Barrelcactusaddict wrote: ↑Tue Nov 02, 2021 8:47 pm
So what is exactly the definition of "nominal aperture"? Is it just the lowest, unaffected aperture on a lens, or is it any kind of f-number possible by adjusting the aperture ring without the addition of a second, paired lens?
More like the latter. For example, suppose we have an EL Nikkor 50 mm f/2.8 enlarging lens, which we mount on enough extension to give 0.5X magnification. If we set the aperture ring on the EL Nikkor to f/2.8, then we have nominal aperture f/2.8 and effective aperture f/4.2 (=2.8*(0.5+1)). But if we set the aperture ring to f/8, then we have nominal aperture f/8 and effective aperture f/12 (=8*(0.5+1)).
(I thought that that would have been 5.6, the base aperture of the Symmar lens)
In the case of paired lenses, usually the limiting aperture is provided by one or the other, but not both. Whichever lens does
not provide the limiting aperture, the nominal aperture of that lens does not matter because it does not affect what light gets through the system. In the case of Lomo+Symmar, the hole in the Lomo lens is so much smaller than the hole in the Symmar at f/5.6 that the 5.6 does not matter.
so the "nominal aperture" in the second part of the author's second formula is actually the "effective f-number on the sensor side" or "nominal aperture" of the Lomo objective "sensor" that you mentioned in the formula from your "working backward" section?
I assume you are asking about "Now with the Lomo 3,7x mounted on the Makro Symmar 5.6/120 the effective aperture would be only f/13.2 using the formula effective aperture = nominal aperture * Magnification."
In that formula, "effective aperture" refers to all the optics, as seen by the sensor, and "nominal aperture" refers to the Lomo alone. This formula assumes that the limiting aperture is provided by the front lens, so it is the nominal aperture of that lens which matters.
Please note that there was a typo in my explanation: at one point I wrote "sensor side" when I meant "subject side". I have edited my earlier post so that it now says
it is always true that effective f-number on the camera side = magnification * effective f-number on the subject side [typo: used to say "sensor side"].
So are "effective f-number" and "nominal aperture" interchangeable terms, or was this a typo?
No, these are not interchangeable terms. "Effective f-number" always describes the angle of a cone of light, usually at the sensor. "Nominal aperture" usually refers to the f-number of some part of the system in isolation.
But as always, there are exceptions. Modern Nikon systems have a clever interaction between camera and lens, such that the f-number you set on the camera body is taken as being the effective aperture that you want, and the camera and lens collaborate to adjust the lens iris to make that happen. Canon has no such thing; whatever you set on the camera body just gets transferred directly to the lens. This leads to some huge confusions, because "1X and f/11" on a Canon system then ends up meaning about the same thing as "1X and f/22" on a Nikon system. When a Nikon shooter marvels at how his Canon friends get more DOF with the same settings, it's because of this confusion.
So there are 3 different formulas, but please correct me if I'm wrong:
one involving numerical aperture to find effective aperture in microscope objectives
[effective aperture = magnification ÷ (2 * numerical aperture)]
one to find effective aperture in microscope objectives using the objective's nominal aperture and magnification
[effective aperture = nominal aperture * (Magnification + 1)]
These two look correct. More precisely, the second formula applies to lenses on empty extension, so it works for finite objectives but not for infinite objectives.
Then
one to find effective aperture on the camera side using magnification and effective aperture on the sensor/outermost lens or objective side
[effective f-number "cameraS" = magnification * effective f-number "sensorS"]
This appears to have inherited my typo that substituted "sensor" for "subject". The proper formula would be
[effective f-number "cameraS" = magnification * effective f-number "subjectS"]
--Rik