Question

A monochromatic light is incident from air on a refracting surface of a prism of angle ${75}^{\circ }$ and refractive index $n_0=\sqrt{3}.$ The other refracting surface of the prism is coated by a film of material of refractive index $n$ as shown in figure. The ray of light suffers total internal reflection at the coated prism surface for an incidence angle of $\theta \le {60}^{\circ }.$ The value of $10n^2$ is

Answer 1

For the total internal reflection at face $\text{PR,}$ we can write,
$\sin {\theta }_c=\frac{n}{\sqrt{3}}$

Where,
${\theta }_c\to$ Critical angle as shown in the figure.

Applying Snell's law at face $\text{PQ}$ :

${\mu }_{air}\sin \theta ={\mu }_{prism}\sin r_1...\left(1\right)$

where,
${\mu }_{air}\to$ Refractive index of air.
${\mu }_{prism}\to$ Refractive index of prism.

According to problem, ${\mu }_{air}=1$ and ${\mu }_{prism}=n_0=\sqrt{3}$

Substituting these in equation $\left(1\right)$, we get

$\Rightarrow 1\times \sin \theta =\sqrt{3}\times \sin r_1..........\left(2\right)$

Since for the case of total internal reflection at the surface $\text{PR}$,
$r_1=(A-{\theta }_c)$

where,
$A\to$ Prism angle$={75}^{\circ }\left(\text{given}\right)$

$\therefore r_1=({75}^{\circ }-{\theta }_c)$

Now, from equation $\left(2\right)$,

$\Rightarrow \sin \theta =\sqrt{3}\sin ({75}^{\circ }-{\theta }_c)$

For $\theta ={60}^{\circ }$ ,

$\sin {60}^{\circ }=\sqrt{3}\sin ({75}^{\circ }-{\theta }_c)$

$\Rightarrow \frac{\sqrt{3}}{2}=\sqrt{3}\sin ({75}^{\circ }-{\theta }_c)$

$\therefore {\theta }_c={45}^{\circ }$

According to problem,

$\sin {\theta }_c=\frac{n}{\sqrt{3}}$

$\Rightarrow \sin {45}^{\circ }=\frac{n}{\sqrt{3}}$

$=\frac{1}{\sqrt{2}}=\frac{n}{\sqrt{3}}$

$\Rightarrow n=\sqrt{\frac{3}{2}}$

$\therefore n^2=\frac{3}{2}=1.5$

From this, the value of $10n^2=15$

Accepted Answer: 15

Why this Question : This question tests the concept of multiple refractions and $\text{TIR}$ in a Prism.