It seems odd to me that Canon EF 100-400mm f/4.5-5.6L does away with a front element of only about 63 mm, as reported by @jrista - which would be enough for only f/6.3 at 400mm, missing the spec by third of a stop.

It makes me wonder if it's possible to measure what aperture is actually used during taking a photo. It'd be useful both in the described case and exploring how exact stopping down to a smaller aperture actually is.

So my question is - how to measure what aperture is actually used to take a photo? It's okay if the scene has to be specially constructed/measured for performing the test.

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    \$\begingroup\$ Manufacturers sometimes take liberties when it comes to specifications. I've used lenses where the stated focal length is out by at least 10%, when you account for a slightly shorter focal length, a slightly smaller max aperture and a few millimetres in Jrista's measurement, balance is restored! \$\endgroup\$
    – Matt Grum
    Commented Jun 23, 2011 at 12:15
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    \$\begingroup\$ And that's why I think it's silly to say "ISO 256,000" instead of "ISO 250k". \$\endgroup\$
    – mattdm
    Commented Jun 23, 2011 at 13:04

3 Answers 3


You can probably calculate this by rearranging the DOF formula to solve for c, or circleOfConfusion, as @MattGrum stated. I haven't tried to rearrange a formula as complex as DOF for a while, so I hope my math is correct here:

DOF = (2 Ncƒ²s²)/(ƒ⁴ – N²c²s²)

The terms of that equation are as so:

DOF = depth of field
N = f-number
ƒ = focal length
s = subject distance
c = circle of confusion

For simplicity sake, I'm going to reduce the DOF term to just D.

Now, the term for c appears twice in this equation, one of them to the power of two, so were probably looking at a polynomial of some sort in the end. To rearrange:

D = (2Ncƒ²s²)/(ƒ⁴ – N²c²s²)
D * (ƒ⁴ – N²c²s²) = (2Ncƒ²s²)
Dƒ⁴ – DN²c²s² = 2Ncƒ²s²
0 = 2Ncƒ²s² + DN²c²s² – Dƒ⁴
DN²c²s² + 2Ncƒ²s² – Dƒ⁴ = 0 <-- QUADRATIC!

As Indicated, rearranging terms produces a quadratic polynomial. That makes it pretty strait forward to solve, since quadratics are a common type of polynomial. We can simplify for a moment by substituting some more general terms:

X = DN²s²
Y = 2Nƒ²s²
Z = –Dƒ⁴

That gives us:

Xc² + Yc + Z = 0

Now we can use the quadratic equation to solve for c:

c = (–Y ± √(Y² – 4XZ) ) / (2X)

Replacing the X, Y, and Z terms with their originals and reducing:

c = (–2Nƒ²s² ± √(4N²ƒ⁴s⁴ + 4D²N²ƒ⁴s²) ) / (2DN²s²)

(Whew, thats pretty nasty, and I hope I got all the right terms replaced and typed in correctly. Apologies for discrepancies.)

My brain is a bit too fried right now to figure out exactly what it means for the circleOfConfusion to be quadratic (i.e. having both a positive and negative result.) My first guess would have to be that c grows both when you move towards the camera from the focal plane (negative?), as well as away from the camera and focal plane (positive?), and since quadratic equations grow to infinity pretty quickly, that would indicate the limit on how large or small the circle of confusion could actually become. But again, take that analysis with a grain of salt...I scratched out the solution to the formula and that took the last bit of brainpower I had left today. ;)

If that is the case, then you should be able to determine a maximum CoC for a given aperture and focal length, which would, hopefully, be (or allow deriving) the diameter of the aperture (entrance pupil.) I am willing to bet, however, that this is not actually necessary. My analysis on the linked answer of @Imre's question was rather rough...I don't quite have the ability to observe my 400mm lens' aperture at "infinity", so I am probably seeing the entrance pupil incorrectly. I would be willing to bet that at a sufficient distance that you could call "infinity", the 100–400mm lenses f/5.6 aperture at 400mm would indeed appear to be the same diameter as the front lens element, so at least 63mm in diameter. My measurement of the diameter of that lens was a bit rough too, and it could be off by ±3mm as well. If Canon's patent for a 100–400mm f/4-5.6 lens is telling, the actual focal length of the lens is 390mm, and the actual maximum aperture at "f/5.6" is really f/5.9. That would mean the entrance pupil would only need to appear 66mm in diameter "at infinity", which is within margin of error for my measurements. As such:

I believe the EF 100–400mm f/4.5–5.6 L IS USM lens from Canon is probably spot-on as far as aperture goes, with a 390mm actual focal length and a 66mm entrance pupil diameter, all of which would jive with my own actual measurements of this lens.

  • \$\begingroup\$ Complicated... Looks like 7th grade algebra LOL \$\endgroup\$
    – J. Walker
    Commented Mar 30, 2012 at 23:00
  • \$\begingroup\$ The process is algebra, its just a lot of ugly terms to stuff into a quadratic equation, which can make it difficult to keep everything strait (especially when you've already been up and working for over 12 hours, and haven't had to solve the quadratic equation for...years). I already see that I forgot a closing parenthesis, which could have lead to misinterpretation. -.- \$\endgroup\$
    – jrista
    Commented Mar 30, 2012 at 23:05
  • \$\begingroup\$ I know this is several years later, but a couple comments. 1., a little algebra and factoring will yield 2 solutions to your final equation (separating the ±√... into separate solutions), c₁ = (ƒ²/DN) * (√(1 + D²/s²) – 1) and c₂ = –(ƒ²/DN) * (√(1 + D²/s²) + 1). But note that c₂ is strictly negative, so it can be ignored, so only c₁ is a real-world solution. But regardless, comment 2., this is not feasible. DoF is not really a "free variable" that can be measured in order to determine the size of the blur circle. Note that DoF is defined in terms of ƒ, N, c, and s. ... \$\endgroup\$
    – scottbb
    Commented Mar 24, 2019 at 2:40
  • \$\begingroup\$ ... and more importantly, when s is equal to or greater than the hyperfocal distance, DoF is infinite. So any value of s > H is useless when using this equation. But collimated light parallel to the lens's optical axis and focusing at infinity is how focal length, and entrance pupil size, is really measured in a lab. \$\endgroup\$
    – scottbb
    Commented Mar 24, 2019 at 2:46

If you have a point lightsource at a known distance and you know the focal distance (the distance to which the lens is focussed) then you can calculate the aperture based on the size of the circle of confusion (the round blob you get when a highlight is OOF).

I don't know the formula off the top of my head but it could be rearranged from the depth of field formula (might have a go at this when I have time).

You also have to know the exact focal length, which I suspect might be partially to blame for the discrepancy.

  • 1
    \$\begingroup\$ I guess it'd be best to have two point lightsources at known distances: focus to one, measure the OOF blob from the other. \$\endgroup\$ Commented Jun 23, 2011 at 12:18
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    \$\begingroup\$ @jukka That's what I was thinking also. Still leaves the problem of calculating the correct focal length, which can be done with known object sizes at known distances... \$\endgroup\$
    – Matt Grum
    Commented Jun 23, 2011 at 13:31
  • \$\begingroup\$ @Jukka Measure the width of your LED (or other point light source)? \$\endgroup\$
    – nchpmn
    Commented Jun 25, 2011 at 11:06
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    \$\begingroup\$ @Crashdown for accuracy you want to measure something larger than that. A triangle of LEDs would be a good test setup. The distance between two of the LEDs gives you your angle of view and hence focal length, the distance to the two to the camera gives your focal distance (having them both in focus ensures the camera is square on) and finally the size of the blur disc lets you calculate the aperture. \$\endgroup\$
    – Matt Grum
    Commented Jun 25, 2011 at 12:10

The aperture f-number describes the amount of light that passes trough the lens, for a theoretical single element lens this is also the ratio between focal length and physical size of the entrance pupil - but no camera lens sold today is a single element lens.

In 1874, John Henry Dallmeyer wrote that the only way to get the "intensity ratio" (that was before the term f-number was coined) of a lens with more than two elements is to measure the amount of light that goes trough the lens (search for "effective aperture" in the wikipedia article on f-numbers).

note: I believe it is possible to calculate it today, but not with my math expertize

So, what you should be measuring is the amount of light that passes trough the lens, this would have been easy if we had a good reference point -

Take a picture of a solid color surface under constant light with the same ISO and shutter speed, once with the reference lens at the reference aperture and once with the test lens at the test aperture - calculate the light intensity difference between photos to get the aperture difference in stops.

In real life you don't have a good reference point but you can just take a lens that should have no trouble opening up to f/5.6 (a 50mm f/1.8, a kit lens at the wide end or the 100-400 at 100mm).

You don't even have to do anything fancy with the image data, if the histogram in the two pictures is the same both were taken with the same aperture.

If you want to get fancy and don't have a lens you can "trust" you can probably shoot a gray card and use a light meter so you know the expected intensity or the result photo.

And remember to repeat the experiment multiple times - the mechanical aperture on most lens in notoriously inaccurate.

  • \$\begingroup\$ When using another lens for reference, you can calculate difference of T-stops based on exposure; to get F-stops, you'd also need to know transmission difference of the lenses. \$\endgroup\$
    – Imre
    Commented Mar 30, 2012 at 21:40
  • \$\begingroup\$ @Imre - no I dont, think about light meters - they let you enter ISO and shutter speed and give you an accurate aperture value without knowing what lens I used \$\endgroup\$
    – Nir
    Commented Mar 31, 2012 at 6:01

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