Question

A 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 $\theta$ with the normal. For $n=\sqrt{3},$ the value of $\theta$ is ${60}^{\circ }$ and $\frac{d\theta }{dn}=m.$ The value of $m$ is.

Answer 1

Given,
Angle of incidence, $i_1={60}^{\circ }$
Refractive index of material of the prism, $n=\sqrt{3}$
Prism angle, $A={60}^{\circ }$

Applying Snell's law at the face $\text{AB}$,

$n_{\text{air}}\sin i_1=n\sin r_1$

$1\sin {60}^{\circ }=n\sin r_1\Rightarrow \sin r_1=\frac{\sqrt{3}}{2n}$
also, $\cos r_1=\sqrt{1-\frac{3}{4n^2}}$


Applying Snell's law on the face $\text{AC}$ we get,

$n\sin r_2=n_{\text{air}}\sin \theta$

Substituting the given values we get,

$n\sin r_2=1\sin \theta$

$\Rightarrow \sin \theta =n\sin r_2=n\sin ({60}^{\circ }-r_1)$
$\Rightarrow \sin \theta =n[\sin {60}^{\circ }\cos r_1-\cos {60}^{\circ }\sin r_1]$
$\Rightarrow \sin \theta =n[(\frac{\sqrt{3}}{2}\times \sqrt{1-\frac{3}{4n^2}})-(\frac{1}{2}\times \frac{\sqrt{3}}{2n})]$
$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}\times \frac{\sqrt{4n^2-3}}{2}-\frac{\sqrt{3}}{4}$

Differentiating w.r.t. $n$ on both sides, we get

$\cos \theta \frac{d\theta }{dn}=\frac{\sqrt{3}}{4}\times \frac{1}{2\times \sqrt{4n^2-3}}\times 8n$
$\Rightarrow \frac{d\theta }{dn}=\frac{\sqrt{3}n}{\cos \theta \times \sqrt{4n^2-3}}$

Given, $\theta ={60}^{\circ },n=\sqrt{3}$

$\therefore \frac{d\theta }{dn}=\frac{\sqrt{3}\times \sqrt{3}}{\frac{1}{2}\times \sqrt{4\times (\sqrt{3}{)}^2-3}}=\frac{6}{3}=2$

From the data given in the question, $\frac{d\theta }{dn}=m$

By comparing we get, $m=2$