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

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Question

(C) The refractive index of the material of the prism is given by the formula:$\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}$G

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$A$ 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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