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(B) Given: Prism angle $A = 75^o$,refractive index $\mu = \sqrt{2}$.Let $r_1$ be the angle of refraction at the first face and $r_2$ be the angle of i

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$45$

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(B) Given: Prism angle $A = 75^o$,refractive index $\mu = \sqrt{2}$.Let $r_1$ be the angle of refraction at the first face and $r_2$ be the angle of incidence at the second face. Since the ray is incident at the critical angle $C$ on the second face,$r_2 = C$.We know that for a prism,$A = r_1 + r_2 = r_1 + C$.The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$. Thus,$C = 45^o$.Substituting the values: $75^o = r_1 + 45^o$,which gives $r_1 = 30^o$.Applying Snell's law at the first face: $\mu = \frac{\sin i}{\sin r_1}$.$\sqrt{2} = \frac{\sin i}{\sin 30^o} = \frac{\sin i}{0.5}$.$\sin i = \sqrt{2} 	imes 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = 45^o$.

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