I have a Canon FD 28mm f/2.8 that I shoot on my entry-level Canon DSLR using an adapter. You have to set the aperture manually, but only every second aperture is marked: between ƒ/2.8 and ƒ/4, for example, there’s an unmarked stop on the aperture ring.

This got me curious: how can I determine the ƒ number of that middle stop? The obvious answer is that it’s ƒ/3.2, and of course the number doesn’t really matter; I should use that stop if it will allow me to make the photograph I’m trying to make.

For the sake of curiosity, though, is there a way to experimentally determine the ƒ number or aperture size when that stop is in use? (I’d prefer a software-based solution, not something that requires equipment beyond the camera and the lens.)


One method is to use the camera to make the measurement. This is actually surprisingly accurate, and is how most T stop measurements are done with a single reference measurement using a spectrophotometer.

  1. Take an exposure at a known (or even "known") aperture, like f/2.8

  2. Open the raw image in a software like RawDigger, or process it with DCRaw to get the unmanipulated image data. Find the intensity of the image in the center

  3. Stop the lens down and repeat, the ratio of the brightness of the two informs you about the aperture.

If you get e.g. 1024 at f/? and 2048 at f/2.8, you know you have exactly halved the exposure and f/? = f/4. The new aperture will be (old aperture) + log2(ratio of exposures).

You can also do this with Jpeg files, or images processed (opening counts as processing in this case) by e.g. Adobe Camera Raw. You just have to linearize them first, when the raw image is almost always already quite linear.

  • This is a really good suggestion. Question: when finding the intensity of the image in the center, how much of the center do you suggest? A 1x1 pixel in the center, or a certain radius...? – scottbb Jul 29 '16 at 22:47
  • @scottbb it doesn't really matter. The requirement of being in the center of the image is only to be free from the effects of vignetting. There is a trade off between sampling a larger area and getting better accuracy with averaging and sampling a larger area and getting worse accuracy because of vignetting. – Brandon Dube Jul 29 '16 at 23:12

You could calculate it by hand. The ƒ# is actually a formula N = ƒ/D where N is is the ƒ#, ƒ is the focal length and D is the diameter of the entry pupil. So, if you take your focal length of 28mm and measure the diameter of the pupil, D, you should get 8.75mm for ƒ/3.2.

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    Measuring the diameter of the entrance pupil would be quite difficult without an optical method for measuring it (e.g. nodal bench). Since it's virtual and you can't actually touch it. – Brandon Dube Jul 29 '16 at 21:17
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    @Brandon Dube I guess one could simply project the shape onto a piece of paper and measure that. It wouldn't be the exact size, but that shouldn't be necessary either, as only the ratio to the neighbouring apertures is of interest. – null Jul 30 '16 at 11:27

The f/number (focal ratio) is derived by dividing the focal length by the working diameter of the aperture. The working diameter of a 28mm lens set to f/2.8 is 10mm. The working diameter of a 28mm lens set to f/4 is 7mm. The traditional ½ stop between the two is f/3.4. What will be the working diameter for this f/stop? Answer = 28mm ÷ 3.4 = 8.2mm Suppose the unmarked stop is one third f/stop smaller. This is f/3.2. The working diameter is thus 28mm ÷ 3.2 = 8.75mm Now the problem is this: how to find the actual working diameter? Not that easy but --- dismount the lens. Using a flashlight that is adjustable to a spot (parallel light rays exit). Shine the flashlight into the lens from the front. Hold a piece of paper about 25mm from the back of the lens. You will see a circle of light projected on the paper. This circle will be what is called the exit pupil. Measure the diameter of this circle and divide this value into the 28mm. You can check your work by setting the lens to some of the marked f/number settings.

You could also aim the camera at a uniform (mundane) target. At f/4 allow the camera automation to select an appropriate shutter speed. Now set the lens to the unmarked aperture setting. Did the shutter speed change? If so, what is the multiplying factor? Using this method you can figure out if the unmarked increment is 1/2 or 1/3 f/stop.

  • "Hold a piece of paper about 25mm from the back of the lens." How did you come up with 25mm? The focal length should be measured from the sensor plane (which is 44mm from the lens's bayonet base on Canons). Thus the thin-lens equivalent distance is 44-28 = 16mm from the base of the bayonet mount. Also, doesn't that technique require the lens to be focused at infinity? – scottbb Jul 29 '16 at 22:56
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    This is very wrong. The exit pupil need not be behind the lens at all. In the case of a 28mm retrofocus lens, it is most likely actually in front of the lens to begin with. The spot that appears on a piece of paper 25mm behind the lens is just an in (or out of) focus image of whatever the lens is pointed at. It gives you no information as to the exit (or entrance) pupils. – Brandon Dube Jul 29 '16 at 23:14
  • @BrandonDube I think we need to clarify "in front" and "behind" the lens. What Alan is describing is measuring the diameter of the "virtual" or "apparent" diameter of the thin-lens equivalent (This diagram is adequate to describe what he's talking about) (replace 35mm in the diagram with 28mm in this case). You're right that the spot will be out of focus, but at 28mm from the image sensor (or 16mm from the bayonet base of the lens), that's the thin lens equivalent. – scottbb Jul 29 '16 at 23:50
  • @scottbb the diagram is at least idealistically correct for locating the rear principal plane. This is not the exit pupil. I use in front literally. For a retrofocus lens, the exit pupil is often in front of the front lens' front vertex. The exception is very long retrofocus lenses, such as planetarium projection lenses, for which it is buried somewhere inside the system. In any case, with a retrofocus lens the exit pupil will never be behind the lens. – Brandon Dube Jul 30 '16 at 0:33
  • @BrandonDube actually, yes it is the exit pupil (to the extent that the edges of the virtual lens are distinct and not blurred). 28mm divided by the this diameter is the f-number of the equivalent thin-lens. – scottbb Jul 30 '16 at 0:38

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