$ in eq (iv) we get
$\frac{d\theta }{dn}=\frac{1}{cos60^{\circ }}[sin30^{\circ }+cos30^{\circ }tan30^{\circ }]=2$