How do telescopes see many billion light years distant object in our universe?

As an individual with limited expertise in the field of astronomy, my current understanding suggests that the observation of the furthest points in the universe using telescopes requires an unobstructed line of sight. However, with billions of galaxies, stars, and dust particles present in our universe, the presence of these obstacles could create a spherical dome of obstructions, significantly decreasing the probability of observing the farthest regions of space. Despite this potential challenge, modern telescopes are capable of observing objects located billions of light years away. Therefore, I would like to know how telescopes manage to overcome these obstacles and successfully observe objects situated billions of light years away in the universe.

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    $\begingroup$ This a version of Olber's Paradox (en.wikipedia.org/wiki/Olbers%27_paradox) and the resolution is the same as that. $\endgroup$
    – mike stone
    Commented Apr 7, 2023 at 12:29
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    $\begingroup$ Note that we don't see everything at those distances, just the things that are bright enough. We also do some of the "seeing" using light outside the visible spectrum, like radio waves. $\endgroup$ Commented Apr 7, 2023 at 13:03
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    $\begingroup$ The universe is much more empty space than it is stuff. Much much more. $\endgroup$
    – DKNguyen
    Commented Apr 7, 2023 at 14:16
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    $\begingroup$ Also note that the rest of the Milky way is dense enough and close enough we can't see through plane of the disc to the other side. $\endgroup$
    – DKNguyen
    Commented Apr 7, 2023 at 15:38
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    $\begingroup$ for a smaller scale analogy you can think of something like the asteroid belt. if you're used to watching sci fi, you get the impression that they are fairly dense, and navigation is difficult (cough star wars). in reality to have any decent chance of hitting an asteroid in the belt you pretty much have to aim for it -- if you just fly in a random direction through it you'll most likely not have a single interaction. $\endgroup$
    – eps
    Commented Apr 7, 2023 at 23:22

5 Answers 5


In a galaxy, there is a lot of space between stars. The diameter of the sun is $1.4 \times 10^6$ km. The distance to the nearest star is $4.0 \times 10^{13}$ km, or about $30$ million solar diameters. When two galaxies collide, they pass through each other. No stars collide.

From this, you would think you should be able to see through a galaxy. But galaxies also contain a lot of dust. This is a region where there are more atoms or molecules than most places in the galaxy. Not that many. A nebula is still a better vacuum than we can produce on earth. But they are big. There are so many atoms that visible light scatters off one before passing all the way through. Infrared light scatters less. The James Webb space telescope can see through them better than visible telescopes. But still, galaxies do block the view.

So you can see between galaxies. Galaxies are big, typically $3000$ to $300,000$ light years across. The distance to the nearest galaxy is typically 20 galactic diameters. For example, the Andromeda galaxy is $150,000$ light years across, and $2,500,000$ light years away.

But galaxies occur in groups. The local group has $2$ big galaxies and perhaps $80$ dwarf galaxies spread over $10,000,000$ light years. They don't block that much of the sky.

Clusters are larger structures, containing $100$ to $1000$ galaxies over $5$ to $15$ million light years. Again, they are so spread out that nearby galaxies don't block much of the sky.

The largest structures are filaments. These can be shaped like superclusters, walls, or sheets. The space between filaments is huge largely empty areas. Our galaxy is part of the Virgo Supercluster. Again, so spread out that nearby galaxies don't block much of the view.

Still, some interesting views are blocked. The center of our galaxy is hidden behind gas. So is the Great Attractor.


To put it into numbers, lets take as an example Hubble Space Telescope which covers about $11~\text{arcminutes}^2$ of solid angle in the Sky. Now, probability that say a pair of random galaxies will hit the same monitoring telescope area is:

$$ P_{overlap} = \left(\frac {11~\text{arcminute}^2}{4~\pi~\text{sr}}\right)^2 \approx 10^{-15} \tag 1 $$

Hence this probability of overlapping objects is so small that you don't need to worry about. Actually real overlapping probability in the line of sight is even smaller, because in (1) formula it's just a probability that a pair of random galaxies will hit same telescope viewing area. But within that area pair of galaxies can hit any part of image.

Of course given vast cosmos emptiness, some galaxies do hit same telescope viewing angle. For example, with my calculations if HST can monitor in total $200~\text{billion}$ galaxies in the sky, then about:

$$ N_{per-image} = (200~\text{billion})\times \left(\frac {11~\text{arcminute}^2}{4~\pi~\text{sr}}\right) \approx 15K \tag 2 $$ of them should be captured on average per 1 HST image. Of course again, some of them will get into the exact same line of sight and due to that will not be seen. But still, there remains hundreds or even thousands of galaxies which doesn't overlap in the same image.

As about cosmic dust, it's only more actual in the inside of galaxy where there is plenty of interstellar dust and gas material. As soon as photon gets out of a galaxy into cosmic space, there is no big deal, because empty outer-space has only about a few Hydrogen atoms per cubic meter. Chances for a photon to hit a Hydrogen atom per cubic meter is about:

$$ P_{hydrogen} = \frac {4/3 ~ \pi ~ (0.037 nm)^3} {1~ m^3} \approx 10^{-31}, \tag 3$$

where the numerator is the approximate volume of a Hydrogen atom. So as we see, vast outer space is not an issue to a photon.

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    $\begingroup$ That's actually interesting. I knew the probability was low, but $10^{-31}$ is a lot smaller than what I could have imagined. $\endgroup$
    – joseph h
    Commented Apr 9, 2023 at 1:41

If your question is why we can see faraway things, in the sense that they are not covered by closer objects, it all comes down to angular resolution. You can imagine that telescopes have arrays of 'pixels' and that these arrays cover a selected patch of sky. What happens then is that light "hits" one of the cells of the array and we get an "image". If you have a fine enough grid, you can separate points that are close together in the sky. So, the finer the array grid (smaller 'pixels') the closer the objects needing distinguished can be. Obviously some objects will be covered, like stars behind the moon, but the sky is mainly empty space and the chance that 2 objects lie in the same line of sight to small angular degrees is very narrow. It surely happens, but there are a lot of objects which are not covered and that we can see.


You are right, you can not see all of the universe, since some parts are behind others.

It even happens in our own solar systems; if the moon is between us and the sun there are several minutes when you cannot see it, or the moons of Jupiter go behind Jupiter. And if you look in the direction of the center of our galaxy (so in the direction of the Milky Way) you will not see any distant galaxies in this direction... and so on

You will not see a faraway galaxy if the light from it has to go through another galaxy or through a nearby star. But there are still so many gaps, you can see enough.

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    $\begingroup$ It is worthwhile to note the cosmological principle: "...the notion that the spatial distribution of matter in the universe is homogeneous and isotropic when viewed on a large enough scale, ..." So even if I can't see the whole universe, I can see a representative sample. $\endgroup$ Commented Apr 8, 2023 at 0:37

Hide & Seek

Turn the question around. Let's say you are God, and can place galaxies anywhere you want. Let's say that Satan comes to you and says: "Hide as many stars/galaxies/whatever as you can from the prying eyes of humans on earth...if you can hide more than X% of them, I'll give you a cookie." How much of the visible universe could God hide in this way? What is the highest achievable value of X?

Well, X is obviously bounded by the number of stars we can see with the naked eye. You can't make any of those invisible, but you could hide everything else behind them. But then you have stars visible with small telescopes, which greatly increases your "hiding places". Even so, after trying a few configurations, ask yourself: "How likely is this universe to occur on its own?" And you'll find even if you try this exercise in 2 dimensions in a parking lot with a bunch of rocks, it is very difficult to accidentally occlude far objects with near ones, unless they are mostly surrounding the "viewing center".

As long as the nearest objects block a relatively small portion of the total viewing angle (and even the brightest stars in the night sky qualify), there will be an enormous amount of empty space left to fill with other points.

Ring of Fire

To come at it from a different angle, let's think about what it would take to make a continuous ring of light surrounding the earth, using average-sized stars. A typical star has a diameter around $10^6 m$, and they are spaced about 5 ly ($10^{16} m$) apart, on average. That means if you take two nearby stars, and try to fill up a line between them with more stars to make a continuous segment of starlight, you need about 10 billion stars! That's about 100 galaxies' worth of stars!!

Now, the circle having a radius of 5 ly from earth has a diameter around 16 ly. So just to make this ring of stars requires about 300 galaxies of stars. Trying to fill up the entire night sky with stars is going to be significantly harder. In this scenario, I've made the star placement as easy as possible: putting stars very close to earth, where they subtend the largest viewing angle. If we distribute them across distances that reflect the actual matter distribution in the observed universe, we will need many more stars, because most of them will get much smaller due to being billions of ly away, rather than 5.

So let's scale this up to the size of the universe: about 100 billion ly. Let's say the average stellar distance (distance from earth to a random star) is 30 billion ly (just to have a round number to make the math easier). A ring at a distance of 30 billion ly has a diameter of about 100 billion ly, or about $10^{26}m$. That means the ring could "host" about $10^{20}$ stars set side-by-side. Now, there are about $10^{24}$ in the visible universe. That means we could make about 10,000 of these rings if we lined up all the stars! That sounds like a lot, but 10,000 stars won't even fill up the average gap between 2 normally spaced stars. A band 10,000 stars thick would not even be as large as a dwarf galaxy (mainly because real galaxies don't have stars sitting side-by-side).


If we try to line up all the stars in the observable universe side-by-side, they take up a pitifully small portion of the visible viewing angle. If you then "shake the snow globe" and let them settle in random locations, it should become a little more obvious why it is unlikely that very many of them will end up occluding each other.

One final comparison: if we take take all the sand on earth, and turn all the mountains to sand, we will have roughly a number of sand grains equal to the number of stars in the observable universe, give or take 3 grains. At this scale, the observable universe would be about 100 ly across (there's about 60,000 stars in that volume near earth). That's the size of our "snow globe". Now shake it as hard as you can, so the grains of sand spread out through that entire space. After some vigorous shaking, how many of those grains do you think will line up exactly so that one hides another? I think you'd have much better odds of winning the PowerBall while getting struck by lightning and getting bitten by a shark simultaneously. Of course, this assumes all the stars are free-floating. In reality, the stars are bound into galaxies, and we should really only have counted the number of galaxies. But that is a more complicated calculation that I'll leave as an exercise for the reader.


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