Thank you for checking this thread.
I have some confusion about calculating effective F aperture number for objective NA and magnification.
I know F effective = F x (1+m)
Where m = magnification.
And F = 1/ (2 x NA)
Where NA is numeric aperture of microscope objective
Is that m on-sensor magnification or visual magnification (total microscopy magnification, meaning eyepiece power times objective power)?
As an actual example,
if I use an 1x NA 0.05 160TL objective, at 160TL with 10x eyepiece, together with afocal micro four thirds sensor and 30mm m3/4 lens. What effective F aperture am I using the objective at? F/22 or F/110? My gut feeling is that F/22 may be the right answer.
I did experiment and on-senor magnification is about 1.2x. Microscopy total magnification is 10x (1x objective with 10x eyepiece).
Also when you reverse a camera lens, its F number does not change, though effective F number change with sensor size/extension/magnification, correct?
Thank you again!
Effective F aperture for objective NA
Moderators: rjlittlefield, ChrisR, Chris S., Pau
Effective F aperture for objective NA
Selling my Canon FD 200mm F/2.8 lens
This follows some PMs, so I'll chime in early to see if I have it right!
Magnifications for these purposes are just On Sensor.
What you DO with it, is up to you. If you use an afocal setup you're magnifying then shrinking. For diffraction considerations, you aren't changing anything by the enlargment.
If you can cover your sensor without overly enlarging the image circle, that may be your best bet, on your 18x 13.5mm sensor.
I've not seen a "normal" 1x objective with decent coverage.
Only the extension - and hence magnification, changes the effective aperture.
Even if I'm 100% right, Rik would express it differently.
Magnifications for these purposes are just On Sensor.
Just 1/(2x 0.05) = eff/10if I use an 1x NA 0.05 160TL objective, at 160TL with 10x eyepiece, together with afocal micro four thirds sensor and 30mm m3/4 lens. What effective F aperture am I using the objective at?
What you DO with it, is up to you. If you use an afocal setup you're magnifying then shrinking. For diffraction considerations, you aren't changing anything by the enlargment.
If you can cover your sensor without overly enlarging the image circle, that may be your best bet, on your 18x 13.5mm sensor.
I've not seen a "normal" 1x objective with decent coverage.
Its effective F number changes with the PMF, but with the lens I think you have in mind, SC 50mm f/2.8, the PMF will be close to 1.Also when you reverse a camera lens, its F number does not change, though effective F number change with sensor size/extension/magnification, correct?
Only the extension - and hence magnification, changes the effective aperture.
Even if I'm 100% right, Rik would express it differently.
Chris R
Thank you, ChrisR.
PMF = pupillary magnification factor?
That is new and complex to a macro beginner like me again
Yes, I have the Schneider Componon S 50mm F/2.8 in mind.
It seems that, as you pointed out, the SCS50's effective aperture value will be better than my macro objectives (1x NA 0.05 and 2x NA 0.09).
At 3x on sensor magnification, it probably loses to my 3x NA 0.10 or 3x NA 0.08? Though I am not positive about my calculations.
I am not sure if F effective = F x (1+m) or F effective = F x m (magnification), in this situations. Please kindly correct me.
PMF = pupillary magnification factor?
That is new and complex to a macro beginner like me again
Yes, I have the Schneider Componon S 50mm F/2.8 in mind.
It seems that, as you pointed out, the SCS50's effective aperture value will be better than my macro objectives (1x NA 0.05 and 2x NA 0.09).
At 3x on sensor magnification, it probably loses to my 3x NA 0.10 or 3x NA 0.08? Though I am not positive about my calculations.
I am not sure if F effective = F x (1+m) or F effective = F x m (magnification), in this situations. Please kindly correct me.
Selling my Canon FD 200mm F/2.8 lens
If you use a camera type lens, you need the "1+ " in the calculations.
If you use an objective, that's already "built in".
So eg for a 4x NA 0.2, which is eff/10
a camera lens would have to be marked f/2 because
"f/2" * (M+1)
= "f/2" * 5
= eff/10
Sorry about the sloppy terminology, I expect you get it...
And a 4x NA 0.1 is like an f/4 lens, giving eff/20.
I wouldn't llike to predict your outcomes. Worth a few cheap tubes or a bellows to find out. I suspect the enlarger lens will be nicer to use.
If you use an objective, that's already "built in".
So eg for a 4x NA 0.2, which is eff/10
a camera lens would have to be marked f/2 because
"f/2" * (M+1)
= "f/2" * 5
= eff/10
Sorry about the sloppy terminology, I expect you get it...
And a 4x NA 0.1 is like an f/4 lens, giving eff/20.
I wouldn't llike to predict your outcomes. Worth a few cheap tubes or a bellows to find out. I suspect the enlarger lens will be nicer to use.
Chris R
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zzffnn wrote: if I use an 1x NA 0.05 160TL objective, at 160TL with 10x eyepiece, together with afocal micro four thirds sensor and 30mm m3/4 lens. What effective F aperture am I using the objective at?
...
I did experiment and on-sensor magnification is about 1.2x.
Probably true!ChrisR wrote: Just 1/(2x 0.05) = eff/10
...
Even if I'm 100% right, Rik would express it differently.
But in this case there's a technical difference also.
The calculation should be 1.2/(2x 0.05) = effective f/12.
That's because it's always the on-sensor magnification that has to be used in the formula that effective f# = mag/(2*NA) .
If you have relay optics that change the magnification, then the effective f# changes also.
It's just like the effect that a teleconverter has on effective f#, where making the image bigger also increases the effective f-number by the same multiplier.
In both cases, there is no effect in the resolution on subject, except for whatever aberrations might be added.
--Rik