Physics/optics: why does angle of incidence = angle of reflection?

Factual Questions
1

If you create a smooth enough surface - whether it’s the free surface of a liquid, or the polished surface of a hard solid material - you can achieve specular reflection, i.e. what most people know as a mirror. The defining feature of specular reflections is that the angle of incidence of the incoming ray of light is equal to the angle of departure of the reflected ray (“angle of incidence equals angle of reflection”).

Thinking more about what a “smooth enough” surface is, and what light is, I’m now wondering why specular reflection works at all. When an inbound photon hits the surface, isn’t it interacting with a single atom of material? The inbound photon (or the atom it hits) would have to “know” what the orientation of the flat/smooth surface is in order to know which way the reflected photon should go. But are mirrors truly atomically flat surfaces? I mean no mirror is perfectly flat over large distances, but for a photon to know which way to go after reflection, doesn’t the target atom of the reflector need to be mostly coplanar with nearby atoms in order to define a plane of reflection for that photon? And even assuming atomic flatness, if one photon interacts with only one atom, how can the position of other nearby atoms influence what that photon does?

2

Reflection is more a wave phenomenon that a particle phenomenon. The light wave is interacting with the atomic lattice at the interface of the media.

3

I’ll just comment that this is an excellent question. We have a number of physicists who might be able to answer this but one of them is @CalMeacham who’s specialty is optics (on top of mythology, monster movies, etc.).

4

The mistake is assuming that the photon is interacting with only one atom.
It actually induces EM fields in the entire surface.
With a bunch of math, it can be shown that the angle or incidence must equal the angle of reflection or you get a discontinuity in the field.

5

Yes, the simplest case should be to assume the EM wave is interacting with an isotropic material, not random atoms. Then you can derive something about what the fields should look like at the boundary.

EG
https://www.feynmanlectures.caltech.edu/II_33.html

6

Popping in to note - this is getting heavily into the territory of Fermat’s principle (taking the path of least time/energy/distance/what have you), which bridges ray and wave based analysis

7

You can recreate these angles with simple geometry and Huygens principle. Waves falling on a point in space can be reimagined as originating from that point and going in all directions (though not necessarily doing so with equal strength). If you look at wavefronts in light hitting a mirror, and bouncing as hemispheres away from all the points on the surface, you will see the hemispheres constructively interfering to create new wavefronts going in the other azimuthal direction but at the same angle of incidence.

8

To amplify the last a bit, at ant other point on the reflecting surface, nearby waves interfere destructively with each other. It is only at a stationary point in the distance function that this doesn’t happen. Notice I said stationary point, not necessarily minimum. If you imagine a hyperbolic reflector, the reflection point will be at the maximum of the sum of the distance. So “least action” is something of a misnomer. Stationary action is closer to the truth.

9

You ask several philosophical questions that don’t necessarily have simple answers.

Although photons are really small items, it’s clear to me that their influence spreads over a large region, just as the lowest S-orbital of a hydrogen atom dies away exponentially. If it didn’t, then you’d have to exp[lain to me how a photon “knows” how wide a slit is when it contributes to the interference pattern it produces. Besides, in the wave picture of light a plane wave extends indefinitely to the sides.

The simple answer to “why does the angle of incidence equal the angle of reflection?” isd that in this was the electromagnetic boundary conditions are met and are consistent. This also preserves the photon’s momentum parallel to the surface.

10

I can say learning this helped my pool (billiards) game.

11

What’s been said so far, while correct, is mainly just the classical view of what’s happening. The quantum picture has similarities but isn’t quite the same.

Feynman’s view is that a particle like a photon takes all possible paths, including ones that seem nonsensical, like where the particle goes backwards for a while and then reverses. So how does the universe “choose” the path that the particle actually took? And why, for a photon bouncing off a mirror, is it the one where the reflection law holds?

The photon has a property called phase. The phase is an angle (i.e., degrees or radians), and increases as it travels. It also has a small amplitude. You can imagine this as a small arrow pointing out from a point, like the hand on a clock, with the clock turning at a constant rate (proportional to the light frequency) as the photon travels.

If two paths have exactly opposite phase–that is, they’re 180 degrees apart–then they cancel and contribute nothing overall. But if paths have the same phase, or similar phase, then they can add up to something significant. Imagine a zillion clock hands, with the center of one put at the tip of the previous one, all forming a chain.

The universe picks paths (probabilistically) with large amplitudes; that is, where the phases were similar and the amplitudes added.

So, we might conclude from this that the classical reflected path is the one where the phases lined up, since that’s the path we observe. And that’s exactly right. The exact reason is a little hard to discern, but it basically comes down to this: if you tweak a random path by even a tiny amount, the phase will change dramatically (basically, because the wavelength of light is small). It changes so wildly that it may as well be random, like flicking a spinner from a board game. And this happens across almost all paths, and so all of these random phases cancel each other out and add to zero.

That is, except for some special paths–like the reflection case. Here, the geometry works out such that a small tweak to the path leads to almost no change in phase. So a large number of almost-identical paths next to each other all contribute their amplitude with the same phase, leading to a large overall amplitude.

Since that’s the only path with a high amplitude (all the others got averaged out to zero), that’s the one the universe picks.

12

Why does a rubber ball bounce off the floor at equal angles? Well, actually it doesn’t unless the floor is frictionless. If it were frictionless it would be able to exchange momentum with the ball only in the z direction, so the ball would need to exit at equal angles to preserve the x/y component of its momentum. For a mirror, being “frictionless” would be being perfectly reflective. I plane of mirroronium would be incapable of absorbing a photon without emitting another photon at equal angles. As those above have explained, a detailed analysis of phase cancelling explains what happens to a wave impinging on mirroronium.
Or course no real surface is a prefect reflector, but some are pretty damn close - and why they are is an interesting question. Why angle of incidence = angle of reflection has an intuitive answer, but why reflection happens at all is to me the bigger mystery. Somehow the electrons in some atoms of a sufficiently regular surface respond to impinging electromagnetic fields in such a way as to reject absorbing any energy from them.

13

A huge fraction of the above is captured in one thing. The principle of least action.

Whether you consider this an emergent property of the system of not, you can cut to the chase of all manner of systems bouncing, reflecting and generally doing stuff with this one idea.

Feynman showed how it falls out as an intrinsic result of his path integral representation of QED.

14

Indeed. It’s a very general principle that shows up everywhere. We might more accurately call it the principle of stationary action, where “least” is just one case.

Let me give an example from my explanation above. Let’s suppose you have a mirror on the x-axis, a source at the point (0, 1) and a receiver at (2, 1). We expect that light will bounce off the mirror in a V-shape.

Let’s consider though the case where we have a distorted V, where light bounces from the point (0, 0). How long is that path? Well, using Pythagoras, we just get \sqrt{0^2 + 1^2} + \sqrt{2^2 + 1^2} = 1 + \sqrt{5} \approx 3.236068 . Now suppose we shift the bounce point slightly–introduce a variation. Shift by 0.01, and we get \sqrt{0.01^2 + 1^2} + \sqrt{1.99^2 + 1^2} \approx 3.227178. That’s a difference of 0.0089, which isn’t much different from the shift of 0.01 we introduced.

Now let’s look at what happens when it bounces in the expected location. In the exact center, the distance is \sqrt{1^2 + 1^2} + \sqrt{1^2 + 1^2} \approx 2.828427. And again shifting by 0.01, we get \sqrt{1.01^2 + 1^2} + \sqrt{0.99^2 + 1^2} \approx 2.828462. Now, the difference is only 0.000035–much smaller than the variation we introduced!

The difference is even more dramatic for smaller variations. So for most paths, tweaking them by some amount X is likely to change the path length by something roughly proportional to X. But some paths are special, where the path length is stationary with respect to small variations. And it’s those paths that light takes, due to the principle of adding quantum amplitudes based on the phase.

15

NB Feynman’s example of total internal reflection when light passes to a medium of smaller index of refraction. When the light hits the glass at a large enough angle, it reflects off. However, introduce a second piece of glass sufficiently close to the surface and some of the light will continue through, even though it is striking the exact same surface atoms that respond in the exact same way.