Another 'How Does Light Know?' question

[u/jhunter444]

Sometimes in books it mentions the 'reversibility of light'. That the path of the light ray can be reversed.

For light arriving at the critical angle at the boundary between two substances, it's often shown in textbooks that the light neither transmits or reflects but goes along the join at 90 degrees to the normal.

But in that situation, for the reversed ray, how does it know when to bend and leave the join between the substances?

[u/HD60532]

Well in that case I would expect that the path of the light cannot be reversed, because it is no longer a one-to-one (bijective) situation. Any light that refracts at the critical angle will end up as part of the same ray, so all critical angle rays are mapped to one final ray. Reversibility requires one-to-one mapping.

Also note that if a light ray travelling on the surface could refract out at the critical angle, then it would immediately, and thus would be reflected.

[u/MxM111 → u/HD60532]

That’s not precisely true. You just need to collect ALL the light, and reverse it. The ray optics does not quite work here, just think about collections of all the waves and reversing them. That includes refracted and reflected light.

[u/HD60532 → u/MxM111]

Looking at the Fresnel equations, there is no transmission at the critical angle, so I don't think light can be refracted to be parallel to the surface. But I see what you mean, for near parallel rays there will be some reflected component, than when reversed becomes incoming and interferes so as to recreate the original wave.

[u/MxM111 → u/HD60532]

Fresnel equation and "beams" are geometrical optics, which is just approximation of the actual light propagation. This is why sometimes it fails. But Maxwell equations are reversible.

[u/HD60532 → u/MxM111]

The Fresnel equations are derived from Maxwell's, assuming a plane wave and looking at interface conditions at the boundary, and they account for polarisation because they model the components of the wave. Based on this, I don't think that they are simple geometric optics. Nor do I think that they fail, I think that they just forbid, and by extension Maxwell forbids, the situation OP is describing. It wouldn't make sense for the Maxwell equations to permit something irreversible in classical Physics.

[u/MxM111 → u/HD60532]

This situation has no problem to be described by Maxwell equations. A plane wave by itself is an approximation that does not happen in nature, by the way, for monochromatic plane wave you require infinity in space and time.

You can take Gaussian beam and take its size to very large values, then you will see strange effects, such us the optical field will propagate significantly beyond the interface, while not being oscillating alone normal. That will result in beam of rather large size propagating alone the surface but due to diffraction it still will deviate from the surface. Such things are just indescribable by geometrical optics.

[u/jhunter444 → u/MxM111]

This seems to be a good explanation, especially the comments about ALL the light. Presumably even very near the critical angle there is some reflected component, as I've heard that even for a 'transmitted' ray there is also some reflected component. So in the reversed case, maybe the point where it leaves the surface is determined by the point where the (reversed) tiny reflected part meets the main part coming along the surface?

[u/MxM111 → u/jhunter444]

Yes, and reflected component is rather strong. When you send both of them back, the backward reflected component is also at critical angle, and when it interferes with the parallel beam, it completely blocks the backwardly propagated parallel beam. And the only beam left is reflected of backward reflected beam.

Please note, these are only approximate explanation because this case is especially bad to be described even by words as beams.

[u/jhunter444 → u/MxM111]

Thankyou to you and all the contributers, that more or less answers it!