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(C) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given $A = 60^{\circ}$ and $\mu =

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$45$

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(C) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.Given $A = 60^{\circ}$ and $\mu = \sqrt{2}$,we substitute these values: $\sqrt{2} = \frac{\sin((60^{\circ} + \delta_m)/2)}{\sin(30^{\circ})}$.Since $\sin(30^{\circ}) = 0.5$,we have $\sqrt{2} 	imes 0.5 = \sin((60^{\circ} + \delta_m)/2)$,which simplifies to $\sin(45^{\circ}) = \sin((60^{\circ} + \delta_m)/2)$.Thus,$45^{\circ} = (60^{\circ} + \delta_m)/2$,which gives $90^{\circ} = 60^{\circ} + \delta_m$,so $\delta_m = 30^{\circ}$.At the condition of minimum deviation,the angle of incidence $i$ is related to the angle of the prism $A$ and the angle of minimum deviation $\delta_m$ by the formula: $i = (A + \delta_m)/2$.Substituting the values: $i = (60^{\circ} + 30^{\circ})/2 = 90^{\circ}/2 = 45^{\circ}$.

$A$ ray of light suffers minimum deviation when incident on a prism having an angle of the prism equal to $60^{\circ}$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is . . . . . . .

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