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The refracting angle of prism is $A$ and refractive index of material of prism is $\cot \frac{A}{2}$. The angle of minimum deviation is $A$
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The correct answer is:
$180^{\circ}-2 A$
Given, angle of prism $=A$
Refractive index of prism, $\mu=\cot \frac{A}{2}$
As, we know, $\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
where, $\delta_{m}=$ angle of minimum deviation.
$\Rightarrow \quad \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \quad \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \sin \left(90^{\circ}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_{m}}{2}\right)$
$\Rightarrow \quad \frac{180^{\circ}-A}{2}=\frac{A+\delta_{m}}{2}$
$\Rightarrow \quad \delta_{m}=180^{\circ}-2 A$
Refractive index of prism, $\mu=\cot \frac{A}{2}$
As, we know, $\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
where, $\delta_{m}=$ angle of minimum deviation.
$\Rightarrow \quad \cot \left(\frac{A}{2}\right)=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \quad \frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \sin \left(90^{\circ}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_{m}}{2}\right)$
$\Rightarrow \quad \frac{180^{\circ}-A}{2}=\frac{A+\delta_{m}}{2}$
$\Rightarrow \quad \delta_{m}=180^{\circ}-2 A$
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