Question

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

Answer 1

According to snell's law at the left face,
$${\mu }_1\sin i_1=\sin r_1...\left(1\right)$$ Where,
${\mu }_1=1$refractive index of air,
${\mu }_2=n=$refractive index of prism
$i_1=$incident angle$={60}^{\circ }$,
$r_1=$refractive angle,

Substituting the values, we get
$1\sin {60}^{\circ }=\sqrt{3}\sin r_1$
$\Rightarrow \sqrt{3}\sin r_1=\frac{\sqrt{3}}{2}$
$\Rightarrow r_1={30}^o...\left(2\right)$

By differentiating equation $\left(1\right)$ with respect to $n$, we get

$\frac{d\left({\mu }_1\sin i_1\right)}{dn}=\frac{d\left(n\right)}{dn}\sin r_1+n\frac{d\left(\sin r_1\right)}{dn}$

$\Rightarrow 0=\sin r_1+n\cos r_1\left(\frac{d{r}_1}{dn}\right)$

Substituting the value of $r_1$,

$\Rightarrow 0=\sin {30}^{\circ }+n\cos {30}^{\circ }\left(\frac{d{r}_1}{dn}\right)$

$\Rightarrow 0=\frac{1}{2}+\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\frac{d{r}_1}{dn}$

$\frac{d{r}_1}{dn}=-\frac{1}{3}...\left(3\right)$

Applying snell's law at the right face of prism, we have
$$n\sin r_2={\mu }_2\sin \theta$$where,
${\mu }_2=1=$refractive index of air.
from geometry we have,
$r_1+r_2={60}^{\circ }$

Substituting the values, we get
$n\sin (60-r_2)=1\sin \theta$
by diffrentiating with respect to $n$,

$\sin (60-r_1)-n\cos (60-r_1)\frac{d{r}_1}{dn}=\cos \theta \frac{d\theta }{dn}$

Substituting the values from equation $\left(2\right)$ and $\left(3\right)$, we get
$\sin ({60}^{\circ }-{30}^{\circ })-\sqrt{3}\cos ({60}^{\circ }-{30}^{\circ })\times (-\frac{1}{3})=\cos {60}^{\circ }\frac{d\theta }{dn}$

$\Rightarrow \frac{1}{2}-\sqrt{3}\times \frac{\sqrt{3}}{2}\times (-\frac{1}{3})=\frac{1}{2}\times \frac{d\theta }{dn}$

$\therefore \frac{d\theta }{dn}=2$