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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Ray Optics

JIPMER 2018 · Physics · Ray Optics

If minimum deviation =30^ , then speed of light in shown prism will...

If minimum deviation =30∘=30^{\circ}, then speed of light in shown prism will be If minimum deviation =30^ , then speed of light in shown prism will be

Options

  1. A32×108 m/s\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}
  2. B12×108 m/s\frac{1}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}
  3. C23×108 m/s\frac{2}{\sqrt{3}} \times 10^8 \mathrm{~m} / \mathrm{s}
  4. D23×108 m/s\frac{2}{3} \times 10^8 \mathrm{~m} / \mathrm{s}

Correct answer

A. 32×108 m/s\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}

Step-by-step solution

Given, A=60∘,δm=30∘\mathrm{A}=60^{\circ}, \delta_{\mathrm{m}}=30^{\circ} According to prism formula, μ=sin⁡(A+δm2)sin⁡(A2)orμ=sin⁡(60∘+30∘2)sin⁡60∘2=sin⁡45∘sin⁡30∘orμ=2\mu=\frac{\sin \left(\frac{A+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} or \mu=\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \frac{60^{\circ}}{2}}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}} or \quad \mu=\sqrt{2}∵μ= Speed of light in air (c)  Speed of light in prism (v) v=cμ=3×1082=32×108m/s\begin{aligned} \because \quad \mu & =\frac{\text{ Speed of light in air (c) }}{\text{ Speed of light in prism (v) }} \\ v & =\frac{\mathrm{c}}{\mu}=\frac{3 \times 10^8}{\sqrt{2}}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{ m} / \mathrm{s}\end{aligned}$$
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