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

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

Applying Snell’s law at $Q$
$sin \,\theta = n\, sin (60^{\circ} - r)\,\, ...(iii)$
Differentiating the above equation w.r.t ‘$n$’ we get
$cos\,\theta \frac{d\theta}{dn} = sin (60^{\circ} - r) + n\,cos(60^{\circ} - r)[-\frac{dr}{dn}]$
$\therefore cos\,\theta \frac{d\theta}{dn} = sin(60^{\circ} - r) - n\,cos(60^{\circ} - r)$
$[-\frac{tan\,r}{n}] \,\,$[from (ii)]
$\therefore \frac{d\theta}{dn} = \frac{1}{cos\,\theta}[sin(60^{\circ} - r) + cos (60^{\circ} - r) tan\,r] ...(iv)$
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
$\frac{d\theta}{dn} = \frac{1}{cos\,60^{\circ}} [sin\,30^{\circ} + cos\,30^{\circ} \,tan\,30^{\circ}] = 2$