At first sight this may seem like a dumb question. Of course you just head off in some direction and go as far as you can go, then you keeping going and doing that, and eventually you'll get to infinity. Except that at a finite velocity (even c) in a finite time one can only go a finite distance, and infinity is much farther than that. So you need either an infinite velocity or an infinite amount of time to get to infinity, which sort of begs the question.

Fortunately, photographers don't have to worry about any of that. From a practical point of view they need to know if, say 1000 feet, is, for all intents and purposes, effectively the same as infinity so far as the focus in a picture is concerned.

It turns out there's an equation for that, and this is where we'll see the dreaded depth-of-field equations, because depth-of-field is central to the question of how far infinity is, from a photographer's perspective.

The crucial equation gives what's called the hyperfocal distance, H. The hyperfocal distance is the distance to the near limit of the depth-of-field when a lens is focused at infinity.

Let me say that again: when the lens is focused at infinity there's a nearer distance representing the limit of the depth-of-field, and that distance is called the hyperfocal distance, H:

H = F² / f·d,

Here F is the lens's focal length (typically in mm's),
f is the focal ratio number (11 for f/11, etc.) that the lens is set at, and
d is the size of the "circle of confusion" (in the same units as F).

It turns out that if the camera is focused at this disance, H, instead of infinity, then everything between H/2 and infinity will be in focus. Half the hyperfocal distance then is essentially the same as infinity. One could say infinity starts at H/2. So that's the answer to the question posed in the page's title. The rest will be by way of explanation.

The circle of confusion diameter is a measure of how sharp (small) something needs to be at the sensor (film or solid state) to qualify as being in 'acceptable' focus. This is obviously getting into subjective territory, as things don't go in jumps from being in focus to being out of focus. It's gradual. There are two logical and sensible ways of coming up with a good number for this quantity that people have used:

1. The simplest is just to take d as a small fraction of the lens's focal length.
   A ratio of F/600-800 gives a loose critera for the sharpness at the limits of the depth-of-field (~1/5 mm for F=135mm), F divided by 800-1000 a tighter one, while 1200 gives a fairly stringent criteria. Higher numbers, like 1500 or 2000, might be called for in extreme circumstances. F/1720 as it turns out corresponds to 2 arc-minutes, which is about the limit of resolution of the unaided eye in typical daylight conditions -- because Tan 2' = 0.00058 = 1/1720. For an F=210 mm lens this is ~1/8 mm. Remember, this is on the film or sensor. So an otherwise pinpoint of light at infinity (like a star) would record as a tiny blob ~1/8 mm across if the lens was focused at the hyperfocal distance.
   [A series of photos made with pinholes rather than a lens and using these fractions of F for the sizes of the pinholes would be instructive if I ever got around to it. The practical difficulty in doing this is the small size of the pinholes required. For a normal focal length for 4x5, it's convenient to take F=172mm. Then F/1720 is just 0.1mm, or four thousandths of an inch. This is too small to drill, but it's the diameter of a #42 gauge wire which could be used to punch a hole in, say, aluminum foil.]
2. With small formats and short focal lengths the ratio approach might yield a size so tiny it's comparable to (or beyond) the resolving power of the system, usually it being the film or pixel size for a solid state sensor which is the limiting factor.
   So the other way of coming up with a number for the circle of confusion size is by considering the sensor's resolution (resolving power), and by then converting this into a linear size by multiplying it by some amount -- typically several times, like 6-8x. One then gets an appropriate quantity.
   It often varied from format to format. For example, Kodak Publication No. O-18 from nearly fifty years ago had a table (pg. 47) where for 35mm film the number is d=0.002" = 0.050 mm's; for an F=50 mm standard lens this would be F/1000 by the ratio approach above. The depth-of-field markings on old Canon 35mm film SLR lenses (FL and FD series) corresponded to a slightly smaller d=0.035 mm, or just 35 microns. I've only measured one newer Canon EF lens, but apparently this same criteria was carried over (by Canon at least) into the DSLR era, where the sensors typically have pixels ~5-6 microns across. So d is ~6-7x the pixel size.

These rule-of-thumb ratios derive from the days when large format photography was the norm, and so degrees of enlargement were small (1x-4x). If it looked sharp (more or less) on the ground glass when the lens was stopped down to the taking aperture, then it wasn't going to be much worse in the final contact print (1x) or minimal enlargement of the negative.

There's also something in these ratios that I deftly skipped over earlier, namely in the part about the definition of the hyperfocal distance. The phrase "...when a lens is focused at infinity" is a bit of circular reasoning if we're trying to figure out how far infinity is, which we need to know to focus the lens at that distance. Well, anything beyond ~2000·F qualifies. For a 210mm lens this is about ¼ mile.

So let's imagine we have the three numbers we need to plug into the equation to calculate H for a given situation. The first thing we notice is that H is a large number. This is because F gets squared, and then gets divided by a small number (d), making the result even larger. f is maybe in the range 2-20 (8 or 11 to maybe 32 for large format), so this brings H back down by a factor of 10-500x, but it will still be pretty big. This makes sense because H is in mm's (or maybe inches, if one wants to work in those units), and it represents a distance out into object space to the thing being photographed, which will typically be a lot of mm's or inches away.

So we'll typically be dividing the result by large conversion factors to get it into meters or feet. Only for very short focal lengths and/or fast lenses (f/2 - f/2.8 or less) will it be only several feet; in fact, this was what was employed in making simple point-n-shoot cameras like the Kodak Instamatic possible, where everthing from a few feet to infinity was within the depth-of-field, so it needed no focus control and was thus that much quicker and easier to use. Many web cams, car dash cams, security cameras, phone cameras, drone cams, and the like also utilize this priniciple.

But for large format the focal lengths are going to be in the dozens or hundreds of mm's, and H will be correspondingly large. For example, for F=210mm and d=0.150mm (a little bit better than the 0.200mm used before), and at a typical f/16, H will be 60 feet (18 3/8 meters), so everything from about 30 feet (H/2) to infinity will be "in focus". This is, of course, without any view camera swings or tilts being used; in fact, these exist in the first place because in some circumstances they make it possible to extend the depth-of-field range considerably.

For astro-photography, focal lengths will range from a fraction of a meter up to several meters. For example, the venerable "C8" (Celestron) telescope has an 8" aperture and is f/8, so F=64", or 1626 mm's; they made bigger models with even longer F's. For the super-scopes F is in the tens of meters.

When H is large it can be constructive to switch perspectives and look at what's going on at the focal plane. After all, a view camera doesn't generally have a focusing distance scale, and we are looking at the image on the ground glass.

The standard lens equation provides a start:

1/F = 1/o + 1/i

Where, F is the lens focal length (typically in mm's),
o is the distance from the lens center to the object, and
i is the distance from the lens center to the image focus.

If we know any two of the three quantities we can solve for the third. Looking at the lens equation, if o is very large then 1/o is very small; if we ignore that term as if it were actually zero then it's easy to see that i=F. For an o which is large but not infinity i will be just a little bit larger than F. This makes it sensible to subtract off F and deal with the distance from where infinity focus is (i=F) to where it actually is:

x = i - F

For the example above of an F=210mm lens focused at ¼ mile (400,000+ mm's ≈ ∞), x=0.11 mm. Note that this is less than the circle of confusion size (d = 0.15 mm) used earlier, so from the focal plane (x=0) a circle of diameter d at a distance x away would subtend an angle >45° -- 68½° to be precise.

A trick we can then do (since we know F) is set o equal to H and solve for i, and thus x.