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(A) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $r = i$.The reflected and refracted rays

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

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(A) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $r = i$.The reflected and refracted rays make an angle of $90^{\circ}$ with each other. Since the sum of angles on a straight line is $180^{\circ}$,we have $r + 90^{\circ} + r^{\prime} = 180^{\circ}$.Substituting $r = i$,we get $i + 90^{\circ} + r^{\prime} = 180^{\circ}$,which implies $r^{\prime} = 90^{\circ} - i$.Using Snell's Law at the interface,$n_1 \sin i = n_2 \sin r^{\prime}$,where $n_1$ is the refractive index of the denser medium and $n_2$ is the refractive index of the rarer medium.Substituting $r^{\prime} = 90^{\circ} - i$,we get $n_1 \sin i = n_2 \sin(90^{\circ} - i) = n_2 \cos i$.Therefore,$\frac{n_2}{n_1} = \frac{\sin i}{\cos i} = 	an i$.The critical angle $C$ is defined by $\sin C = \frac{n_2}{n_1}$.Since $\frac{n_2}{n_1} = 	an i$ and $i = r$,we have $\sin C = 	an r$.Thus,$C = \sin^{-1}(	an r)$.

$A$ ray of light passes from a denser medium to a rarer medium at an angle of incidence $i$. The reflected and refracted rays make an angle of $90^{\circ}$ with each other. The angles of reflection and refraction are respectively $r$ and $r^{\prime}$. The critical angle is given by:

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