Question

A monochromatic light is incident from air on a refracting surface of a prism of angle ${75}^o$ and refractive index $n_0=\sqrt{3}$. The other refracting surface of the prism is coated by a thin film of material of refractive index n as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of $\theta \le {60}^o$. The value of $n^2$ is .

Answer 1

For T.I.R at coating
$\sin c=\frac{n}{\sqrt{3}}$
Appling snell's law at first surface
$\sin \theta =\sqrt{3}\sin (75-c)$

for limiting condition, at $\theta ={60}^o$

$\sin 60=\sqrt{3}\sin (75-c)$

$\frac{\sqrt{3}}{2}=\sqrt{3}\sin (75-c)$

$\frac{1}{2}=\sin (75-c)\Rightarrow \sin 30=\sin (75-c)$

$30=75-c\Rightarrow c={45}^o$

Since, $\sin c=\frac{n}{\sqrt{3}}$

$\frac{n}{\sqrt{3}}=\frac{1}{\sqrt{2}}\Rightarrow n^2=\frac{3}{2}=1.50$