JEE Advanced 2019 · Paper 2 · Physics Q11

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PhysicsNumerical+3Question 11 of 54

A monochromatic light is incident from air on a refracting surface of a prism of angle 75° and refractive index $n_0 = \sqrt3$ . 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 $\theta \le 60^\circ$ . The value of $n^2$ is ______.

$A prism of apex angle 75° and refractive index n₀ = √3. A ray from air meets the left face at incidence angle θ; the right face is coated with a thin film of refractive index n (drawn as a thick black layer).$

Your answer

Answer and worked solution

Official answer: 1.5

At the limiting incidence $\theta=60^\circ$ .
$\sin60^\circ=\sqrt3\sin r_1\implies r_1=30^\circ,\qquad r_2=A-r_1=45^\circ.$
Smaller $\theta$ gives larger $r_2$ , so total internal reflection for all $\theta\le60^\circ$ means $r_2=45^\circ$ is the critical angle at the film.

Film index.
$\sqrt3\sin45^\circ=n\implies n^2=\frac32=1.50.$
Answer: 1.50

Topics:Prism and dispersion Total internal reflection

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JEE Advanced 2019 · Paper 2, Physics question 11