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And how do you know this?
In reply to fft81,
5 months ago
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fft81 wrote:
Diffraction does not quite work like that. Your effective stops of light are indeed high, but that only effects the amount of light falling on the sensor. Diffraction is proportional to the size of opening in the lens shutter. If the opening is equal to f/8 for what ever lens, and then you magically make three quarters of the light disappear your effective f/stop is f/16 but for diffraction purposes it is still f/8.
I beg to differ. Diffraction is a consequence of the interaction of light with the edges of an aperture and is described by the Fourier transform of the aperture:
http://en.wikipedia.org/wiki/Airy_Disk
The intensity described by this transform has only angular dependence. So, for any given aperture, the amount of diffraction is directly proportional to the distance from the aperture. This goes as the effective f/number.
So let's say your lens is set to f/8. From the Rayleigh criterion of 1.22 (wavelength)*(f/number), you'll get about 5 microns of diffraction blur. But if you use the lens at unit magnification, it will be twice the distance from the sensor. So it will have twice the blur, or 10 microns. This correlates exactly with the effective aperture (f/16 in the above situation).
--
Leonard Migliore
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