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Physics · Geometrical Optics

JEE Main 2026 — 5 April, Morning Shift — Question 13

A ray of light passing through an equilateral prism is having velocity 2.12×1082.12 \times 10^8 m/s in the prism material, then the minimum angle of deviation is ______ degrees.

  1. Option A:

    45

  2. Option B:

    30

    Correct
  3. Option C:

    28

  4. Option D:

    58

Answer: B

Step-by-step solution

μ=c/v=3×108/2.12×108=2\mu = c/v = 3\times10^8 / 2.12\times10^8 = \sqrt{2}. For prism, μ=sin⁡((δm+A)/2)sin⁡(A/2)\mu = \frac{\sin((\delta_m+A)/2)}{\sin(A/2)}. With A=60°, sin⁡30∘=0.5\sin30^\circ=0.5, so sin⁡((δm+60)/2)=2×0.5=1/2\sin((\delta_m+60)/2) = \sqrt{2}\times0.5 = 1/\sqrt{2}, thus (δm+60)/2=45∘(\delta_m+60)/2 = 45^\circ, δm=30∘\delta_m = 30^\circ.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Morning Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Apparent Depth, Glass Slab, Prism and Dispersion

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