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A prism is made of a glass having refractive index $\sqrt{2}$. If the angle of minimum deviation is equal to angle of the prism, then the angle of prism is
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Verified Answer
The correct answer is:
$90^{\circ}$
Let $\delta_m$ be the angle of minimum deviation.
As we know that,
$\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}$
Given, $\quad \mu=\sqrt{2}$ and $A=\delta_m$
$\Rightarrow \quad \sqrt{2}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin \frac{A}{\sin \frac{A}{2}}}{\sin }$
$\Rightarrow \quad \sqrt{2}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}[\because \sin 2 A=2 \sin A \cos A]$
$\sqrt{2}=2 \cos \frac{A}{2} \Rightarrow \cos \frac{A}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \quad \frac{A}{2}=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \Rightarrow \frac{A}{2}=45^{\circ} \Rightarrow A=90^{\circ}$
As we know that,
$\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}$
Given, $\quad \mu=\sqrt{2}$ and $A=\delta_m$
$\Rightarrow \quad \sqrt{2}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin \frac{A}{\sin \frac{A}{2}}}{\sin }$
$\Rightarrow \quad \sqrt{2}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}[\because \sin 2 A=2 \sin A \cos A]$
$\sqrt{2}=2 \cos \frac{A}{2} \Rightarrow \cos \frac{A}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \quad \frac{A}{2}=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \Rightarrow \frac{A}{2}=45^{\circ} \Rightarrow A=90^{\circ}$
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