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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?

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  • $\begingroup$ can you please add a drawing? $\endgroup$ – Árpád Szendrei Aug 12 '18 at 2:22
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    $\begingroup$ Possible duplicate physics.stackexchange.com/questions/137581/… $\endgroup$ – Aaron Stevens Aug 12 '18 at 2:56
  • $\begingroup$ If photons fly very close and exactly parallel to the surface, some of them would interact with electrons in the dense medium, thus entering the medium at that point. In quantum mechanics interactions happen randomly according to a certain probability. Since the probability here is equal everywhere, your beam would enter the medium equally in all points until the last photon. This makes perfect sense, because you are reversing the critical reflection that could happen at any point, not just one chosen point, so you get the exact reversal of every possible beam all at once according to the law. $\endgroup$ – safesphere Aug 12 '18 at 4:02
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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.

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  • $\begingroup$ What if we use a monochromatic source to demonstrate the same effect? Now all the rays will actually be parallel to the surface. So, how will we explain principle of reversibility then? $\endgroup$ – Harshdeep Singh Aug 12 '18 at 5:25
  • $\begingroup$ @HarshdeepSingh Purely monochromatic sources don't exist - lasers are close but not perfect. As a result, as mentioned, any given angle of incidence will correspond to a critical angle only for an infinitely small fraction of light in the beam. $\endgroup$ – V.F. Aug 12 '18 at 12:40

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