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The refracting angle of a prism is A, and refractive index of the material of the prism is $\cot(A/2)$. The angle of minimum deviation is
A
$180^\circ - 3A$
B
$180^\circ - 2A$
C
$90^\circ - A$
D
$180^\circ + 2A$
Detailed Solution
$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$
$\sin\left(\frac{A + \delta_m}{2}\right) = \cos\frac{A}{2} = \sin\left(90^\circ - \frac{A}{2}\right)$
$\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$
$\delta_m = 180^\circ - 2A$
$\sin\left(\frac{A + \delta_m}{2}\right) = \cos\frac{A}{2} = \sin\left(90^\circ - \frac{A}{2}\right)$
$\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$
$\delta_m = 180^\circ - 2A$