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Many lenses list a range of maximum apertures. For example, the Nikon kit lens has a maximum aperture of f/3.5 at 18mm, but at 55mm it's only f/5.6. Does that mean that at half-way between 18mm and 55mm, the maximum aperture will be half-way between f/3.5 and f/5.6? Or is it more complicated than that?

If the relationship is simply linear, that would seem to imply that my 18-105mm lens has a superior maximum aperture at 55mm:

Lens A             Lens B
18mm = f/3.5       18mm = f/3.5
55mm = f/5.6       55mm = (mean of f/3.5 and f/5.6) = f/4.55 ?
                   105mm = f/5.6

If my numbers are right, that means the longer zoom is potentially about a stop brighter at 55mm. But am I calculating this right? Does a calculation like this even make sense? It assumes a linear relationship, but I don't know if that's justified.

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3 Answers 3

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No, I doubt most lenses will be linear that way. Here is a Nikon 16-85mm result. It is a f/3.5 to f/5.6 lens.

The results reported in EXIF data are:

16mm zoom - reports 16mm at f.3.5
24mm zoom - reports 24mm at f.4
35mm zoom - reports 35mm at f/4.5
50mm zoom - reports 50mm at f/5
70mm zoom - reports 68mm at f/5.6
85mm zoom - reports 85 mm at f/5.6

16 to 35mm is roughly double, and the difference is 2/3 stops.
24 to 50mm is roughly double, and the difference is 2/3 stops.
35 to 70mm is roughly double, and the difference is 2/3 stops.

But don't assume linear. Modern lenses do all kinds of tricks inside, shifting elements to implement zoom and focus; both modify focal length.

If you want to know intermediate values for your lens, instead of guessing, you can simply do this same check and see what it reports. Do make sure your focused distance is several feet, say 6 or 8 feet anyway, because the focal length also changes with distance. The marked focal length is only valid for infinity focus. It will be different (but not reported) if closer. If up pretty close, the lens probably won't open to maximum aperture number (because the new true focal length computes a different number).

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  • \$\begingroup\$ I hadn't noticed that the Exif data actually has a field that explicitly says "maximum possible aperture", which changes depending on focal length. Should make it quite easy to determine which lens actually has best aperture... \$\endgroup\$
    – MathematicalOrchid
    Sep 17, 2016 at 15:28
  • \$\begingroup\$ Since the aperture values are rounded to the nearest standard 1/3 stop value, if you center each value at the focal length in the middle of the range for each value most lenses become very close to linear if you also account for the fact that wide open may not actually go all the way to f/3.5 (It may be more like f/3.7 wide open). Notice that in your example each doubling of focal length increases the aperture by about 2/3 stops. So in terms of stops, it is roughly linear. Of course there is the difference between measuring the focal length in one dimension and the area of the EP in two. \$\endgroup\$
    – Michael C
    Sep 17, 2016 at 17:27
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    \$\begingroup\$ If the f-number is increased roughly 2/3 stop for each doubling of focal length that is a linear relationship with a 2:3 ratio. Any time you have a constant ratio between two sequences of numbers that is a linear relationship. It doesn't have to be 1:1. \$\endgroup\$
    – Michael C
    Sep 17, 2016 at 20:36
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    \$\begingroup\$ That is merely making up stuff. You need data, and you have no data, but the data is this: The middle point of the posted actual focal length data (16-85mm lens) is 50mm, which is roughly 35mm from either 16 or 85mm ends (16+35=51, 85-35=50). But the maximum aperture at 50mm is in fact f/5, which is 1.33 stop from the f/3.5 16mm end, and only 1/3 stop from the f/5.6 85mm end. How do you imagine that to be linear? LOL. \$\endgroup\$
    – WayneF
    Sep 17, 2016 at 21:43
  • \$\begingroup\$ The same way f/4 is halfway between f/2 and f/8. It's two stops in each direction. 50mm is 3.125X 16mm, 85mm is only 1.7X 50mm. The focal length midpoint in terms of ratios is not 50mm, it is 36mm. 36mm is 2.25X 16mm. 80mm is 2.22X 36mm. At 36mm the f-number is between f/4 and f/4.5 (true f/4 is somewhere around 27-28mm, true f/4.5 is around 40mm), which is almost exactly the same number of stops (about .8 stops) to f/3.5 and f/5.6 in each direction. If you don't understand how logarithmic scales can express exponential relationships in linear terms then you can't truly understand the ... \$\endgroup\$
    – Michael C
    Sep 18, 2016 at 8:38
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The camera lens acts like a funnel in that it gathers light. The larger the working aperture diameter, the more light the lens will gather. The camera lens also acts like a projection lens; it projects an image of the outside world onto the surface of film or digital chip.

The focal length of the lens determines the size of the image produced. The longer the focal length, the larger the image. Suppose you take a picture of a tree using a 50mm focal length lens, the image of the tree measures 5mm in height. Now you mount a 100mm lens, the image of the tree will now measure 10mm. Now you mount a 25mm lens, the image height is now 2 ½ mm. Thus the magnification yield of the lens is a linear function.

What you likely do not know: When the focal length is changed, the image brightness also changes. Each time we double the focal length, the image brightness falls off four fold (4X reduction). This is because the now twice bigger (2X magnified) image must play over 4X more surface area. In other words, double the focal length and the image brightness reduces two f/stops (a stop is a 2X change in light energy). This image brightness change with focal length is true should we keep the working diameter of the lens aperture unchanged. The brightening and dimming of image brightness with focal length changes is chaotic. We need to avoid this at all cost. If we don’t, under and over exposure will result.

Ratio to the rescue: We need to calculate a lens speed (brightness) value that holds regardless of focal length or working aperture diameter. Fortunately a ratio is dimensionless. The focal ratio is computed by dividing the focal length by the working diameter. Thus a 50mm lens with a aperture diameter of 25mm has a focal ratio of 50 ÷ 25 = 2. We write this value a f/2. The beauty of this system is: any lens, regardless of focal length or aperture diameter, functioning at f/2, will deliver the same image brightness. The f/number system intertwines the focal length and the working diameter and the resulting focal ratio takes the chaos away.

The f/number system is based on changing the working diameter of the lens to deliver a 2X (double of half) change in light energy on film or chip. This ratio is founded of the geometry of circles. Now the amount of light that a lens transmits changes as we change the focal length or working diameter of the aperture. Factorial: If we multiply the diameter of any circle by the square root of 2 = 1.414, we compute a revised circle with twice the surface area. For photographic purposes, we can round this value to 1.4. If we apply this value to the working diameter of lens, a number set emerges. The number set (f/numbers): 1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 – 22 – 32 Note: each number going right is its neighbor on the left multiplied by 1.4. Each number going left is its neighbor on the right divided by 1.4. This is linear progression of the f/number system. Each value is called a full stop. The term stops is used because the aperture diameter stops some light energy and passes some light energy. The stop is a 2X change is light brightness as it plays on film or chip.

For some applications, the full stop i.e. 2X change is too course, we can further refine the f/number system into ½ stops and 1/3 stops. The multiplier for ½ stop is 1.18 and the multiple for 1/3 stop is 1.12. This is the linear progression of the f/number system.

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The entire purpose of using f-stops is to make the exponential relationship of the powers of 2 and the square root of 2 expressible in a linear scale. It is the same principle that means a slide rule will allow us to add linear measurements of the logarithms of two numbers together and get the products, rather than the sum, of two numbers.

Since the aperture values at any specific focal length are rounded to the nearest standard 1/3 stop value, if you center each 1/3-stop aperture value at the focal length in the middle of the range for each aperture value most variable aperture lenses become very close to linear when you also account for the fact that wide open may not actually go all the way to the rated aperture. Remember that the actual f-number at any particular focal length isn't exactly the reported number.

Notice that in the example of the Nikon 16-85mm f/3.5-5.6 lens cited in another answer each doubling of focal length increases the aperture by about 2/3 stops. In a sense, doubling the focal length is a sort of stop as well. 36mm is one stop longer and reveals one-half the linear field of view than 18mm. 72mm is one stop longer and reveals one-half the linear field of view than 36mm, etc. So in terms of stops, it is roughly linear at a ratio of 0.67:1 or 2:3. As long as a constant ratio is maintained between a set of two number sequences, the relationship is considered linear. It doesn't necessarily have to be a 1:1 ratio to be considered linear.

Using the example of the Nikon 16-85mm zoom: f/3.5 may be more like f/3.63 wide open at 16mm. Within the 24mm to 35mm focal length range reported as f/4 the actual f-number at 24mm is just enough closer to f/4 than it is to f/3.5 to be labeled f/4 instead of f/3.5. Call it f/3.75. At 34mm the f-number is just short of being closer to f/4.5 than it is to f/4. Call it f/4.21. The lens is only exactly f/4 at somewhere around 28-29mm in focal length. Between 24mm and 28mm it is being rounded up to f/4. Between 29mm and 35mm it is being rounded down to f/4.

Of course there is the difference between comparing the ratio between two focal lengths in one dimension and comparing the area of the entrance pupil in areal dimensions, but that is pretty much the entire point of the f-number/f-stop scale: to make exponential relationships expressible in linear terms.

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  • \$\begingroup\$ Lots of luck with that notion. Zoom and focus are shifting elements back and forth in todays lenses, both affecting focal length. You cannot assume focal length is linear, shifting can even reverse direction in mid zoom (as another shift starts). Also as elements shift, the constant diameter aperture is shifted from the nodes, so its effective area can change. But fstop = focal length / diameter. So doubling focal length should be 2x fstop number, which is two stops, not 2/3 stop. But there are many things happening in there, all at once. \$\endgroup\$
    – WayneF
    Sep 17, 2016 at 18:53
  • \$\begingroup\$ As one example that things are NOT what we might expect, carefully measure exact distance to a subject at three meters. Then with a kit lens today, photograph it at widest zoom and at longest zoom. Then check the reported subject distances in Exif. Then repeat with another kits lens. It's laughable, the lenses have no clue. Too much shifting happening in there. \$\endgroup\$
    – WayneF
    Sep 17, 2016 at 18:53
  • \$\begingroup\$ All focal lengths are only accurate at infinity focus. If you use the actual focal length (instead of the reported FL) of a lens that is focus breathing at short distances, the relationship holds. With very rare exceptions, zoom lenses do all of the movement of zooming elements between the front element and the physical aperture diaphragm, thus the magnification affects the focal length and entrance pupil roughly equally. If this were not the case a 16-85mm zoom with an f/3.5 aperture at 16mm would be f/19 at 85mm with the same 4.6mm aperture diameter. \$\endgroup\$
    – Michael C
    Sep 17, 2016 at 20:19
  • \$\begingroup\$ Even with zoom lenses with "retrograde" zoom at in the wider end of the zoom range the numbers hold pretty closely. \$\endgroup\$
    – Michael C
    Sep 17, 2016 at 20:27
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    \$\begingroup\$ f-number=focal length/*entrance pupil* diameter. The physical diameter of the aperture diaphragm is only relevant in that it is the basis for the size of the entrance pupil after magnification by the lens elements between the physical diaphragm and the front of the lens. \$\endgroup\$
    – Michael C
    Sep 17, 2016 at 20:30

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