A ray of light is incident at an angle i from | vedclass

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(C) 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 ray and

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

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(C) 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 ray and the refracted ray are mutually perpendicular,the sum of the angle of reflection $r$,the angle between the reflected ray and the refracted ray $(90^{\circ})$,and the angle of refraction $r'$ is $180^{\circ}$.Thus,$r + 90^{\circ} + r' = 180^{\circ}$,which implies $r' = 90^{\circ} - r$. Since $i = r$,we have $r' = 90^{\circ} - i$.Applying Snell's law at the interface,the refractive index of the denser medium with respect to the rarer medium is given by $\mu = \frac{\sin r'}{\sin i}$.We know that the critical angle $C$ is related to the refractive index by $\sin C = \frac{1}{\mu}$.Therefore,$\sin C = \frac{\sin i}{\sin r'} = \frac{\sin i}{\sin(90^{\circ} - i)} = \frac{\sin i}{\cos i} = 	an i$.Hence,the critical angle is $C = \sin^{-1}(	an i)$.

$A$ ray of light is incident at an angle $i$ from a denser to a rarer medium. The reflected and the refracted rays are mutually perpendicular. The angle of reflection is $r$ and the angle of refraction is $r'$. Then the critical angle $C$ will be:

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