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(A) 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 give

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$180^{\circ}-2A$

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(A) 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 that $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)}$,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 $\cos(	heta) = \sin(90^{\circ} - 	heta)$,we can write:$\sin(90^{\circ} - A/2) = \sin((A + \delta_{m})/2)$Comparing the angles:$90^{\circ} - A/2 = (A + \delta_{m})/2$$180^{\circ} - A = A + \delta_{m}$$\delta_{m} = 180^{\circ} - 2A$

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

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