![]() I have allowed for the different focal lengths, f, by reference to a 400mm lens as standard and multiplying on the x-axis the lp/mm by a factor of f/400 for each lens (i.e., x2 for 800mm etc). ![]() 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.
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