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In a paraxial thin-lens approximation, two otherwise identical beams from two different (but paraxial) directions produce the same image (but displaced inside the image plane). You can see this by tilting your setup such that first one and then the other source is on-axis: For a thin lense, this process only introduces changes that scale with $1 / \cos ... 2 The shadow of your hand may look crisp to you, but that's because you're not looking closely enough, compared to how short the wavelengths in visible light are. Indeed, if you look at the shadow from a viewing distance of 40 cm, you can't possibly see the difference between a (hypothetical) exactly crisp edge and one where the intensity of illumination ... 1 There is a perfectly valid physical explanation: use Maxwell's equations to find how the wave propagates beyond the aperture. Use as boundary conditions the idea that the wave is completely killed off beyond the aperture. Sound reasonable? This results in an integral. Turns out that the integral can be interpreted, at least to a very good approximation, as ... 0 It's not quite as fictitious as you think! Imagine a circular wave pool. To make perfect circular waves of wavelength 5m, you would need a cylinder of radius 5m to rapidly go in and out of the water, displacing a lot of water for a brief period of time, periodically. Now imagine ocean waves of wavelength 5m going through a slit 5m wide. Inside that 5m wide ... 1 Thus - does the diffracted electron radiate photons? In your question, you use terms acceleration from classical mechanics and photon from quantum theory. Since these theories are not mutually compatible, the question is badly stated. To get a useful and clear answer, you have to state which theory you are asking. If you ask "does the electron radiate ... 5 The diffraction seems to form from the pixels (basically a diffraction grating). The pixels have a translational symmetry in$x$and$y$directions, so the pattern also exhibits this symmetry. On a 15-inch macbook retina display, the pixels are separated by$\$d = \frac{15.6inch}{\sqrt {2880^2+1800^2}} = \frac{0.396 m}{\sqrt {2880^2+1800^2}} = 1.17 \cdot ...