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QJEEAdv19Phe3d5: JEE Advanced 2019 Physics - Class 12 JEEAdv19Phe3d5 Solution: A monochromatic light is incident from air on a refracting…

A monochromatic light is incident from air on a refracting surface of a prism of angle 75 ∘^\circ and refractive index n0=3{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≤60∘\theta \le 60^\circ . The value of n2 is .............
JEE Advanced 2019 Paper 2 Offline Physics - Geometrical Optics Question 49 English

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For TIR at coating,
JEE Advanced 2019 Paper 2 Offline Physics - Geometrical Optics Question 49 English Explanation

Applying Snell's law at first surface sin⁡θ=3sin⁡(75∘−θc)\sin \theta = \sqrt 3 \sin (75^\circ - {\theta _c})

For limiting condition, at θ\theta = 60 ∘^\circ

sin⁡60∘=3sin⁡(75∘−θc)\sin 60^\circ = \sqrt 3 \sin (75^\circ - {\theta _c})

32=3sin⁡(75∘−θc){{\sqrt 3 } \over 2} = \sqrt 3 \sin (75^\circ - {\theta _c})

12=sin⁡(75∘−θc){1 \over 2} = \sin (75^\circ - {\theta _c})

⇒\Rightarrow sin30∘=sin⁡(75∘−θc)sin30^\circ = \sin (75^\circ - {\theta _c})

30∘30^\circ = 75∘−θc75^\circ - {\theta _c} ⇒\Rightarrow θc{\theta _c} = 45 ∘^\circ

Now since θc < 60o, therefore total internal reflection takes place. For total internal reflection at the other refracting surface, using Snell’s law, we have

n3=12⇒n2=32=1.50{n \over {\sqrt 3 }} = {1 \over {\sqrt 2 }} \Rightarrow {n^2} = {3 \over 2} = 1.50

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