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(D) For minimum deviation,the angle of incidence $i$ is related to the angle of prism $A$ and the angle of minimum deviation $\delta_m$ by the relatio

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$60$

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(D) For minimum deviation,the angle of incidence $i$ is related to the angle of prism $A$ and the angle of minimum deviation $\delta_m$ by the relation $i = \frac{A + \delta_m}{2}$.Given that the angle of incidence $i = 45^o$ and the refractive index $\mu = \sqrt{2}$.The formula for the refractive index of a prism is $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(A/2)}$.Substituting $i = \frac{A + \delta_m}{2} = 45^o$,we get $\mu = \frac{\sin(45^o)}{\sin(A/2)}$.Substituting the values: $\sqrt{2} = \frac{1/\sqrt{2}}{\sin(A/2)}$.This simplifies to $\sin(A/2) = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.Therefore,$A/2 = 30^o$,which gives $A = 60^o$.

When light rays are incident on a prism at an angle of $45^o$,the minimum deviation is obtained. If the refractive index of the material of the prism is $\sqrt{2}$,then the angle of the prism will be......$^o$

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