09-29-2011, 05:08 AM
[quote name='wim' timestamp='1317246889' post='11948']
For MTF-50 the number to use is 760 lp/mm for the 0.0005 mm wavelength, and 740 lp/mm for 0.000512 mm.
As a simplified formula, f50 = 0.38/(Nω), which equals approximately 0.46*Rayleigh limit. ω is the wavelength here, and N the f-stop setting. Set the latter to 1, and you get the number you are after. Wavelength to be expressed in mm, BTW.
[/quote]
How great, thanks again. So at a very rough cut, the aerial resolution of a lens (before applying various other degrading resolutions such as non-ideal lens behavior, and sensor resolution) is around 1600 over the F-stop for 9% MTF modulation transfer (reproducing 9% of the perfect target's true intensity changes between white and black), and around 800/f-stop if you choose to declare a line pair to be resolved only when there's at least 50% modulation transfer.
That's really easy to remember. So the 9% aerial MTF of an F8 imaginary-perfect lens is around 200 line pairs per mm, and the 50% aerial MTF is around 100 line pairs per mm. All to be further cut down by the lens aberrations, resolution of your sensor, and all the zillion other blurring factors in the chain of elements between usually-moving subject and usually-moving pixel microlenses.
Hmm this implies that some day, a little 10mm Nikon CX sensor at F5 could with zillions of pixels give quite decent enlargements. Start with the perfect-lens-and-sensor aerial MTF of 160 line pairs/mm, call it 80 lp/mm after going through a futuristically only slightly imperfect lens and sensor, that would end up being around 1600 dots per picture width...or 150 dots per inch under perfect conditions on an 8x10 inch print. How pleasant to be able to wander through that kind of speculation so easily.
For MTF-50 the number to use is 760 lp/mm for the 0.0005 mm wavelength, and 740 lp/mm for 0.000512 mm.
As a simplified formula, f50 = 0.38/(Nω), which equals approximately 0.46*Rayleigh limit. ω is the wavelength here, and N the f-stop setting. Set the latter to 1, and you get the number you are after. Wavelength to be expressed in mm, BTW.
[/quote]
How great, thanks again. So at a very rough cut, the aerial resolution of a lens (before applying various other degrading resolutions such as non-ideal lens behavior, and sensor resolution) is around 1600 over the F-stop for 9% MTF modulation transfer (reproducing 9% of the perfect target's true intensity changes between white and black), and around 800/f-stop if you choose to declare a line pair to be resolved only when there's at least 50% modulation transfer.
That's really easy to remember. So the 9% aerial MTF of an F8 imaginary-perfect lens is around 200 line pairs per mm, and the 50% aerial MTF is around 100 line pairs per mm. All to be further cut down by the lens aberrations, resolution of your sensor, and all the zillion other blurring factors in the chain of elements between usually-moving subject and usually-moving pixel microlenses.
Hmm this implies that some day, a little 10mm Nikon CX sensor at F5 could with zillions of pixels give quite decent enlargements. Start with the perfect-lens-and-sensor aerial MTF of 160 line pairs/mm, call it 80 lp/mm after going through a futuristically only slightly imperfect lens and sensor, that would end up being around 1600 dots per picture width...or 150 dots per inch under perfect conditions on an 8x10 inch print. How pleasant to be able to wander through that kind of speculation so easily.