These are plotted versus lp/mm resolution. The effects of Bayer and AA-filters on sharpness have been ignored for the sensor and won’t matter too much when comparing lenses. I have calculated the contributions of the MTF(diffraction) and the MTF(sensor) pixel size. MTF(system) = MTF(diffraction)*MTF(sensor)*MTF(lens aberration) (4). To do this, we need to know the resolution of the different components, using MTF, which goes from 1, being 100%, down to zero, with MTF = 0.09 being the vale for “just resolvable” (the Rayleigh Criterion). We have to consider the diffraction of the system as a whole. So, what is going on? Equation 3 is just for the diffraction of the lens. (3)īut we know that 2x extenders give us more 2x2 more pixels on a target and usually work so we spend hard-earned money on them. Resolution(diffraction limited) varies as (wavelength of light)*D. In fact, the resolving power of the lens that is determined by diffraction depends simply on the wavelength of light and the diameter D: But, as the image is twice as large, you should see exactly the same amount of detail. So, if, say, you double the focal length of the lens and keep the same D, like adding a 2xTC to the lens, you double the size of the Airy Disk, so halving the resolving power. The size of the image on the sensor increases with f. Equation 2 is independent of the focal length of the lens. This equation is used to calculate the diffraction-limited aperture of a lens, DLA, the widest f-number that makes best use of the size of pixels on the sensor. There is a basic rule that the resolving power of a lens is limited by diffraction to being able to detect two lines separated by the diameter of an Airy Disk, d, or more, where D is the diameter of the front element and f is the focal length according to the equation:ĭ = 1.22*(wavelength of light)*f-number (2) There is an introduction to diffraction for non-experts, which explains what is going on here. But, my calculations are steering me to what lenses to use with the R5 and where the R6 is appropriate. Those who are interested can read further – I spent hours doing the calculations and graphs to confirm what I knew intuitively for my own satisfaction. The f/11 lenses are more suited to the lower resolution R6, and this makes it an attractive lower priced alternative.Further, any additional aberration added by the TCs could well render them even worse. Using the 1.4x and 2xTCs on the f/11 lens hardly increases the resolution on the R5 and only a little more on the R6.The 800mm f/11 outresolves the 100-500mm f/7.1 on the R5 and R6, but adding the 1.4xTC to the zoom brings it closer to the 800mm prime.The100-400mm f/5.6 + 1.4xTC at 560mm f/8 outresolves the 600mm f/11 on the R5 and the two lenses are similar on the R6. The 100-400mm at 400mm f/5.6 is slightly outresolved by the 600mm f/11 on the R5 and more so on the R6.As the number of pixels on the sensor decreases, increasing focal length becomes increasingly advantageous.On an unrealistically high-resolution sensor, the f5.6, f/7.1 zooms and the 800mm f/11 would all have the same resolution and outresolve the 600mm f/11, and the TCs would not increase resolution.The summary based on just the effects of diffraction and sensor Mpx, and ignoring optical aberrations is: What I do here is to calculate the contributions of diffraction and sensor Mpx size (R5 vs R6) to the resolving power of the 400mm f/5.6 and 500mm f/7.1 zooms and the 600mm and 800mm f/11 primes and how resolution is affected by 1.4x and 2x teleconverters. Another of my geek articles, which does have some implications for actual use.
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