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 $r$ and $r'$ are angles of reflection and refraction respectively, then the angle of incidence $i$ is equal to $[\,C$ = critical angle $]$

Let the refractive index of the denser medium be $\mu_1$ and that of the rarer medium be $\mu_2$.

The critical angle $C$ is given by $\sin C = \dfrac{\mu_2}{\mu_1}$.

According to the law of reflection, the angle of incidence $i$ is equal to the angle of reflection $r$, so $i = r$.

It is given that the reflected and refracted rays are perpendicular to each other. From the geometry of the interface, the sum of the angle of reflection $r$ and the angle of refraction $r'$ is $90^{\circ}$.

$r + r' = 90^{\circ}$

Since $r = i$, we have $i + r' = 90^{\circ}$, which gives $r' = 90^{\circ} - i$.

Applying Snell's law at the interface:

$\mu_1 \sin i = \mu_2 \sin r'$

Substituting $r' = 90^{\circ} - i$:

$\mu_1 \sin i = \mu_2 \sin(90^{\circ} - i)$

$\mu_1 \sin i = \mu_2 \cos i$

$\dfrac{\sin i}{\cos i} = \dfrac{\mu_2}{\mu_1}$

$\tan i = \sin C$

$i = \tan^{-1}(\sin C)$

Answer: $\tan^{-1}(\sin C)$