FAQ: How can I calculate effective aperture?

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rjlittlefield
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FAQ: How can I calculate effective aperture?

Post by rjlittlefield »

There are several commonly used formulas for calculating effective aperture.

Single lens, focused by extension:
(1) Feff = Fnominal * (magnification+1)

Coupled lenses, stopped in the front, with the rear lens focused at infinity:
(2) Feff = Fnominal * magnification

Coupled lenses, stopped in the rear:
(3) Feff = Feff of the rear lens [front lens has no effect on aperture]

Microscope objectives, either finite or infinite, when used as designed:
(4) Feff = magnification/(2*NA)

Teleconverter added between camera and all other optics.
(5) Feff_withTeleconverter = Feff_withoutTeleconverter * teleconverterFactor

Camera system handles it all
[6] Feff = whatever is set on the camera (for example modern Nikon camera with Nikon macro lens)

In these formulas:

. Feff is the effective aperture, specified as F-number, from the standpoint of the sensor.
. Fnominal is the F-number that the front (or only) lens would have, if it were used by itself at infinity focus.
. NA is the numerical aperture of the objective, as specified by the manufacturer.

To take some simple examples:

(1) A nominal f/4 lens, extended to give 2X magnification, gives effective aperture f/12. 12 = 4*(2+1)

(2) A nominal f/4 lens, coupled with a suitable rear lens to give 2X magnification, gives effective aperture f/8. 8 = 4*2

(3) A rear lens that gives effective aperture f/8 by itself, coupled with a wide front lens to give 2X magnification, still gives effective aperture f/8. This explains the common statement that "You don't lose light by adding a closeup lens."

(4) A 10X NA 0.25 microscope objective gives effective aperture f/20 when used at its rated 10X. 20 = 10/(2*0.25). If the objective is infinite, and you push it down to only 5X by using a short tube lens, then the effective aperture becomes f/10, 10=5/(2*0.25).

(5) Optics that would give effective f/8 by themselves, give f/11.2 with a 1.4X teleconverter added. 11.2 = 8*1.4

It should be clear from formulas (1) and (2) that coupling gives a wider effective aperture than extension, when using the front or only lens at the same nominal F-number. In practice the coupled lens approach is used only at magnifications of 1X or greater, so the gain can be as much as two F-stops (a factor of 2 in Feff). In general the advantage is a factor of (magnification+1)/magnification, so there is less advantage at higher magnifications.

Sometimes people get confused about formulas (1) and (2), when thinking about microscope objectives. The key is to concentrate on formula (4), which basically says that the manufacturer labels an objective by what it does, not what it is. Suppose for example that we have two objectives, one finite and one infinite, both rated at 10X NA 0.25. Because of formula (4), both objectives will give effective f/20 when used as designed, the finite on empty extension and the infinite coupled with its tube lens. Then working backward we can see that the finite objective would be Fnominal = f/1.818... if it were labeled like an ordinary lens, while the infinite objective would be Fnominal = f/2 if labeled like an ordinary lens. In this case the advantage of coupled lenses manifests as getting the same Feff from a narrower front lens, instead of getting a wider Feff from the same front lens.

Note that Formula (1) also assumes that the lens in question has pupil ratio = 1. This is seldom exactly correct, but it's often close enough for practical purposes. See FAQ: What is "pupil ratio" and why would I care? for more information about that.

Note also that some lenses defy simple application of any of these formulas. This is because turning the focus ring changes the relative spacing of elements inside the lens, so that it's really more like a different lens, than the same lens focused differently. In these cases the best approach is to determine lens behavior by experiment, then if you must calculate, do that using an approximate formula fitted to measurements from the experiments.

--Rik

Edited 23 April 2022, adding formulas 5 and 6.

rjlittlefield
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Re: FAQ: How can I calculate effective aperture?

Post by rjlittlefield »

The following question was asked in a recent telecon:
Is there a good reference source for the formulas to calculate effective aperture given various extension tubes, diopter filters, and teleconverters?
I think that most of the relevant formulas are contained in the post above.

The specific hardware being discussed in the telecon was a Canon MP-E 65, with added extension tubes, mounted in front of a 1.4X teleconverter, the whole setup giving a total magnification of 13X, and operated at a nominal aperture of f/5.6.

It turns out that Canon MP-E 65 follows the classic formula pretty well, but the addition of a teleconverter makes the calculation a bit more complicated.

Backing out the teleconverter for a moment, the MP-E 65 with added extension must be producing an optical magnification of 13X / 1.4 = 9.3X. So those optics, again without the teleconverter, would be operating at an effective f# of 5.6*(9.3+1) = f/58 [by Formula (1)]. Then adding the teleconverter, f/58 at 9.3X turns into effective f/81 at 13X as finally seen by the camera [by Formula (5)].

That configuration is far into diffraction territory, even on the 21 megapixel full-frame camera that I believe was in use.

However, the application for that lens setting is to shoot in-the-field images of live tiny creatures such as springtails, either as single shots or as typically short stacks that still show natural fall-off instead of the classic stacked "end-of-the-world" effect. The images speak well for themselves: https://www.chaosofdelight.org/ .

--Rik

Lou Jost
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Re: FAQ: How can I calculate effective aperture?

Post by Lou Jost »

I think it is sometimes very worthwhile to have formulas that include the pupillary magnification factor, particularly when comparing different set-ups. I'd love to see them all in one place.

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