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This question already has an answer here:

The HUDF used to be the deepest image of the universe ever taken by the Hubble telescope, the furthest star in this image is 59000 light years away.

The star in question: The star in question

Now Imagine a light emitting object like a star and an observer like a telescope, 59000 light years apart. I represented everything in 2D, but in 3D I suppose it would be the same but with spheres:

Distance

Now picture the photons that the object is emitting, in particle form. They will all have one direction and they will go towards this direction at C, given that there are no massive objects in between their paths:

Emitting

Notice how the photons have separate paths, because they have different directions, now obviously there should be a lot more photons but for simplicity I drew it like that. My point is, given enough distance shouldn't the probability of hitting a photon be so small that seeing an object becomes nearly impossible? Like so:

Light hits

Here the observer can detect the photons, but if it was slightly off it wouldn't:

Light doesn't hit

So how can we detect objects so far away? Is there something about light's wave form that I'm not considering, is this how light is emitted at all?

Update:

Looking at other duplicates questions I've seen two answers:

1- There are just too many photons being emitted that the "gaps" are just too small to make any difference.

2- This isn't really how photons are emitted, instead a quantum field is spread out radially and we only detect the photon when we interact with the field.

Does this mean that there are objects that we can't see, even if given enough time for the light to reach us (presumably outside of the current observable universe)?

Or do the gaps not exist at all because of answer number 2?

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marked as duplicate by John Rennie, Emilio Pisanty, Kyle Oman, Kyle Kanos, Martin Jun 25 '15 at 17:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Hi Victor, your question is indeed a duplicate of the one Steve B suggests. Check the link and you'll have your answer. $\endgroup$ – John Rennie Jun 25 '15 at 16:09
  • $\begingroup$ @JohnRennie except the OP's star is much farther away, bringing a reduction of $(5900)^2\approx 3\times10^{-8}$. Nevertheless, 0.01 photons per square centimeter per second sounds perfectly reasonable for Hubble. $\endgroup$ – Emilio Pisanty Jun 25 '15 at 16:12
  • $\begingroup$ Sorry for not posting sources, it's indeed a red dwarf, I found it on this wikipedia article: link $\endgroup$ – victormeriqui Jun 25 '15 at 16:21
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There are photons traveling in all directions, not just the dozen or so you show. The further from the source the telescope is, the smaller the amount of solid angle it covers and the fewer photons it will gather. A $1 m^2$ telescope pointed at the sun will receive about $1.4 kW$. Taking a typical photon energy of $2 eV$ that is about $4.2E21$ photons/second. Move the sun to $1,000,000$ light years and it is $6.3E10$ times farther, so we will get a factor $(6.3E10)^2$ less photons, but that is still about one per second.

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  • $\begingroup$ And the objects we see at such distances are typically galaxies with billions of stars, so we still get billions of photons per second. $\endgroup$ – Kyle Oman Jun 25 '15 at 16:12
  • $\begingroup$ I see, Looking at this answer and other duplicates of my question, the problem is that there are a lot of photons being emitted. However, does this mean that there are objects that we simply can't see because they are so far away? $\endgroup$ – victormeriqui Jun 25 '15 at 16:14
  • $\begingroup$ @victormeriqui: no, read my answer to the linked question. The photons emitted are delocalised so they are in effect spread over a wide area. $\endgroup$ – John Rennie Jun 25 '15 at 16:19

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