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(D) For an equilateral triangular prism,the angle of the prism $A = 60^{\circ}$.Given the refractive index $\mu = \sqrt{3}$.The formula for the refrac

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$60$

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(D) For an equilateral triangular prism,the angle of the prism $A = 60^{\circ}$.Given the refractive index $\mu = \sqrt{3}$.The formula for the refractive index of a prism is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Substituting the values: $\sqrt{3} = \frac{\sin((60^{\circ} + \delta_m)/2)}{\sin(60^{\circ}/2)}$.$\sqrt{3} = \frac{\sin((60^{\circ} + \delta_m)/2)}{\sin(30^{\circ})}$.Since $\sin(30^{\circ}) = 0.5$,we have $\sqrt{3} 	imes 0.5 = \sin((60^{\circ} + \delta_m)/2)$.$\frac{\sqrt{3}}{2} = \sin((60^{\circ} + \delta_m)/2)$.We know that $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$,therefore $(60^{\circ} + \delta_m)/2 = 60^{\circ}$.$60^{\circ} + \delta_m = 120^{\circ}$.$\delta_m = 60^{\circ}$.

Calculate the angle of minimum deviation for an equilateral triangular prism of refractive index $\sqrt{3}$. (in $^{\circ}$)

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