JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Geometrical Optics

A monochromatic light is incident from air on a refracting surface of a prism of angle 75° and refractive index

n_{0} = 3

. The other refracting surface of a 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 6 0^{\circ}

. The value of

n^{2}

is ____.

Answer: 1.5

Solution

θ = 60°

, the ray hits the second surface at the critical angle.

Using Snell's law at first surface:

sin 60° = 3 sin r_{1}

\frac{3}{2} = 3 sin r_{1}

r_{1} = 30°

Since the prism angle is 75°:

r_{2} = 75° - 30° = 45°

At critical angle:

3 sin 45° = n sin 90°

3 \times \frac{1}{2} = n

n = \frac{3}{2}

n^{2} = \frac{3}{2} = 1.50

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer

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