Total internal reflection from a prism

JEE Advanced 2019 Paper 2, Question 11

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 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 \leq 60^{\circ}. What is the value of n^{2}?

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First we trace the given incident ray through the prism to find the angle at which it is incident on the opposite face (see figure below). For instance, we can extend the normals at each face (the dashed lines) to complete a quadrilateral with vertex A sitting opposite the prism angle 75^\circ. But opposite angles of a quadrilateral are supplementary, which means \angle A + 75^\circ = 180^\circ \implies \angle A = 105^\circ.

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It is then apparent that the angle of incidence at the n_0n interface is 75^\circ - \phi, as indicated. Here \phi is the angle of refraction and it is related to \theta by Snell’s Law,

(1)   \begin{equation*}   \sin \theta = n_0 \sin \phi .  \end{equation*}

As \theta is increased, 75^\circ-\phi shrinks, till it drops below the critical angle of the n_0n interface, \Theta_c = \sin^{-1}(n/n_0). At this point total internal reflection no longer occurs. This is why the problem states that \theta has an upper bound of 60^\circ.

If \theta is exactly 60^\circ the ray r graces the n_0n interface and 75^\circ-\phi is the critical angle \Theta_c, which means

(2)   \begin{equation*}   \sin(75^\circ-\phi) = \sin \Theta_c = \frac{n}{n_0} .  \end{equation*}

The angle \phi here can be obtained from (1) by setting \theta=60^\circ and n_0 = \sqrt{3},

(3)   \begin{equation*}   \sin \phi = \frac{\sin 60^\circ}{\sqrt{3}} \implies \phi = 30^\circ . \end{equation*}

Substituting this back in (2), we find

(4)   \begin{equation*}   n = n_0 \sin 45^\circ \implies \boxed{ n^2 = \frac{3}{2} } . \end{equation*}

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