An f-number is determined as the ratio of the focal length to the diameter of the entrance pupil. But the actual aperture diaphragm is usually not a perfect circle. Its shape is determined by the number of diaphragm blades and their edge (curved or straight). Depending on those the aperture might be a pentagon, hexagon, octagon... and with straight or curved edges. So how is the diameter actually determined? Is it for example twice the distance between the center and the widest point at an intersection of the blades? Twice the distance between the center and the nearest point on an edge?

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    \$\begingroup\$ Very likely the diameter of a circle with the same area but I will let someone who really know answer \$\endgroup\$
    – lijat
    Nov 28, 2018 at 17:58
  • \$\begingroup\$ @lijat Interesting idea. Since the interest is mostly in the actual surface area of the aperture opening as that determines exposure I suppose the diameter of an equivalent circle would make sense. \$\endgroup\$
    – G_H
    Nov 28, 2018 at 18:33

1 Answer 1


If I had to figure out the answer to this problem, I would start with the open aperture which will be circular as the blades of the iris are retracted.

That would be the f/wide-open. I'd take the diameter of that aperture and divide the focal length by it to get the first "stop." Let's say it's an f/4 lens. Scratch marks the spot opposite the focussing "index."

Close the aperture until the light intensity that passes through the lens is half of that (using a light intensity measuring device of some sort). The engraved mark on the lens aperture would be f/5.6 as each stop is equal to a multiple of 2 (half if stopped down and twice if opened.)

Note: The light measuring device can be calibrated using the inverse-square law: Intensity varies with the square of the distance, in another experiment before this one.

I would continue in this manner until I had full, half, and even third stops if I wanted and if space on the lens barrel permitted.

In short, the mathematical relationship of light intensity to f/ stops is the "tool" I'd use for the task.

  • \$\begingroup\$ That's interesting in that it would be an actual practical approach to determine how far to close the blades to get the correct reduction in exposure. I don't know how camera manufacturers determine this, and possibly they do this for calibration of a lens during manufacture. But I do suppose in the design phase they'll use an actual diameter, so I'm wondering if there's a standard for that relative to the aperture's shape. lijat's comment would make sense in that case. \$\endgroup\$
    – G_H
    Nov 28, 2018 at 18:32
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    \$\begingroup\$ @ G_H This is, in fact, the manner in which "T" (for transmission) stops are determined. T/ stops are used in many Cine cameras for accuracy with variable focal length lenses as the f/#s change with focal length at the same aperture settings. It's also rather trivial to equate regular geometric shapes with circular areas for f/ #s \$\endgroup\$
    – Stan
    Nov 28, 2018 at 18:59
  • \$\begingroup\$ Looking back at this answer it does say how to determine how much to stop down the aperture to reduce light intensity by a given amount, but it doesn't really answer the question regarding what the actual diameter is considered to be. Then again, maybe that doesn't really matter or they do use the diameter of the equivalent circle. I'm waiting a bit to see if someone in the know crops up with a more concrete answer, but I won't forget to accept if that doesn't happen. \$\endgroup\$
    – G_H
    Feb 7, 2019 at 15:13

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