JEE Challenger
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PhyMedAdv 2016 • Paper 1

Angle of Right Angled Triangular Prism Total Internal Reflection

A parallel light beam is incident from air at an angle α\alpha onto the side PQPQ of a right-angled triangular prism having refractive index n=2n = \sqrt{2}. Light undergoes total internal reflection inside the prism at face PRPR when α\alpha takes a minimum value of 45∘45^\circ. Determine the prism angle θ\theta.

Question Diagram 1

Options

A

15∘15^\circ

Correct
B

22.5∘22.5^\circ

C

30∘30^\circ

D

45∘45^\circ

Topics & Concepts

Ray Optics and Optical InstrumentsPrismsReflection and Refraction

Step-by-Step Solution

To determine the prism angle θ\theta, we analyze the path of the light beam step-by-step using Snell's Law and the geometry of the prism.

Step 1: Refraction at Face PQPQ

Applying Snell's Law at the air-prism interface on side PQPQ: 1⋅sin⁡α=n⋅sin⁡r11 \cdot \sin \alpha = n \cdot \sin r_1

Given that the refractive index of the prism is n=2n = \sqrt{2}: sin⁡r1=sin⁡α2\sin r_1 = \frac{\sin \alpha}{\sqrt{2}}

Step 2: Relation Between Angle of Incidence at PRPR and Geometry

Consider the triangle formed by the apex PP, the point of incidence on side PQPQ, and the point of incidence on side PRPR.

  • The angle at vertex PP is θ\theta.
  • The normal to face PQPQ is perpendicular to PQPQ, so the angle between the refracted ray and side PQPQ inside the triangle is 90∘+r190^\circ + r_1.
  • Thus, the angle of incidence i2i_2 at face PRPR is related to the prism angle θ\theta and the angle of refraction r1r_1 by: i2=θ+r1i_2 = \theta + r_1

Step 3: Critical Angle for Total Internal Reflection (TIR)

For total internal reflection to occur at the prism-air interface PRPR, the angle of incidence i2i_2 must be greater than or equal to the critical angle θc\theta_c: sin⁡θc=1n=12  ⟹  θc=45∘\sin \theta_c = \frac{1}{n} = \frac{1}{\sqrt{2}} \implies \theta_c = 45^\circ

Thus, the condition for TIR is: i2=θ+r1≥45∘i_2 = \theta + r_1 \ge 45^\circ

Step 4: Minimum Incident Angle α\alpha

Since r1r_1 increases monotonically as α\alpha increases, i2i_2 also increases with α\alpha. Therefore, total internal reflection occurs for all α≥αmin⁡\alpha \ge \alpha_{\min}.

At the threshold value αmin⁡=45∘\alpha_{\min} = 45^\circ, the angle of incidence at PRPR equals the critical angle (i2=45∘i_2 = 45^\circ).

First, find r1r_1 when α=45∘\alpha = 45^\circ: sin⁡r1=sin⁡45∘2=1/22=12  ⟹  r1=30∘\sin r_1 = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \implies r_1 = 30^\circ

Now, substitute r1=30∘r_1 = 30^\circ and i2=45∘i_2 = 45^\circ into i2=θ+r1i_2 = \theta + r_1: 45∘=θ+30∘  ⟹  θ=15∘45^\circ = \theta + 30^\circ \implies \theta = 15^\circ

Thus, the prism angle is θ=15∘\theta = 15^\circ.

Correct Option: (A)

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