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

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

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

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(C) The refractive index of the prism material is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:$\mu = \frac{\sin((A + A)/2)}{\sin(A/2)}$$\mu = \frac{\sin(2A/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)}$.Using the trigonometric identity $\sin A = 2 \sin(A/2) \cos(A/2)$,we get:$\mu = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)}$.Simplifying the expression,we have $\mu = 2 \cos(A/2)$.Therefore,$\cos(A/2) = \mu/2$.Taking the inverse cosine on both sides,$A/2 = \cos^{-1}(\mu/2)$.Thus,$A = 2 \cos^{-1}(\mu/2)$.

$A$ 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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