rmalarz wrote:
Well, actually it is, or is it?
Let's get our brains working this morning with this poser. I've seen a number of comments on this site regarding using small f-stops and diffraction. The typical, don't use small apertures because that causes diffraction. So, what is diffraction? Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength. I've added the bold to emphasize the size required. So how big are those sizes?
Visible light has a range of wavelengths of 400 - 700 nanometers. Whoa, how big is a nanometer? It's .000000001 meters or .000000039370 inches. So 400 - 700 nanometers is .0000004 - .0000007 meters or .000015748 - .000027559 inches. These dimensions are quite a bit smaller than any apertures we're using with our cameras.
So getting back to our original statement, f/16 is actually equal to f/16. But, that's because f-stops are ratios. What isn't the same is the diameter of the aperture from one lens focal length to another. For example, let's take two lenses, In this case, we'll examine two Schneider-Kreuznach lenses of focal lengths 150mm and 210mm. At f/16 the aperture opening is:
150 - 9.375mm
210 - 13.125mm
Obviously, a large difference in aperture diameters, but the same f-stop. However, neither is close to the wavelength range of visible light.
Since diffraction occurs as stated above, neither of these measurements are close to the dimensions required to meet the above conditions. So, how would light know which lens it's passing through? Oh, and if you want to make an issue of the "passing by a corner", well, that corner exists at every f-stop, other than perhaps the greatest opening.
The conclusion of this could be that we're parroting some misinformation, making a blanket statement that doesn't cover all situations, or we may be concerning ourselves needlessly. If diffraction does occur, is it observable in our photographs?
--Bob
Well, actually it is, or is it? br br Let's get o... (
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Your explanation is patently false. "
" is pure fantasy as the attached gif's show. The first gif shows diffraction of a scalar wave passing through a 1-wavelength-wide slit, while the second gif shows diffraction of a scalar wave passing through a 4-wavelength-wide slit, thus diffraction does NOT require an opening "the approximate size or even smaller than the wavelength", as four wavelengths is not considered the approximate size in wavelengths in any coherent thought system.
A slit which is wider than a wavelength produces interference effects in the space downstream of the slit. These can be explained by assuming that the slit behaves as though it has a large number of point sources spaced evenly across the width of the slit. The analysis of this system is simplified if we consider light of a single wavelength. If the incident light is coherent, these sources all have the same phase. Light incident at a given point in the space downstream of the slit is made up of contributions from each of these point sources and if the relative phases of these contributions vary by 2π or more, we may expect to find minima and maxima in the diffracted light. Such phase differences are caused by differences in the path lengths over which contributing rays reach the point from the slit.
You are 100% correct about "parroting some misinformation" however.