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Not sure how to word my question.

Picture a light source in vacuum, so nothing disturbs the light (or similar conditions), 2d.
If I move very, very far away, will it happen that some of the light hits right beside me (both left and right), but not exactly at where I am?
Seeing as there will be a fixed number of photons sent away from the light, and a huge circumference for them to spread, they cannot possibly hit every point of the circle?

If it was a person kicking footballs in all directions instead of the light, the footballs would hit everywhere on a circle with small radius (red circle). If the radius is bigger, the distance between the balls traveling will be greater, and there will be gaps between the balls where I can stand without being hit (red arc).

footballs in space

I'm unsure how this would work with light, being both like footballs but also a wave.
Thanks

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You may like this video about dispersed light behavior in space: youtube.com/watch?v=cztocbHiiqQ –  BoppreH Apr 3 '13 at 21:08
    
Thanks, explained it clearly. –  Matsemann Apr 4 '13 at 9:28
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2 Answers 2

up vote 11 down vote accepted

It isn't a good model of light propagation to imagine it as a swarm of particles headed out from the light source. The light is delocalised i.e. it's energy is not localised into individual photons. When the light interacts with your detector it does so in discrete units of energy, and these are what we see as photons.

So in your example no matter where you stand you have a probability of interacting with the light at that point i.e. being hit by a photon. This is why you get the situation jkej describes in his answer. The photons appear at random at a rate equal to the energy flux of the light divided by the photon energy.

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Yes, there will be "gaps" between the photons. But where the gaps appear and their width is random, just as the direction of each photon is random. The higher the intensity of light is and the longer time you wait, the more photons will have been emitted and the shorter the gaps will be on average.

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Unless it's coherent emission. –  Kitchi Apr 3 '13 at 21:21
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