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:
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.
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