# Why is angle of incidence equal to the angle of reflection? [duplicate]

A photon strikes an atom in a chrome surface. An electron is elevated to a higher energy level then dissipates the energy through emitting a new photon. How does the electron remember the angle of incidence so that it can fire that new photon off in the angle of reflection? I don't even understand how the electron can know its orientation relative to the surface.

Maybe specular reflections don't involve electrons in this way? Does electrons emitting photons as they drop to lower energy levels only what happens with diffuse reflections? If so, how do specular reflections deal with these angles?

Background to question: I'm an artist and want to understand the physics of what I see but find the literature on the topic cumbersome and choked with equations I don't know how to interpret. Thank you for your kind help!

• While specular reflection does involve electrons and photons in this way, it's not a particularly useful way to look at things. At the photon/electron level, there is no such law that angle of incidence is equal to the angle of reflection: this effect only arises due to the interference of many photons. – Chris Jan 5 '17 at 21:08
• I agree with Sammy's identified duplicate; however, rather than the "accepted" answer, I recommend you read this answer in particular, as it is probably most suitable for your needs / level of understanding of physics. – Floris Jan 5 '17 at 21:10
• There is no law? What about the Law of Reflection. – Lambda Jan 6 '17 at 1:26
• Chris, if it was due to interference wouldn't we see a greater reduction in intensity. Many mirrors reflect nearly 100%. – Lambda Jan 6 '17 at 1:52

So a single atomic electron that gets excited in this way indeed does not have this property, and indeed a vast number of them can do this in parallel without functioning this way: this is part of the classical explanation for why the sky is blue.

Furthermore it matters that the surface is flat: if you cut little parallel lines in the surface then you get a spectrometer; this is part of the explanation for why you see rainbows in the "data track" of a CD or DVD.

Furthermore you see the same reflection when you analyze things like reflection from a glass, even though there is also a transmitted wave.

Our modern understanding is that a photon is able to sense every path that it could possibly take from A to B. Each path can be thought of as a little arrow rotating in 2D at a constant rate with respect to time (the photon's "frequency"). We add up all of these little arrows by connecting them tip-to-tail and then ask how far away the final point is from the initial point, which is a measure of the probability that the photon goes in that direction.

Now if there is a path, and there are other possible paths "nearby" that one, like on a mirror, we have to consider if they take a different time or not. If they all have a different time then each arrow is tilted a little more relative to the last one, and as you add them together, they describe something very circle-like and it will not get far away from the initial point. But if the path is either the fastest or slowest path from point A to point B, then something magical happens: all of the nearby paths take about the same time, so all of the arrows line up, so you get a very large probability that the object moves in that way.

This is known as "Fermat's law of refraction," and it says that light seems to magically find the shortest-time path between two points. And a direct consequence is that the angle of incidence equals the angle of reflection.

• I respect your knowledge on this. It sounds like you have studied this, but sometimes the explanation becomes so complicated and includes the words like "sense" and "magical" it might be best to say simply, "Good question and we really don't know the answer." – Lambda Jan 6 '17 at 1:36
• No, I disagree with Lambda. – Michael Hunter Jan 6 '17 at 19:29
• We see specular reflections the way we do because 1) most photons go that way because it's the shortest time-path they then constructively interfere and 2) any that don't go that way are canceled out by destructive interference? I thought that the photons scattered in random directions became diffuse reflections. – Michael Hunter Jan 6 '17 at 19:41
• Well first off notice that it does not matter that it is the "shortest" per se, just that it's a local extremum. This is why you will sometimes see a mountain and also see its reflection from a placid lake between; the light doesn't all just say "screw you, lake, I know a faster way." Second off remember that diffuse reflections happen because of very rough surfaces at the level of the wavelength of light: can you see why that roughness would randomize phases and lead to amplitudes in all directions? – CR Drost Jan 6 '17 at 20:22