A ray of light enters from a rarer to a dense | 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 and refr

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

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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 and refracted rays are mutually perpendicular,the sum of the angle of reflection $r$,the angle between them $(90^{\circ})$,and the angle of refraction $r'$ is $180^{\circ}$.Thus,$r + 90^{\circ} + r' = 180^{\circ}$.Since $r = i$,we have $i + 90^{\circ} + r' = 180^{\circ}$,which implies $r' = 90^{\circ} - i$.Applying Snell's Law at the interface: $\mu_r \sin i = \mu_d \sin r'$.Substituting $r' = 90^{\circ} - i$,we get $\mu_r \sin i = \mu_d \sin(90^{\circ} - i) = \mu_d \cos i$.Therefore,$\frac{\mu_d}{\mu_r} = \frac{\sin i}{\cos i} = 	an i$.For the critical angle $C$,we know that $\sin C = \frac{\mu_r}{\mu_d}$.Thus,$\sin C = \frac{1}{	an i} = \cot i$.Hence,the critical angle is $C = \sin^{-1}(\cot i)$.

$A$ ray of light enters from a rarer to a denser medium. The angle of incidence is $i$. Then the reflected and refracted rays are mutually perpendicular to each other. The critical angle for the pair of media is

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