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(D) According to Snell's law,the ratio of refractive indices is given by $\frac{\mu_R}{\mu_D} = \frac{\sin i}{\sin r}$.Given that the angle of inciden

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$	an^{-1}(\sin 	heta_C)$

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(D) According to Snell's law,the ratio of refractive indices is given by $\frac{\mu_R}{\mu_D} = \frac{\sin i}{\sin r}$.Given that the angle of incidence $i = A$ and the angle between the reflected and refracted rays is $90^o$,the angle of refraction $r$ can be determined. Since the reflected ray makes an angle $A$ with the normal,the angle between the reflected ray and the interface is $90^o - A$. The angle between the reflected and refracted rays is $90^o$,so the angle between the refracted ray and the interface is $180^o - 90^o - (90^o - A) = A$. Thus,the angle of refraction $r = 90^o - A$.Substituting these into Snell's law: $\frac{\mu_R}{\mu_D} = \frac{\sin A}{\sin(90^o - A)} = \frac{\sin A}{\cos A} = 	an A$.We know that the critical angle $	heta_C$ satisfies $\sin 	heta_C = \frac{\mu_R}{\mu_D}$.Therefore,$	an A = \sin 	heta_C$,which implies $A = 	an^{-1}(\sin 	heta_C)$.

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