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 $θ \leq 60^{\circ} .$ The value of $10 n^{2}$ is

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Solution

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

Where,
$θ_{c} \to$ Critical angle as shown in the figure.

Applying Snell's law at face $PQ$ :

$μ_{a i r} sin θ = μ_{p r i s m} sin r_{1} . . . (1)$

where,
$μ_{a i r} \to$ Refractive index of air.
$μ_{p r i s m} \to$ Refractive index of prism.

According to problem, $μ_{a i r} = 1$ and $μ_{p r i s m} = n_{0} = \sqrt{3}$

Substituting these in equation $(1)$ , we get

$\Rightarrow 1 \times sin θ = \sqrt{3} \times sin r_{1} . . . . . . . . . . (2)$

Since for the case of total internal reflection at the surface $PR$ ,
$r_{1} = (A - θ_{c})$

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

$∴ r_{1} = (75^{\circ} - θ_{c})$

Now, from equation $(2)$ ,

$\Rightarrow sin θ = \sqrt{3} sin (75^{\circ} - θ_{c})$

For $θ = 60^{\circ}$ ,

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

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

$∴ θ_{c} = 45^{\circ}$

According to problem,

$sin θ_{c} = \frac{n}{\sqrt{3}}$

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

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

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

$∴ n^{2} = \frac{3}{2} = 1.5$

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

Accepted Answer: 15

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

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