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(B) The formula for the refractive index $\mu$ of a prism in terms of the refracting angle $A$ and the angle of minimum deviation $\delta_m$ is given

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$\pi-2 A$

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(B) The formula for the refractive index $\mu$ of a prism in terms of the refracting angle $A$ and the angle of minimum deviation $\delta_m$ is given by:$\mu = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$Given $\mu = \cot(A / 2) = \frac{\cos(A / 2)}{\sin(A / 2)}$,we substitute this into the formula:$\frac{\cos(A / 2)}{\sin(A / 2)} = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$Canceling $\sin(A / 2)$ from both sides,we get:$\cos(A / 2) = \sin((A + \delta_m) / 2)$Using the trigonometric identity $\cos 	heta = \sin(90^{\circ} - 	heta)$ or $\sin(\pi/2 - 	heta)$:$\sin(\pi/2 - A/2) = \sin((A + \delta_m) / 2)$Equating the angles:$\pi/2 - A/2 = (A + \delta_m) / 2$Multiplying by $2$:$\pi - A = A + \delta_m$$\delta_m = \pi - 2A$

The refracting angle of a prism is $A$ and the refractive index of the material of the prism is $\cot (A / 2)$. The angle of minimum deviation of the prism is

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