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(D) For a prism,the condition for minimum deviation is given by the formula: $\mu = \frac{\sin i}{\sin(A/2)}$,where $i$ is the angle of incidence and

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$60$

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(D) For a prism,the condition for minimum deviation is given by the formula: $\mu = \frac{\sin i}{\sin(A/2)}$,where $i$ is the angle of incidence and $A$ is the prism angle.Given: $\mu = \sqrt{2}$ and $i = 45^o$.Substituting the values into the formula:$\sqrt{2} = \frac{\sin 45^o}{\sin(A/2)}$Since $\sin 45^o = \frac{1}{\sqrt{2}}$,we have:$\sqrt{2} = \frac{1/\sqrt{2}}{\sin(A/2)}$$\sin(A/2) = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$Since $\sin 30^o = 1/2$,we get:$A/2 = 30^o$$A = 60^o$.

$A$ light ray is incident on a prism at an angle of $45^o$ and produces minimum deviation. If the refractive index of the prism is $\sqrt{2}$,then the prism angle is .......... $^o$.

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