JEE Challenger
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PhyHardAdv 2019 • Paper 2

Total Internal Reflection in Coated Prism for Maximum Incidence Angle

A monochromatic light beam travels from air and strikes a refracting surface of a prism with angle 75∘75^\circ and refractive index n0=3n_0 = \sqrt{3}. The second refracting surface of the prism is covered with a thin film of a material having refractive index nn as shown in the diagram. The light beam undergoes total internal reflection at the coated surface for any angle of incidence θ≤60∘\theta \le 60^\circ. The value of n2n^2 is _____.

Question Diagram 1
Official Numerical Answer1.5

Topics & Concepts

Ray Optics and Optical InstrumentsPrismsReflection and Refraction

Step-by-Step Solution

To find the value of n2n^2, we analyze the path of the light ray through the prism step-by-step.

1. Refraction at the First Surface

The light beam is incident from air (refractive index 11) onto the first surface of the prism at an angle of incidence θ\theta.

By Snell's law at the first surface: 1⋅sin⁡θ=n0sin⁡r11 \cdot \sin\theta = n_0 \sin r_1

Given that the refractive index of the prism is n0=3n_0 = \sqrt{3}: sin⁡r1=sin⁡θ3\sin r_1 = \frac{\sin\theta}{\sqrt{3}}

2. Angle of Incidence at the Second Surface

For a prism with apex angle A=75∘A = 75^\circ, the angles of refraction r1r_1 and r2r_2 inside the prism are related by: r1+r2=A=75∘r_1 + r_2 = A = 75^\circ r2=75∘−r1r_2 = 75^\circ - r_1

3. Condition for Total Internal Reflection (TIR)

For Total Internal Reflection to occur at the second (coated) surface, the angle of incidence r2r_2 must be greater than or equal to the critical angle θc\theta_c between the prism material (n0n_0) and the thin film material (nn): r2≥θcr_2 \ge \theta_c

The critical angle θc\theta_c is given by: sin⁡θc=nn0=n3\sin\theta_c = \frac{n}{n_0} = \frac{n}{\sqrt{3}}

4. Determination of Critical Angle

As the angle of incidence θ\theta increases:

  • sin⁡θ\sin\theta increases, which implies r1r_1 increases.
  • Consequently, r2=75∘−r1r_2 = 75^\circ - r_1 decreases.

Thus, r2r_2 attains its minimum value at the maximum incidence angle θ=60∘\theta = 60^\circ.

Since TIR occurs for all θ≤60∘\theta \le 60^\circ, the boundary condition (limiting case for TIR) is reached when θ=60∘\theta = 60^\circ. At this angle, r2r_2 must equal the critical angle θc\theta_c.

At θ=60∘\theta = 60^\circ: sin⁡r1=sin⁡60∘3=323=12\sin r_1 = \frac{\sin 60^\circ}{\sqrt{3}} = \frac{\frac{\sqrt{3}}{2}}{\sqrt{3}} = \frac{1}{2}

This gives: r1=30∘r_1 = 30^\circ

The corresponding angle of incidence at the second surface is: r2=75∘−30∘=45∘r_2 = 75^\circ - 30^\circ = 45^\circ

Therefore, the critical angle is: θc=45∘\theta_c = 45^\circ

5. Calculation of n2n^2

Using the definition of the critical angle: sin⁡θc=nn0\sin\theta_c = \frac{n}{n_0} sin⁡45∘=n3\sin 45^\circ = \frac{n}{\sqrt{3}} 12=n3  ⟹  n=32\frac{1}{\sqrt{2}} = \frac{n}{\sqrt{3}} \implies n = \sqrt{\frac{3}{2}}

Squaring both sides gives: n2=32=1.5n^2 = \frac{3}{2} = 1.5

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