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(B) According to the law of reflection,the angle of incidence '$i$' is equal to the angle of reflection '$r$'. So,$i = r$.Given that the reflected and

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

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(B) According to the law of reflection,the angle of incidence '$i$' is equal to the angle of reflection '$r$'. So,$i = r$.Given that the reflected and refracted rays are perpendicular to each other,the sum of the angle of reflection and the angle of refraction is $90^{\circ}$.$i + r_1 = 90^{\circ} \implies r_1 = 90^{\circ} - i$.Using Snell's Law,the refractive index '$\mu$' of the denser medium with respect to the rarer medium is given by $\mu = \frac{\sin i}{\sin r_1}$.Substituting $r_1 = 90^{\circ} - i$,we get $\mu = \frac{\sin i}{\sin(90^{\circ} - i)} = \frac{\sin i}{\cos i} = 	an i$.We also know that the refractive index is related to the critical angle '$C$' by $\mu = \frac{1}{\sin C}$.Equating the two expressions for '$\mu$': $	an i = \frac{1}{\sin C}$.This implies $	an i = \csc C$,which is not among the options. Let us re-evaluate: $\mu = \frac{\sin r_1}{\sin i}$ is for light going from rarer to denser. Since light goes from denser to rarer,$\mu = \frac{\sin r_1}{\sin i} = \frac{\sin(90^{\circ}-i)}{\sin i} = \cot i$.Since $\mu = \frac{1}{\sin C}$,we have $\cot i = \frac{1}{\sin C}$,so $	an i = \sin C$.Therefore,$i = 	an^{-1}(\sin C)$.

$A$ ray of light travels from a denser medium to a rarer medium. The reflected and the refracted rays are perpendicular to each other. If '$i$' and '$r_1$' are the angle of incidence and angle of refraction respectively and '$C$' is the critical angle,then the angle of incidence is

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