If a prism having refractive index sqrt{3} ha | vedclass

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(C) Given that the angle of minimum deviation $\delta_{m}$ is equal to the refracting angle of the prism $A$,so $\delta_{m} = A$.The formula for the r

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$60$

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(C) Given that the angle of minimum deviation $\delta_{m}$ is equal to the refracting angle of the prism $A$,so $\delta_{m} = A$.The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$.Substituting $\delta_{m} = A$ into the formula:$\mu = \frac{\sin((A + A)/2)}{\sin(A/2)} = \frac{\sin(A)}{\sin(A/2)}$.Using the trigonometric identity $\sin(A) = 2 \sin(A/2) \cos(A/2)$:$\mu = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$.Given $\mu = \sqrt{3}$,we have:$\sqrt{3} = 2 \cos(A/2)$$\cos(A/2) = \frac{\sqrt{3}}{2}$.Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we get:$A/2 = 30^{\circ}$$A = 60^{\circ}$.

If a prism having refractive index $\sqrt{3}$ has an angle of minimum deviation equal to the refracting angle of the prism,then the refracting angle of the prism is......$^o$

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