Photon reflection off a mirror.

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Does the exact same photon of light bounce back or is it absorbed and a new one with the same properties emitted? how does it remember its path? And if it bounces back, does its velocity comes to near zero for a small-time before being reflected back in a straight line at same angle.

In: Physics

3 Answers

Anonymous 0 Comments

The individual wavefunctions corresponding to the photon interacting with each particle in a perfect mirror are in superposition. They only constructively interfere along the path where the angle of incidence is equal to the angle of reflection.

Anonymous 0 Comments

This question makes little sense.

Suppose you have a wave on a lake bouncing off the shore. Is the wave that bounced the same wave or a different wave with the exact same properties?

IIRC there is evidence that all photons are the same single photon, just in different places, but I really don’t have enough knowledge about this topic to talk about that.

Anonymous 0 Comments

Say that we have a laser at point A and a detector at point B. We aim the laser so that the light coming out of it bounces off a mirror and goes into the detector. The actual physics of what’s going on at the point where the light bounces off the mirror is extremely complicated and depends on material properties and the frequency of light, etc. But in short, what *probably* will happen is that the individual photons are absorbed by electrons in atoms/molecules which then go into an excited state. The electron then wants to lose this energy to be in a more relaxed state, so it spits a photon out, typically with the same frequency at the absorbed photon. The direction that the photon is most probable to be emitted is going to be determined by specific properties of the atoms/molecules that make up the mirror and this absorbing/emitting process might happen multiple times.

Okay, now for the fun question. How does it “remember” its path, and how does it know to reflect at the same angle it came in at? The short answer is that it doesn’t. In fact, as we understand it, the photons (both incoming and outgoing) take *every possible path simultaneously.* Each path gets assigned a “probability” (really, it is an amplitude which you can think of as a kind of “square root” of a probability) based on the fundamental properties of nature and these either add together or subtract from each other. It just so happens that the paths very near the path of shortest distance (the one where the light bounces off the mirror at the same angle it came in) all add together, while paths far away from this subtract out. So you are left with a net effect of light traveling the shortest distance between the two points, never once having to know or remember its path.