A 50mm lens at f/8 has the same sized aperture as a 100mm lens at f/16, or a 200mm lens at f/32. All things being equal would those 3 lenses exhibit the same amount of diffraction ?
Hello Ken
I found in the archives this discussion where you can probably find some useful elements.
http://www.largeformatphotography.in...ffraction-quot
A 50mm lens at f/8 has the same sized aperture as a 100mm lens at f/16, or a 200mm lens at f/32. All things being equal would those 3 lenses exhibit the same amount of diffraction ?
This a very good question regarding how diffraction effects scale with physical aperture sizes and focal length, and the answer can be
1/ no if you look at the diffraction spot size or diffraction cut-off spatial frequency in the image plane;
2/ definitely yes if you look at the angular resolution in object space.
point 1/ can be summarized as follows
- consider a thin positive lens element of focal length f with an iris of diameter a located at is center, hence its f-number is N=f/a, and make a dream: this lens element is aberration-free! Of course this does not exist in the real world but does exist in optical design softwares, and this purely theoretical element is very useful to understand about 95% af all our photographic and theoretical questions regarding depth of field, illumination in the image plane, bellows factors, and of course: diffraction effects.
- consider, as a starting point, that you look at the center of the field, on the optical axis and assume that N is not too small, not f/1 or f/2 but f/8 or more like in real life of LF photography. This hypothesis of N not being too small greatly simplifies the expressions for the diffraction limits. And assume, this is even more unrealistic for photographers (except those who insists on take pictures only with sodium street lights) that the wavelength of light is unique, equal to lambda.
Then the absolute, non negociable, non digitally post-processable diffraction limit can expressed either in terms of diffraction spot size in the image plane, or diffraction cut-off spatial frequency:
diffraction spot-size ~= 1.2 N . lambda
diffraction cut-off spatial frequency = 1 / (N . lambda) (no fuzzy 1.2 factor here)
Now, from those exceedingly simple formulae (** note#1), we get the answer to point 1/, the relative aperture size N=f/a directly determines the diffraction limit in terms of spot size or cut-off frequency in the image plane. If we are operating not in the infinity-focus position but at a certain magnification ratio M, simply replace N in the above formulae by Neff=N(1+M).
And regarding problem /2, this is a typical problem for surveillance cameras, where you have to take high-resolution pictures of objects located at a fixed distance D; e.g. 10 meters for one of those machines that flourish along French roads and that try to automatically decipher your licence plate if you are caught overspeeding; or 100 km, if you are operating a satellite-based surveillance camera.
In problem /2, you want to been able to get the smallest possible resolution at a fixed object distance. Starting from the diffraction spot size in the image plane, which is located in the focal plane for those far-distant objects, you can convert the image diffraction limit into an object diffraction limit simply by muliplying by a basic geometrical magnification factor D/f.
Hence the diffraction limit on the object itself is simply lambda . D / a; in terms of angular resolution, this is simply lambda / a.
And in this case, the 50mm lens at f/8, the 100mm lens at f/16, or a 200mm lens at f/32 yield exactly the same (angular) diffraction limit (in object space)!
** note#1: the Holy "Marxist" legend says that once, somebody tried, in vain, to explain something to Groucho; "Eh, a 8-year old boy could understand this!"
And Groucho answered: "So, please, bring here immediately a 8-year old boy and ask him the question".
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