Q104 / 180 · JIPMER 2018 · Physics

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

  • Correct answer
    A
    $\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}$
  • B
    $\frac{1}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}$
  • C
    $\frac{2}{\sqrt{3}} \times 10^8 \mathrm{~m} / \mathrm{s}$
  • D
    $\frac{2}{3} \times 10^8 \mathrm{~m} / \mathrm{s}$

Solution

Given, $\mathrm{A}=60^{\circ}, \delta_{\mathrm{m}}=30^{\circ}$ According to prism formula, $\mu=\frac{\sin \left(\frac{A+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}$ 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 $\quad \mu=\sqrt{2}$ $\begin{aligned} \because \quad \mu & =\frac{\text { Speed of light in air (c) }}{\text { Speed of light in prism (v) }} \\ v & =\frac{\mathrm{c}}{\mu}=\frac{3 \times 10^8}{\sqrt{2}}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$

Hence, the correct answer is option: 'A'.

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