Coming late to this discussion and since Ken's original question is "diffraction and scanning" it should be added that the original table implies that the f-number is the
effective f-number = N(1+M) where N is the engraved f-number as usual, and M the magnification ratio.
Scanner lenses probably do not operate in the infinity-focus position.
Imagine that you use a lens in infinity-focus position, stopped down to f/11 as engraved, when focusing down to the 1:1 magnification ratio, the effective f-number becomes 11x(1+1) = 22, hence the diffraction-limited resolution limit is twice as big with respect to the infinity-focus position, if the aperture ring is unchanged.
See the attached pdf file with official MTF charts for the 240 mm apo-ronar.
The MTF curve at 1:1 ratio for N=22 is plotted above the MTF curve at 1:20 for N=22 and it is obvious if you look at the 20 cy/mm MTF curve that the apo ronar is better at 1:20 than at 1:1, a kind of a paradox for a lens optimized for 1:1 use
The reason is simply that the effective number for N=22 "engraved", is N
eff = 22x(1+1/20) = 23.1 at 1:20 and N
eff = 22 x 2 = 44 (actually = 45) at 1:1.
The MTF charts for the 240 mm apo ronar also show us that the curves are very flat up to 360 mm of image circle at 1/1 and only half this value, 180 mm at 1:20.
On one hand, the minimum detectable optical period at constant N is doubled when changing from infinity-focus to 1:1, because N
eff is doubled; but on the other hand, the image circle is doubled, in consequence the total number of resolved points remains constant !!
2) To effectively scan that (hypothetical) image, would we need a scanner capable of twice that resolution ?
Yes, but in order to avoid to be confused with various factors like 1.2 (in the formula of Rayleigh's criterion for the diffraction limit) or any fudge factor around 1.2 taking into account the minimal acceptable contrast, plus the factor 2x for the number of samples required per one optical / analog period, I prefer to express the diffraction limit in terms of the absolute analogue cut-off period in a diffraction-limited optical image
cut-off period = N
eff λ
cut-off frequency = 1/(N
eff λ) in cycles per mm or mm
-1
where N
eff = N(1+M) as explained above;
λ = wavelength of light
now we can take a safety margin of 80% for the cut-off frequency of 1.2 for the cut-off period in order to maintain a minimum contrast, and with the 1.2 factor we get Rayleigh's resolution criterion = 1.2 N λ
The sampling theorem simply states that you need two samples per analogue period.
imagine that you have a diffraction-limited image behind an optical system with N
eff = 11, the cut-off period for, say, red light λ=0.65 micron is 11x0.65 = 7.2 microns - cut-off frequency = 140 cycles/mm, in principle you should use a sampling rate at 280 cy/mm i.e. about 7100 samples per inch ...
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