A monochromatic light is incident from air on a refracting surface of a prism of angle and refractive index . The other refracting surface of the prism is coated by a thin film of material of refractive index , as shown in figure. The light suffers total

Question

A monochromatic light is incident from air on a refracting surface of a prism of angle $75 °$ 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 $θ \leq 60 °$ . The value of $n^{2}$ is _______. [2019]

Smart Achievers · Accepted Answer

(1.50)

In $∆ X Y Z,$ $90 ° \times r + 90 ° - C + 75 ° = 180 °$

$∴$ $r + C = 75 °$ $\Rightarrow r = 75 ° - C$ $\dots (i)$

Applying Snell's law at $Z$ ,

$\sqrt{3} \sin C = n \sin 90 °$

$\sqrt{3} \sin C = n$ $\dots (i i)$

Applying Snell's law at $Y$ ,

$1 \times \sin θ = \sqrt{3} \sin r$

$= \sqrt{3} \sin (75 ° - C)$

From eq. (i),

For $θ = 60 °$

$\frac{\sqrt{3}}{2} = \sqrt{3} \sin (75 ° - C)$ $∴$ $C = 45 °$

From eq. (ii), $n = \sqrt{3} \sin 45 ° = \frac{\sqrt{3}}{\sqrt{2}}$

$∴$ $n^{2} = 1.50$

Solution

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Solution

Ans. (1.50) In ∆XYZ, 90°×r+90°-C+75°=180° ∴ r+C=75°⇒r=75°-C ⋯(i) Applying Snell's law at Z, 3sinC=nsin90° 3sinC=n ⋯(ii) Applying Snell's law at Y, 1×sinθ=3sinr =3sin(75°-C) From eq. (i), For θ=60° 32=3sin(75°-C) ∴ C=45° From eq. (ii), n=3sin45°=32 ∴ n2=1.50

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