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(A) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((A + \delta_{\min})/2)}{\sin(A/2)}$.Given $\mu = \cot(A/2) = \

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$\pi - 2A$

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(A) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((A + \delta_{\min})/2)}{\sin(A/2)}$.Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)}$.Equating the two expressions: $\frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin((A + \delta_{\min})/2)}{\sin(A/2)}$.This simplifies to $\cos(A/2) = \sin((A + \delta_{\min})/2)$.Using the identity $\cos 	heta = \sin(\frac{\pi}{2} - 	heta)$,we get $\sin(\frac{\pi}{2} - A/2) = \sin((A + \delta_{\min})/2)$.Comparing the angles: $\frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_{\min}}{2}$.Multiplying by $2$: $\pi - A = A + \delta_{\min}$.Therefore,$\delta_{\min} = \pi - 2A$.

If the refractive index of the material of a prism is $\cot(A/2)$,where $A$ is the angle of the prism,then the angle of minimum deviation will be:

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