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It is possible to perform the Double-Slit Experiment using electrons. The resulting diffraction pattern matches that predicted by quantum mechanics, with characteristics defined by the deBroglie wavelength of the electrons. When I was in college, I watched a Professor perform this experiment in class. No other theory ever devised, aside from QM, explains ...


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When we decrease the size of scattering crystal, the diffractogram peaks broaden. The broadening $\beta$ is related with the crystal size $\tau$ by the Scherrer equation: $$ \tau = {{K\lambda}\over{\beta \cos \theta}}, \beta = {{K\lambda}\over{\tau \cos \theta}} $$ ($K$ depends on the form of the crystal, ~0.9 for spheres). So, the first note would be: ...


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Here is a two slit calculator / simulator I designed in Excel. https://www.dropbox.com/s/kkopfv4xbratc9r/Alsept%20%202%20Slit%20Calculator.xlsx?dl=0 Download,save, open, enable editing The green cells are where you input distance, wavelength, slit width and separation.The yellow cells give you the fringe pattern spacing. The graph represents the right half ...


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Calculate the Fraunhofer diffraction integral of your slits and take the absolute square to get intensities.


2

For each wavelength you have been given (exactly) two values $\theta$, so $$\begin{align}n\lambda &= d\sin\theta_1\\ (n+1)\lambda &= d\sin\theta_2\end{align}$$ subtracting these two equations, we get $$\lambda = d\left(\sin\theta_2-\sin\theta_1\right)$$ You should be able to figure it out from there...


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As a diagram screen $A$ is closer than screen $B$ and so the fringes are closer on screen $A$: Fringe separation $\Delta x = \dfrac {\lambda \; D}{d}$ where $d$ is the slit separation, $\lambda$ is the wavelength of light and $D$ the distance from the slits to the screen.


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Of course, it all depends on what you are after, and what you can tolerate. But you can get an estimate by asking a similar question: over what range of defocus does the spot size not change appreciably? Or equivalently, what is length of the near field of the focused spot? That distance can be estimated (order of magnitude!) in the following way. One ...


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Nothing about light traveling through a slit or opening actually makes spatially incoherent light become coherent. What occurs is that, where you have an incoherent light-source, i.e. a non-point or extended source, and you place in its path a small enough opening, you're isolating light that was emitted, relatively speaking, from a single point on that ...


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Each of the two edges of a slit act as point sources and all light coming from them is coherent. This can easily be proven mathematically. Each edge forms a single edge fringe pattern on the screen with alternating dark and light fringes calculated as m=the square root of wavelength times distance times (m+3/4) where m = the number of spaces out from the ...


3

$k$ is the incoming beam, $k'$ is the reflected beam, expressed as wave vectors in the reciprocal lattice, which makes $Q=k-k'$ represents a particular plane in reciprocal space. If you assume that the diffracting beam is essentially a plane wave when it elastically scatters off of multiple sites within the crystal, the kinematics of the Laue equation ...


0

You're mostly correct, about how you expect both patterns together to produce uniform intensity corresponding to a plane wave (this is described as Babinet's principle. What you're forgetting is that it's the sum of the amplitude, including a phase. A circular hole and the complementary shape will produce identical patterns but opposite phase, except for ...


1

I think you are missing things in both points. 1) There is no reason why the sum of diffracted patterns would give a uniform intensity. Why? Because waves are made of amplitudes, but patterns are made of intensities, that is: squared amplitudes. Two waves that would interfere destructively, give non-zero intensities when taken individually. In fact, they ...


2

According 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. From the question: In other words, when the Fraunhofer criteria are met, complementary obstacles produce complementary diffraction patterns. If I believe the OP, and also this ...


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As all the other commentators have said, both diffraction and interference are manifestations of the same thing: the fact that waves superpose. The two words are slightly different, but it's clunky to work with two concepts when you only need one. You never do a "diffraction calculation" or an "interference calculation", you just do wave mechanics and ...


1

These high reputation members aren't confused about the difference between diffraction and interference. The two terms are synonymous in the physics community (we're not arguing on the basis of linguistics. I can agree the terms are linguistically different just because people sometimes prefer one over another in certain contexts). The way the physics ...


1

Whether the amount of diffraction is 'negligible' depends on how you define this criterion. The first order minimum in the diffraction pattern from a single slit occurs where $\sin\theta = \lambda/d$ where $d$ is slit width, $\theta$ is diffraction angle and $\lambda$ is wavelength. If $d = \lambda$ the central lobe of the diffraction pattern will spread ...


2

Yes, it is generally true. Imagine that your source is composed of many many point sources. Each point source produces a diffraction spot. The complete image is the sum of all of those spots. The points at the edge of an object will "spill" some intensity into the interior of the geometric image, but the diffracted intensity will "spill" outside of the ...



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