Ray Optics

JEE Main level, single correct, medium

241 of 295

Question 241 of 295

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.

If refractive index of the material of prism is 2\sqrt{2}, the angle θ\theta of prism is.

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Answer: (A) 60∘60^{\circ}

From the geometry of the prism, the angle of the prism is A=180∘−(45∘+θ)A = 180^\circ - (45^\circ + \theta).
The incident ray is parallel to the base, so the angle of incidence ii at the first surface is equal to the base angle 45∘45^\circ. Thus, i=45∘i = 45^\circ.
Applying Snell's law at the first surface: 1⋅sin⁡(45∘)=μsin⁡(r1)1 \cdot \sin(45^\circ) = \mu \sin(r_1).
Given μ=2\mu = \sqrt{2}, we have 12=2sin⁡(r1)  ⟹  sin⁡(r1)=12  ⟹  r1=30∘\frac{1}{\sqrt{2}} = \sqrt{2} \sin(r_1) \implies \sin(r_1) = \frac{1}{2} \implies r_1 = 30^\circ.
At the second surface, the emergent ray grazes the surface, so the angle of emergence e=90∘e = 90^\circ.
Applying Snell's law at the second surface: μsin⁡(r2)=1⋅sin⁡(90∘)\mu \sin(r_2) = 1 \cdot \sin(90^\circ).
2sin⁡(r2)=1  ⟹  sin⁡(r2)=12  ⟹  r2=45∘\sqrt{2} \sin(r_2) = 1 \implies \sin(r_2) = \frac{1}{\sqrt{2}} \implies r_2 = 45^\circ.
The relation between prism angle and internal angles is A=r1+r2A = r_1 + r_2.
A=30∘+45∘=75∘A = 30^\circ + 45^\circ = 75^\circ.
Substituting AA in the first equation: 75∘=180∘−(45∘+θ)75^\circ = 180^\circ - (45^\circ + \theta).
75∘=135∘−θ  ⟹  θ=135∘−75∘=60∘75^\circ = 135^\circ - \theta \implies \theta = 135^\circ - 75^\circ = 60^\circ.

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