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JEE Advanced 2016 · Paper 1, Physics question 4

A parallel beam of light is incident from air at an angle α\alpha on the side PQ of a right angled triangular prism of refractive index n=2n=\sqrt2. Light undergoes total internal reflection in the prism at the face PR when α\alpha has a minimum value of 45°. The angle θ\theta of the prism is

Right-angled triangular prism PQR with the right angle at Q: PQ is the vertical left face, QR the horizontal base and PR the hypotenuse; the apex angle at P is θ. A ray in air strikes face PQ at angle α to the normal (dashed horizontal line). The prism is labelled n = √2.
Choose one option
Answer and worked solution

Official answer: A

Refraction at PQ. sin⁡45°=2sin⁡r\sin45°=\sqrt2\sin r, so r=30°r=30°: inside, the ray runs 30° below the normal to PQ.

Angle at PR. In the triangle formed by P and the two points where the ray meets PQ and PR, the angle at PQ is 90°+30°=120°90°+30°=120°. So the angle at PR is 60°−θ60°-\theta, and the angle of incidence on PR is 90°−(60°−θ)=30°+θ90°-(60°-\theta)=30°+\theta.

Critical angle. For n=2n=\sqrt2, C=45°C=45°. TIR starts exactly when α=45°\alpha=45°, so 30°+θ=45°30°+\theta=45° and θ=15°\theta=15°.

(A)

Topics:Total internal reflectionRefraction at plane surfaces
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