In the case where light is moving from denser medium to a rarer medium, at the critical angle, the refracted ray will be parallel to the surface. Now according to the principle of reversibility of light, if we reverse the direction of light, it will follow the same path. But in this case, if we reverse the direction of light, that is, send a ray parallel to the surface, how will it retrace its path since the ray will never cross the surface?
[Physics] Principle of reversibility of light in case of refraction at the critical angle
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Related Solutions
When one is exactly at the critical angle, the light behaves in a way that may be interpreted as "something in between" refraction and reflection: it continues in a direction that is tangent to the boundary of the mediums.
When the angle is smaller than the critical angle, we get refraction. At the critical angle, $\theta_2$ of the refraction becomes 90 degrees, so we get the tangent propagation. At angles larger than the critical ones, there is a discontinuity: the equation for $\theta_2$ (arcsine of something) has no solutions which is why we get a total internal reflection.
There is nothing about these facts that would contradict reversibility or time-reversal symmetry of the laws of physics. If we time reverse the behavior at the critical angle, it indeed looks like the light must "randomly pick" a moment at which it enters the medium with the higher refraction index and there isn't any unique way to pick the preferred moment.
But that's not a problem because the probability that the direction of light is "exactly" tangent to the boundary is zero. In a real-world situation, the light will be a superposition of beams with angles $\theta_2=\pi/2-\epsilon$ for various small values of $\epsilon$, and for any nonzero $\epsilon$, the light will know very well when it hits the boundary. So your problem only occurs at a negligible, "measure zero" portion of the situation, so it is at most a "measure zero" problem. When one adds the appropriate degree of realism and specifies the precise angles and deviations from the "idealized model", the problem goes away.
Best Answer
Light beam is not an ideal ray and is not perfectly collimated, so, in real life, only its small fraction will be exactly at the critical angle for any given incidence angle.
Another factor, making the transition between refraction and reflection gradual is the dispersion, which makes the refractive index and, therefore, the critical angle, dependent on the frequency. Since the energy of a non-ideal light beam is spread over some finite frequency range, any incidence angle would be exactly critical only for a negligible fraction of light in the beam, even is the beam was perfectly collimated.
This point is illustrated in this video.
Note how the beam gets separated into different colors, each color reaching the boundary at a different incidence angle.
So, a scenario commonly shown on diagrams illustrating total internal reflection, where an incident light beam hit the boundary and then just keeps moving along it, is not realistic.
Therefore, if we direct a beam of light along the boundary between two media with different refractive indices, we are not exactly reversing what happens in the critical angle scenario.