A prism of refractive index mu and angle A is | vedclass

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(C) The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.Given that the angle

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$2\cos^{-1}(\frac{\mu}{2})$

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(C) The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:$\mu = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$.Using the trigonometric identity $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$,we get:$\mu = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$.Rearranging for $A$:$\cos(\frac{A}{2}) = \frac{\mu}{2} \Rightarrow \frac{A}{2} = \cos^{-1}(\frac{\mu}{2}) \Rightarrow A = 2\cos^{-1}(\frac{\mu}{2})$.

$A$ prism of refractive index $\mu$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is $A,$ then the value of $A$ in terms of $\mu$ is

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