photon density and inverse square law

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killium

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I wondered about that for a while. When a light source emits photons, those photons leaves it at C in all directions in an expanding sphere. Let looks at a small scale of time, where we consider only one "layer" of photons (like a giant photon balloon). The surface of that balloon is stretched as the sphere is growing.

I wouldn't have any problems with that if the energy was continuous. But photons are quantized. So each individual photons in that layer, are receding sideways from each other.

So as the distance from the light source grows, the lateral separation between photons grows too.

What happens when a light detector, placed at a sufficient distance from the light source, is smaller than the separation distance between two side-by-side photons ?

Is it possible that one photon passes to the right of the detector and the other one passes on the left so the detector detects nothing ?
 
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theridane

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It is. Those imaginary 'layers' leave one after another however, so if you miss a photon from layer x, you're likely to catch a photon from layer x+1, or x+2, ... and so on.

It's not that uncommon - take the Apollo 11's Lunar Laser Ranging Experiment as an example. Out of 10[super]17[/super] photons sent from the Earth to the Moon only one returns back. It amounts to just a few single photons every second, in good conditions. So even though the mirror on the moon is continuously reflecting energy from the laser, we only get hit by a photon so often.
 
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Kessy

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This is why observations of very distant or faint objects require long exposure times. For example, the Hubble Ultra Deep Field required more then 11 days of exposure to make - the image came in one photon at a time.

Another way to think about is that photons in free space propagate as probability waves, which are continuous. As the light travels, the probability of seeing a photon goes down according to the inverse square law.
 
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