Angle of minimum deviation is equal to the ha | vedclass

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(C) For an equilateral prism, the prism angle $A = 60^\circ$.Given that the angle of minimum deviation $\delta_m = A/2 = 60^\circ / 2 = 30^\circ$.The

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$\sqrt{2}$

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(C) For an equilateral prism, the prism angle $A = 60^\circ$.Given that the angle of minimum deviation $\delta_m = A/2 = 60^\circ / 2 = 30^\circ$.The formula for the refractive index $\mu$ is given by $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.Substituting the values, we get $\mu = \frac{\sin((60^\circ + 30^\circ)/2)}{\sin(60^\circ/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$.Since $\sin(45^\circ) = 1/\sqrt{2}$ and $\sin(30^\circ) = 1/2$, we have $\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is . . . . . . .

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