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

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(B) The formula for the refractive index of a prism is given by $\mu = \frac{\sin((A+D)/2)}{\sin(A/2)}$, where $D$ is the angle of minimum deviation a

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

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

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

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