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Physics · Geometrical Optics

JEE Main 2026 — 28 January, Evening Shift — Question 43

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index nn of the prism satisfies.

  1. Option A:

    2<n<22\sqrt{2}<\mathrm{n}<2 \sqrt{2}

  2. Option B:

    1<1< n <2<2

    Correct
  3. Option C:

    n≥2n \geq 2

  4. Option D:

    2<n<2\sqrt{2}<\mathrm{n}<2

Answer: B

Step-by-step solution

δmin =2i−A⇒i=δmin =A\delta_{\text {min }}=2 \mathrm{i}-\mathrm{A} \Rightarrow \mathrm{i}=\delta_{\text {min }}=\mathrm{A} Also, μ=sin⁡(δmin +A2)sin⁡(A2)\mu=\frac{\sin \left(\frac{\delta_{\text {min }}+\mathrm{A}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} ⇒μ=sin⁡Asin⁡A2=2cos⁡( A2)\Rightarrow \mu=\frac{\sin \mathrm{A}}{\sin \frac{\mathrm{A}}{2}}=2 \cos \left(\frac{\mathrm{~A}}{2}\right) Therefore, 1<μ<21<\mu<2.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Apparent Depth, Glass Slab, Prism and Dispersion

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