![]() ![]() Two waves that would interfere destructively, give non-zero intensities when taken individually. Why? Because waves are made of amplitudes, but patterns are made of intensities, that is: squared amplitudes. I think you are missing things in both points.ġ) There is no reason why the sum of diffracted patterns would give a uniform intensity. So there is certainly no contradiction in this limit. the diffraction pattern in the Fraunhofer limit does not have a spot in the middle. I think the confusion might simply come from that the OP used an argument about the interference pattern in Fraunhofer diffraction and added the Arago spot into the mix, which does not actually appear there, i.e. We don't even need to consider the applicability of Babinet's principle though. Therefore I will only claim that this is a possible answer, which relies upon Babinet's principle only applying in the Fraunhofer limit. However I don't completely understand which limits Babinet's principle applies in. If I believe the OP, and also this article seems to assume the Fraunhofer limit, the contradiction is resolved. In other words, when the Fraunhofer criteria are met, complementary obstacles produce complementary diffraction patterns. Anyone has input on this ?Īccording to wikipedia's article about the Poisson-Arago spot it is a phenomenon occurring in Fresnel diffraction, which is a different limit from Fraunhofer diffraction. What am I missing ? I somehow have the feeling that patterns are complementary "everywhere except the center" because there is something ill-defined about using Huygens wavelets to predict the center of the pattern. Therefore these patterns are NOT complementary. ![]() ![]() In other words, when the Fraunhofer criteria are met, complementary obstacles produce complementary diffraction patterns.Ģ) Experiment shows that a circular stop and a circular aperture indeed produce opposite fringes, but the circular stop's diffraction pattern has the Poisson-Arago spot in the middle, while the circular aperture doesn't have a corresponding dark spot in the middle. In other words, the SUM of the diffracted patterns is expected to be a uniform intensity on the screen (all the wavelets = same as what we have when there is NO obstacle). My theoretical analysis leads to contradictory conclusions about such situations.ġ) Assuming incident light is a plane wave consisting of N sources of Huygens wavelets, obstacle 1 keeps a subset of those wavelets (blocking the others) while the obstacle 2 keeps the complementary subset of wavelets. For example, a circular aperture and a circular stop of same radii. This may be done, for example, by collecting the diffracted wave with a “positive” (converging) lens and observing the diffraction pattern in its focal plane.Assume two obstacles are complementary. Note also that the Fraunhofer limit is always valid if the diffraction is measured as a function of the diffraction angle \(\ \theta\) alone. by measuring the diffraction pattern farther and Of course, this crossover from the Fresnel to Fraunhofer diffraction may be also observed, at fixed wavelength \(\ \lambda\) and slit width \(\ a\), by increasing \(\ z\), i.e. The resulting interference pattern is somewhat complicated, and only when a becomes substantially less than \(\ \delta x\), it is reduced to the simple Fraunhofer pattern (110). (107), is just a sum of two contributions of the type (111) from both edges of the slit. The resulting wave, fully described by Eq. If the slit is gradually narrowed so that its width a becomes comparable to \(\ \delta x\), 42 the Fresnel diffraction patterns from both edges start to “collide” (interfere). is complies with the estimate given by Eq. ![]()
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