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by Bart Arondson

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This widely praised Nikon 105mm macro lens has a maximum f-stop value of 2.8. I've held it in my hands, it's a big lens. Meanwhile, this 50mm Nikon lens can go up to f/1.2, despite being 25% cheaper and a LOT smaller. So at least with these two lens, there is not a very direct correlation between price and f-stop ability.

What is it that determines a given lens' f-stop range? Why can't that 105mm go sub-2?

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3 Answers 3

up vote 12 down vote accepted

The pupil (aperture opening) area is proportional to the square of the focal length (at the same f-stop). So 105mm being about twice the focal length of the 50mm, it would need 4x the pupil to be f/1.2.

In other words f/1.2, or any f-stop, doesn't correspond to a fixed diameter - it increases for larger focal lengths.

That also assumes both lenses gather and transmit the same amount of light to the aperture. Given the 50mm has a wider field of view, it will tend to gather more light, so it has a further advantage there.

The maximum aperture area is clearly restricted by the format of the camera - it can't be bigger than the lens mount. A lens can compensate by gathering more light, which is why big 300mm f/2.8 and 600mm f/4 lenses have enormous front lens elements.

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In the Wikipedia article about f-stop (f-number) it is mentioned how big the size of the pupil of your lens can be. Found in the "Notation" section. –  Johan Karlsson Jan 26 '12 at 10:16
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Wider field of view doesn't come into it as a separate — that advantage is directly and completely accounted for by the smaller physical size of the aperture as you explain in the first part of your answer. –  mattdm Jan 26 '12 at 12:09
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Um... the f-number is a ratio of focal length to aperture; the relationship is linear, not quadratic. So a 100mm lens would need an entrance pupil twice the diameter of a 50mm lens to achieve the same f-number. (If the design is scaled but otherwise identical, that means 4x the area and 8x the volume of glass, of course.) I'll assume the "square" got in there by force of habit -- it almost always comes up in any discussion of aperture. –  user2719 Jan 26 '12 at 17:14
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It is a linear ratio to the aperture diameter (f/D), but quadratic relationship to the area of the entrance pupil, which is the light gathering ability. The area being (D/2) squared, times pi. –  MikeW Jan 26 '12 at 18:35
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The same folk that think a 24" pizza is twice the size of a 12" one I imagine ;) –  MikeW Jan 27 '12 at 2:14

There's quite a bit more to it than a longer lens needing a larger diameter to maintain the same relative aperture.

For a couple of examples, spherical aberration and coma are both proportional to the square of the lens' aperture. If we took the design of the 100/2.8 lens and doubled the diameter of each element, we could expect to get a 100 f/1.4 lens -- but with four times as much spherical aberration and coma as the f/2.8 version. That would reduce quality from "a little soft wide open" to "Uh, nice bokeh I guess, but wouldn't be better if something was sharp?"

Okay, I'm exaggerating a little, but you get the idea. Along with just needing more material (e.g., doubling the diameter of an element requires about 8 times as much glass) the design needs much better correction for some of the common aberrations to maintain the same (or similar) quality at a larger aperture.

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To answer the title question, there are two limiting factors. A soft limit is design economics. That 105mm lens could be designed larger (smaller ratio). But is there enough demand for it to cover the costs of making such a lens? Apparently the manufacturer thinks not. A hard limit is the maximum size of the system mount. Typical SLR/DSLR systems can go down to about f/1.2. Cinema camera mounts can go beyond even f/1.0.

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