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(C) For an equilateral prism,the angle of the prism $A = 60^\circ$.The formula for the refractive index $\mu$ in terms of the angle of minimum deviati

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$60$

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(C) For an equilateral prism,the angle of the prism $A = 60^\circ$.The formula for the refractive index $\mu$ in terms of the angle of minimum deviation $\delta_m$ is given by:$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}ight)}{\sin\left(\frac{A}{2}ight)}$Given $\mu = \sqrt{3}$ and $A = 60^\circ$,we substitute these values:$\sqrt{3} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}ight)}{\sin(30^\circ)}$Since $\sin(30^\circ) = 0.5 = \frac{1}{2}$,we have:$\sqrt{3} 	imes \frac{1}{2} = \sin\left(30^\circ + \frac{\delta_m}{2}ight)$$\frac{\sqrt{3}}{2} = \sin\left(30^\circ + \frac{\delta_m}{2}ight)$We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$,so:$60^\circ = 30^\circ + \frac{\delta_m}{2}$$\frac{\delta_m}{2} = 30^\circ$$\delta_m = 60^\circ$

If the refractive index of a material of an equilateral prism is $\sqrt{3}$,then the angle of minimum deviation of the prism is......$^o$

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