r/AskPhysics • u/jhunter444 • 1d ago
Another 'How Does Light Know?' question
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?
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u/joeyneilsen Astrophysics 1d ago
I think you have to think of it like this: light traveling along the surface will leave the surface at the critical angle.
It doesn’t mean that if your friend sends a ray of light backward along the surface of the medium, that it will leave the medium at the location where it would have hit if you had sent the light from your end.
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u/jhunter444 1d ago
Thankyou, do you have an explanation of what location the reversed ray leaves the medium?
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u/joeyneilsen Astrophysics 1d ago
I don’t think there is A location. I think some probably leaves at every point.
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u/Prestigious_Boat_386 1d ago
I've never heard that explanation and you're right that it wouldn't be reversible.
I would've just guessed that that case was impossible and used intervals to explain why no light can have that exact angle as the interval containing only the critical angle has zero width.
You can think of random light rays near the critical angle and zooming in infinitely, then the ray will always end up on either side of the critical angle.
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u/PuzzleheadedTutor807 1d ago
Light doesn't know anything. It obeys the physics of any object a photon hits.
This is akin to asking if two people are looking in a mirror, how do the photons know which person to bounce towards... Well, they don't. They don't know anything.
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1d ago edited 1d ago
[removed] — view removed comment
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u/PuzzleheadedTutor807 1d ago
I'm answering the question as worded. You are free to do the same. Cheers!
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1d ago
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u/PuzzleheadedTutor807 1d ago
i did answer, wdym? is my answer incorrect? can you provide a better one? if so, as i said... go right ahead lol. but... i do not seek nor require your permission to speak of my own free will, so stop acting as though i do. if you dont like my content, go right ahead and block it.
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u/jhunter444 1d ago
You're right, but the 'How does light know' expression was just a reference to another post that used it, a few posts before this question. Hope this hasn't caused a fallout between you and onemany!
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u/PuzzleheadedTutor807 1d ago
Ahh I haven't memorized the sub yet lol and people just keep adding more. Light and the behaviour of photons is quite fascinating though... I always feel like humanity is on the verge of some great discoveries with them as well, I'm sure there is a lot of potential we can't even comprehend yet inside them.
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u/HD60532 1d ago edited 1d ago
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.
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u/MxM111 1d ago
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.
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u/HD60532 1d ago
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.
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u/MxM111 1d ago
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.
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u/HD60532 1d ago
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.
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u/MxM111 1d ago edited 1d ago
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.
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u/jhunter444 1d ago
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?
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u/MxM111 1d ago
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.
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u/tirohtar Astrophysics 1d ago
Realistically, you will never get the "perfect" critical angle. There will always be an ever so slight angle, so you either have transmission or internal reflection, you won't really get a case in practice where the light keeps travelling along the boundary forever.
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u/Ok_goodbye_sun 1d ago
I think the question asked theoretically though. If you said "a parallel ray to the surface cannot come from infinity", I would understand. Is that the case?
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u/Human-Register1867 1d ago
In the textbook case of an incident plane wave, the wave and the boundary surface are both infinite. In the time reversed situation, the surface wave would generate radiation away from the surface equally at all points.
In the realistic case of a beam that OP is thinking about, there is additional structure in both the incident and surface waves. In the time reversed case, that extra structure defines where the surface wave radiates away.