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Can a magnetic ball “diffract” around a metal object? Yes. An eddy current separator (see the picture) works on this principle. The difference is that the magnet there can not have a translation movement. It is the nonferrous object the one wwhich is deflected. However, according to Newton's third law, the same force that deflects the nonferrous ...

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Both diffraction and interference occur in the double slit experiment. The wavefront is diffracted as it passes through each of the slits. The diffraction causes the wavefronts to spread out as if they were coming from light sources located at the slits. These two wavefronts overlap, and interference occurs. This is what give the diffraction pattern. I ...

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Diffraction is the phenomenon of the change of the movement from the straight line (in a flat, not curved space) in the cases, that it is not a reflection. For the expression "change of the movement from the straight line" it would be better to say "deflection", but this seems not to be so ok because, due to Wikipedia it could be misunderstood in every day ...

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Interference and diffraction are the same thing. In fact so is refraction. The propagation of light is conveniently described using the Huygens-Fresnel principle. The amplitude of the EM wave at a point is calculated by summing up the amplitudes of all the EM waves reaching that point, taking the relative phases into account. This describes the phenomena we ...

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If by constant you mean equal spacing between fringes then yes the spacing of a double slit is constant and so is any slit including diffraction grating's. Sometimes if you are not using monochromatic light the rainbows or wide spectrum will give the appearance of unequal spacing. Not counting the center fringe as far as I know the only Fringe pattern with ...

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I asked myself the exact same question. This PDF file from Harvard says that indeed diffraction gratings should be instead called interference gratings (page 13, first paragraph of "Remarks"). EDIT : Here is the paragraph from the link, as suggested by @Kyle Kanos : A diffraction grating should more appropriately be called an “interference grating,” ...

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In case someone else have this question, I finally found that Goodman (Introduction to Fourier Optics, ISBN 9780974707723) explicitly states that the Fresnel approximation is indeed valid in the far-field.

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Your laser cavity is a Fabry-Pérot interferometer. The free spectral range tells you how close two neighboring laser modes can get: $\Delta \nu=\frac{c}{2nl}$ (for a linear resonator, length l, refractive index n). The resolution of your spectrometer needs to be smaller than this free spectral range. You can increase the free spectral range by either ...

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Your de Broglie's wavelength is (assuming your weight is around 70 kg and you are moving 1 m/s) $\lambda = h/mv = 6.6\times10^{-34}/70 \approx 10^{-35}$ which is less then the Plank length - the least sensible distance. And this is much less than the diameter of protons/neutrons so even decomposed on elementary particles you wouldn't fit into such a door.

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As Kevin Zhou pointed out in his comment, behind every edge light will be distributed in fringes. As long as one expose curved edges or with light from a point like source or with light from parallel rays there will appear an detectable intensity distribution behind edges. Using monochromatic light and a point like source will give the best results. The ...

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I did a quick calculation of the single-slit diffraction patterns for three different slit widths of 100, 60, and 40 microns. Wavelength was that of green light (about 5000 Angstroms). Intensity plot is shown below for a progression of decreasing slit widths from 100 microns width (blue curve), to 60 microns width (green curve), and then to 40 microns width ...

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Short answer: On axis or central lobe field increases with the area of the slit (and thus the intensity $I\propto area^2$). Long answer: If you have a monochromatic plane wave traveling along the $z$ axis incident on a diffraction slit of width $w$ (as shown in the image below), then the initial field is $E(r,t) = E_0 e^{ik_0 z- \omega t}$. Now ...

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From an asymmetrical peak one can determine both a- and c- lattice constant. \begin{align}a=\frac{\lambda\cdot h}{(q_\text{par} \cdot \sqrt3)} \\c=\frac{\lambda\cdot l}{(q_\text{ort} \cdot 2)}\end{align} Where $\lambda$ is the wave length of your X-rays, $q_\text{par}$ (q_parallel) is the peak position alongside the in-plane direction ($q_x$ or $q_y$) ...

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I'll rephrase the question as I interpret it: Electromagnetic waves are drawn like this: source: https://commons.wikimedia.org/wiki/File:Onde_electromagnetique.svg Suppose this wave comes up to a vertical slit (a slit in the z-direction). What if the red arrows are longer than the slit? Then the wave won't fit through. But if the arrows ...

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remember that the amplitude is not an amplitude in space, it is an amplitude in the sense of the intensity of the electromagnetic field. The spatial amplitude is given by the wavelenght

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1) The ends of single slit act as two sources of light waves 2) the situation is similar to double slit interference experiment 3) thus in single slit also, interference pattern is seen

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You used the energy-wavelength relation for photons $$E=h\nu$$ which is valid for neutrons as well, but with some implicit conversions. Here, since neutrons don't move with speed c, you can't use that. Instead, you use the velocity implicit in the de Broglie equation. In the future, try this: For neutrons, use the energy approximation like this: ...

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The function describing the interference pattern is $f(x)=(\sin (x)/x)^2$, where $x=sin(\pi a/ \lambda)$ to find the extremes you need to derive and equal to zero. For the minima is easy, you find the condition $\sin x=0$, but for the maxima you get $\tan x=x$ which has to be solved numerically.

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