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The refractive index of material of glass is $\sqrt{3}$. If the angle of minimum deviation is equal to the angle of prism, the angle of prism is
$\left(\cos 30^{\circ}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}, \sin 30^{\circ}=\frac{1}{2}=\cos 60^{\circ}\right)$
Options:
$\left(\cos 30^{\circ}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}, \sin 30^{\circ}=\frac{1}{2}=\cos 60^{\circ}\right)$
Solution:
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Verified Answer
The correct answer is:
$60^{\circ}$
Given, $\mu=\sqrt{3}$ and $\frac{\sin \left(60^{\circ}\right)}{\sin \left(30^{\circ}\right)}=\sqrt{3}$
When light passes through a prism of refracting angle A, it suffers minimum deviation $\delta$ given by
$\mu=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} \Rightarrow \sqrt{3}=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}$
$\therefore \frac{\mathrm{A}}{2}=30^{\circ}$ and $\frac{\mathrm{A}+\delta}{2}=60^{\circ}$ are the solutions
$\therefore A=60^{\circ}$
When light passes through a prism of refracting angle A, it suffers minimum deviation $\delta$ given by
$\mu=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} \Rightarrow \sqrt{3}=\frac{\sin \left(\frac{\mathrm{A}+\delta}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}$
$\therefore \frac{\mathrm{A}}{2}=30^{\circ}$ and $\frac{\mathrm{A}+\delta}{2}=60^{\circ}$ are the solutions
$\therefore A=60^{\circ}$
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