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A monochromatic light is incident from air on a refracting surface of a prism of angle $75^{0}$ 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 $θ \leq 60^{0}$ . The value of $n^{2}$ is x.Then 2x is

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answer is 3.

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Detailed Solution

$I n Δ X Y Z, 90^{0} \times r + 90^{0} - C + 75^{0} = 180^{0} ∴ r + C = 75 \Rightarrow r = 75^{0} - C ..... (i)$

Applying Snell’s law at Z $\sqrt{3} \sin C = n \sin 90^{0} \sqrt{3} \sin C = n$

Applying Snell’s law at Y $1 \times s i n θ = \sqrt{3} s i n r = \sqrt{3} s i n (75^{0} - C)$

From eq.(i) For $θ = 60^{0}$ $\frac{\sqrt{3}}{2} = \sqrt{3} \sin (75^{0} - C) ∴ C = 45^{0}$

From eq (ii) $n = \sqrt{3} \sin 45^{0} = \frac{\sqrt{3}}{\sqrt{2}}$ $∴ n^{2} = 1.50$

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