I have a sigma 180mm f/2.8 lens, which I bought because it was the largest aperture I could find that focuses around 0.5m See this question for why this was important.. Upon arrival, however, the lens cannot maintain the aperture as I bring the focal plane close to the camera. The maximum aperture the camera (D850) reports as I bring the focal plane from infinity to the closest focus is in the table below. Is there any way to force the lens to stay open at a wider aperture? I'm guessing the image quality or something must deteriorate else the manufacturer would not limit it. Can anyone describe these effects?

Focal length (m) Max aperture (f#)
Infinity 2.8
3 3.2
1.5 3.2
1.2 3.3
1 3.5
0.85 3.8
0.8 4
0.65 4.2
0.575 4.5
0.55 4.8
0.525 5

3 Answers 3


Is there any way to force the lens to stay open at a wider aperture?

Your question implies a misconception: as you focus closer and the reported aperture is decreasing, the physical aperture is not actually closing down. The camera/lens is reporting the effective aperture.

The effective aperture is given by:

$$ E = N\left(M+1\right) $$

where \$E\$ is the effective aperture, \$N\$ is the aperture setting f-number, and \$M\$ is the magnification of the lens at a particular focus distance.

See also:

The magnification \$M\$ is the ratio of the image distance \$d_\text{i}\$ (the distance from rear principal plane of the lens) to the object distance \$d_\text{o}\$ (the distance from the front principal plane of the lens to the subject in focus): \$M = d_\text{i}\,/\,d_\text{o}\$.

See also:

Practically, you usually don't know the image distance. You usually don't know the subject distance accurately (especially at close focusing distances), but you can estimate it fairly closely. If you rewrite the magnification ratio in terms of focal length \$f\$ and subject distance using the thin lens equation, you get an equivalent expression:

$$ M = \frac{f}{d_\text{o} - f} $$

So from those equations, you can see that the effective aperture is equal to the real aperture (the focal length of the lens divided by the entrance pupil diameter) only when the lens is focused at infinity. At infinity focus (or even just very large subject distances, such that \$d_\text{o} \gg f\$), the magnification is basically zero, so the effective aperture \$E\$ equals \$N\$.

But when \$d_\text{o}\$ is not significantly larger than \$f\$, the magnification \$M\$ becomes appreciable enough to significantly impact the effective aperture.

As a matter of interest, but not directly pertaining to your question, let's compute where the front principal plane is for your lens. Using the reported effective aperture \$E = 5\$, we find from the effective aperture equation your magnification is about \$M = 0.79\$.

Solving the magnification equation for \$d_\text{o}\$, I get a subject focus distance of about 409 mm. But your table reports a focusing distance of 525 mm. How can this be? Firstly, both the reported aperture of f/5 and the actual set aperture of f/2.8 are probably not highly accurate numbers, so some error most likely propagated there.

However, even if we could account for the inaccuracies in the f-numbers, the subject distance \$d_\text{o}\$ would still be less than your reported working distance, simply by the very nature of your Sigma 180 mm being a telephoto lens: telephoto lenses move the principal planes out in front of the lens. Thus, the optical subject focus distance is less than the actual subject-to-front-of-lens distance for a telephoto lens.

For more on telephoto lenses, see:

  • \$\begingroup\$ Excellent! Thank you. None of your formulae mention the actual aperture? What is the relationship between actual aperture, effective aperture, and magnification? \$\endgroup\$
    – James
    Commented Jul 10, 2018 at 4:29
  • 1
    \$\begingroup\$ @James The first equation refers to the set (what you call the “actual”) aperture number, N (i.e., 2.8, that you set in the camera). I prefer the terms “set” and “effective” aperture, as opposed to “actual”, because in a very real sense, the Effective aperture is the Actual aperture for real calculations (especially so in macro situations). That’s why the camera reports the effective aperture, because that’s what it really is. The simple f/2.8 is just an approximation, only actually precisely applicable at long focal distances. \$\endgroup\$
    – scottbb
    Commented Jul 10, 2018 at 4:50
  • \$\begingroup\$ I see. For my application I am trying to maximise the collection angle, hence my questions about the aperture. \$\endgroup\$
    – James
    Commented Jul 10, 2018 at 5:37
  • \$\begingroup\$ @scottbb Have you written a book that I can just buy. That would save me a lot of time compared to going through and copying and pasting all the excellent answers you have posted. 😊 \$\endgroup\$ Commented Jul 26, 2023 at 16:01
  • \$\begingroup\$ @GrantRobertson lol, no, but thanks. =) There used to be a blog functionality here, and for "textbook"-style stuff, I wish it were still around. It could have really worked well for Photo-SE eventually, I think. \$\endgroup\$
    – scottbb
    Commented Jul 26, 2023 at 22:53

No, not possible. That is simply what all macro lenses do. They can only focus closer by moving the lens farther forward (so to speak, it's probably done internally today). Because, it is necessary to increase the focal length to increase the magnification. And f/stop is defined as focal length / aperture diameter, so the f/number increases somewhat. The chart you show is very normal for that macro lens.

Marked focal length is only valid at infinity, but focusing closer changes it a bit. Holding f/stop constant to avoid that change is one reason regular lenses are not allowed to increase magnification to more than about 1:10.

  • \$\begingroup\$ Thanks for the answer. When you say f/stop = focal length / aperture, what is the definition of focal length? I assumed that would be the 180mm of the lens. \$\endgroup\$
    – James
    Commented Jul 9, 2018 at 3:04
  • \$\begingroup\$ @James It is the "180mm of the lens", but that nominal value is measured when the lens is focused at infinity. As you focus closer, the actual focal length increases. \$\endgroup\$
    – mattdm
    Commented Jul 9, 2018 at 3:50
  • \$\begingroup\$ Okay one more question then. Is the actual aperture still 180/2.8=64mm when focusing at minimum focal length? \$\endgroup\$
    – James
    Commented Jul 9, 2018 at 4:04
  • 1
    \$\begingroup\$ @James Likely not - aperture in this context is measured as a diameter of physical aperture, seen from front of the lens (entrance pupil en.wikipedia.org/wiki/Entrance_pupil ) - this may change when you change focal distance. \$\endgroup\$
    – Arvo
    Commented Jul 9, 2018 at 10:09

The aperture is already staying all the way open when the lens is set to f/2.8. But the f-number is not staying the same because the focal length is changing as the focus distance is reduced.

Most lenses focal length is measured when the lens is focused at infinity. As the lens is focused at shorter distances, the focal length tends to change.

There are some lenses with front elements that do not move and the focusing elements located in certain points in the optical formula that actually lose focal length as they are focused closer. The Nikon AF-S 70-200mm f/2.8G VR II set at 200mm has an approximate field of view of a 140mm lens when focused at its minimum focusing distance.

But most lenses, including pretty much all macro lenses, increase their focal length as they are focused closer. Since the size of the entrance pupil (the aperture diaphragm as viewed through the front of the lens) remains the same at the same time as the focal length effectively increases, the effective f-number goes up. This is also true with many non-macro lenses, but the effect is not too noticeable until the much shorter focus distances that only macro lenses are capable of are used.


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