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← JEE Advanced 2015 · Paper 2
PhysicsInteger+4Question 6 of 60

JEE Advanced 2015 · Paper 2, Physics question 6

A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index nn and emerges from the opposite face making an angle θ(n)\theta(n) with the normal (see the figure). For n=3n=\sqrt3 the value of θ\theta is 60° and dθdn=m\dfrac{d\theta}{dn}=m. The value of mm is

An equilateral prism; a ray strikes the left face at 60° to the normal, travels parallel to the base inside, and emerges from the right face at angle θ to the normal.
Answer and worked solution

Official answer: 2

Snell's law at both faces, with r1+r2=60°r_1+r_2=60°:
sin⁡60°=nsin⁡r1,sin⁡θ=nsin⁡r2.\sin60°=n\sin r_1,\qquad\sin\theta=n\sin r_2.
At n=3n=\sqrt3: r1=r2=30°r_1=r_2=30° and θ=60°\theta=60°.

Differentiate.

  • First face: 0=sin⁡r1+ncos⁡r1dr1dn0=\sin r_1+n\cos r_1\dfrac{dr_1}{dn}, so dr1dn=−tan⁡30°3=−13\dfrac{dr_1}{dn}=-\dfrac{\tan30°}{\sqrt3}=-\dfrac13. Hence dr2dn=+13\dfrac{dr_2}{dn}=+\dfrac13.
  • Second face: cos⁡θdθdn=sin⁡r2+ncos⁡r2dr2dn=12+3⋅32⋅13=1\cos\theta\dfrac{d\theta}{dn}=\sin r_2+n\cos r_2\dfrac{dr_2}{dn}=\dfrac12+\sqrt3\cdot\dfrac{\sqrt3}2\cdot\dfrac13=1.

Result. dθdn=1cos⁡60°=2\dfrac{d\theta}{dn}=\dfrac1{\cos60°}=2.

Answer: 2

Topics:Prism and dispersionRefraction at plane surfaces
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