# If we shoot to photons to a surface, in which conditions we can differentiate the points where the photons collided with the surface?

Now, I heard this example from one of my professor 4 years ago when I first entered university, and at that I had some objections (and I still do), because, I think, if we flick our wrist fast enough, the spot on the moon will not be a continuous spot, but that spot will be like "$$. \quad .\quad . \quad . \quad .$$", as if we as turning on and off the light while moving it, and these ideas led me to the following question:

Lets shoot a two photon onto a reflective surface, and observe where the photons hit the surface;

What conditions should be satisfied so that we can differentiate the points where the photons collided with the surface ?

## 1 Answer

Look at this experiment where photons hit a sensitive "screen".

On the left it is the footprints of individual photons in a dxdy interval. The condition that is satisfied is that the individual photon's energy raises a photosensitive interaction which can be recorded. The accumulation gives the usual classical interference pattern of light in this geometry.

In this experiment the photon energy is absorbed. One could use a different one where the photon would compton scatter and again leave a foot print but there is no reason to do so.

Light is composed out of zillions of photons and forms a continuum . To get at your ". . ." on the moon one would have to use single photons as in the experiment above

• Yes light contains a zillions of photons, but instead of using mars, we can use another celestial object, so that $R$ can get big as much as we want; these are just experimental technicalities. This means even a small change in the angle that the photons are sent from the source can collide with that celestial objects far apart, so at some point, that motion of that spot on that celestial object cease to have a continuos motion. – Our Jan 23 '19 at 14:15
• as you see in the experiment above the source is point like from the laser at a low density of photons, and light like at the end on the right. – anna v Jan 23 '19 at 14:41