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Physics · Geometrical Optics

JEE Main 2024 — 27 January, Shift 1 — Question 31

If the refractive index of the material of a prism is cot⁡(A2)\cot \left(\frac{\mathrm{A}}{2}\right), where A is the angle of prism then the angle of minimum deviation will be

  1. Option A:

    π−2 A\pi-2 \mathrm{~A}

    Correct
  2. Option B:

    π2−2 A\frac{\pi}{2}-2 \mathrm{~A}

  3. Option C:

    π−A\pi-A

  4. Option D:

    π2−A\frac{\pi}{2}-\mathrm{A}

Answer: A

Step-by-step solution

cot⁡A2=sin⁡(A+δmin 2)sin⁡A2\cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\text {min }}}{2}\right)}{\sin \frac{\mathrm{A}}{2}}

⇒cos⁡A2=sin⁡(A+δmin⁡2)\Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)

A+δmin 2=π2−A2\frac{\mathrm{A}+\delta_{\text {min }}}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2}

δmin =π−2 A\delta_{\text {min }}=\pi-2 \mathrm{~A}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 1
Subject
Physics
Chapter
Geometrical Optics
Topic
Apparent Depth, Glass Slab, Prism and Dispersion

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