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If minimum deviation $=30^{\circ}$, then speed of light in shown prism will be
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Verified Answer
The correct answer is:
$\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}$
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 $\quad \mu=\sqrt{2}$
$\begin{aligned} \because \quad \mu & =\frac{\text { Speed of light in air (c) }}{\text { Speed of light in prism (v) }} \\ v & =\frac{\mathrm{c}}{\mu}=\frac{3 \times 10^8}{\sqrt{2}}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$
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 $\quad \mu=\sqrt{2}$
$\begin{aligned} \because \quad \mu & =\frac{\text { Speed of light in air (c) }}{\text { Speed of light in prism (v) }} \\ v & =\frac{\mathrm{c}}{\mu}=\frac{3 \times 10^8}{\sqrt{2}}=\frac{3}{\sqrt{2}} \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$
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