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 $\theta \left(n\right)$ with the normal. For $n=\sqrt{3}$ the value of $\theta$ is ${60}^o$ and $\frac{d\theta }{dn}=m$. The value of $m$ is

Answer 1

From the geometry we have $r_1+r_2={60}^o$
Using Snells law at the left side we have $\frac{sin{60}^o}{sin{r}_1}=n$
or
$sin{r}_1=\frac{\sqrt{3}}{2n}$
Using Snells law at the right end
$\frac{sin{r}_2}{sin\theta }=\frac{1}{n}$
or
$sin\theta =nsin{r}_2$
$=nsin({60}^o-r_1)$
$=n(\frac{\sqrt{3}}{2}cos{r}_1-\frac{1}{2}sin{r}_1)$
or
$=n(\frac{\sqrt{3}}{2}\sqrt{1-si{n}^2{r}_1}-\frac{1}{2}sin{r}_1)$
or
$=n(\frac{\sqrt{3}}{2}\sqrt{1-\frac{3}{4n^2}}-\frac{1}{2}\frac{\sqrt{3}}{2n})$
Solving we get
$sin\theta =\frac{\sqrt{3}}{4}\sqrt{4n^2-3}-\frac{\sqrt{3}}{4}$
Differentiating
$cos\theta d\theta =\frac{\sqrt{3}}{4}\times \frac{1}{2\sqrt{4n^2-3}}\times 8ndn$
or
$\frac{d\theta }{dn}=\frac{\sqrt{3}n}{\sqrt{4n^2-3}}\times \frac{1}{cos\theta }$

$=\frac{\sqrt{3}\times \sqrt{3}}{\sqrt{4\times 3-3}}\times \frac{1}{1/2}$
$=2$