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(C) Given,the prism is equilateral,so the angle of the prism $A = 60^{\circ}$.Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.Th

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$\sqrt{3} 	imes 10^8 \ m/s$

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(C) Given,the prism is equilateral,so the angle of the prism $A = 60^{\circ}$.Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.The refractive index $\mu$ of the prism is given by the formula:$\mu = \frac{\sin \left(\frac{A + \delta_m}{2}ight)}{\sin \left(\frac{A}{2}ight)}$Substituting the values:$\mu = \frac{\sin \left(\frac{60^{\circ} + 60^{\circ}}{2}ight)}{\sin \left(\frac{60^{\circ}}{2}ight)} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.We know that the refractive index $\mu$ is related to the speed of light in vacuum $c$ and the speed of light in the medium $v$ as $\mu = \frac{c}{v}$.Therefore,$v = \frac{c}{\mu} = \frac{3 	imes 10^8 \ m/s}{\sqrt{3}} = \sqrt{3} 	imes 10^8 \ m/s$.

If the angle of minimum deviation is equal to the angle of an equilateral prism,then the speed of light inside the prism is .....

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