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The answer is the two are very close, but there are theoretical differences. It is VERY hard to construct actual scenarios where linear acoustic diffraction differs substantially from the diffraction of light. As noted already, linear acoustic waves are scalar pressure fields, whereas electromagnetism is a vector phenomenon, i.e. there is a polarisation ...


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When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves ...? The answer is no. The diffraction pattern of sound and of light behind a slit is not similar and so the mathematical description is not complete. This is because the real patterns on an observation screen ...


0

When sound waves diffract through a single slit, do they produce an interference pattern which is mathematically identical to that of light waves in the corresponding experiment? I think the answer is almost, but not quite. It's a little complicated. First off, there's the question of whether you're talking about the far field (Fraunhofer diffraction) ...


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If you can ignore the effects of sounds attenuation in air, I believe the answer is "yes". Note that when the slits are electrically conducting at the frequency of interest for the electromagnetic case, then certain effects relating to polarization come into play as explained in Ben Crowell's answer, and the sound and light cases become more and more ...


1

Yes. There are more complete developments based on electromagnetic theory. Exact solutions are known for a few cases. Almost all applications rely on approximations, many of which are very good approximations under the right conditions. But the more complete explanations end up looking like Huygens Principle, and serve to justify it. Huygens' original ...


2

The positions of the diffraction peaks do not depend on the size of the atoms. The peak positions are determined by the spacings of the crystal lattice and it does not matter what the atoms are or how big they are. That's why the atom size does not appear in the Bragg formula. However the intensities of the lines, both absolute and relative, depend very ...


1

It turns out that there's a pretty detailed analysis of this in Jackson, Classical Electrodynamics, sections 3.13, 5.13, and 9.5. Although Jackson does it in excruciating detail using Bessel functions and infinite series, it's actually pretty easy to pull out the basic ideas. Start by considering a simpler problem. A thin, conducting sheet in the $x-y$ ...


0

This problem is intimately bound up with tunnelling (wholly analogous to quantum tunnelling) and evanescent waves. The results you get from analyses like those below depend critically on how one assumes the hole interacts with the incident field. I shall look at the problem with scalar optical theory, which is analogous to sound waves, or acoustics: ...


0

As the question arose from a question about water I give an answer first according to water waves. In a slit you observe the wave on the boundary between two mediums. As you want to "dive" to a hole you will see the related movement of particles and the compression and decompression of these particles to each over from a wave in this area. What happens in ...


2

What you have to imagine is the fact, that waves when are (ideally) reflected from a wall lead to a standing wave. If in the wall is a tiny slit the components of the reflected waves left and right the slit dissipate in the direction of the slit (spherical waves in each point according to Huygens description). These components undermine the incoming wave at ...


3

I hope you'll find this paper useful: J Weiner, Rep. Prog. Phys. 72 (2009) 064401. Although it mainly discusses light waves, the idea (and even the results) may be generic. For very small apertures (sub-wavelength apertures), strong reflection is expected to happen. According to Bethe's theory of light transmission through small holes, the transmission ...


1

I cannot speak for how to match with a mechanical resonator but you can match arbitrarily well any EM radiator that is small relative to its wavelength at one (1) frequency. This means that the radiator (loop or dipole, monopole, etc.) will absorb all radiation coming from directions in which it can radiate as a source without any reflection. The rest is ...


0

Well - it depends what you mean by "diffraction". Any point along a wave front will act as a point source of wavefronts travelling radially outwards - so while a plane wave will be arriving at the aperture, the waves that come out from the other side will almost look circular. To me, that's diffraction. Although there will be no regions where extinction ...


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Diffraction is the tendency of a wave emitted from a finite source or passing through a finite aperture to spread out as it propagates. Diffraction results from the interference of an infinite number of waves emitted by a continuous distribution of source points. If light is incident onto an obstacle which contains two very small slits a distance d ...


7

Feynman has come from heaven to answer your question! Listen to him: No one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a quest of usage, and there is no specific, important physical difference between them. The best we can do is, roughly speaking, is to say that when there are only a few ...


3

Two separate wave fronts originating from two coherent sources produce interference. Secondary wavelets originating from different parts of the same wave front constitute diffraction. Thus the two are entirely different in nature. The region of minimum intensity is perfectly dark in interference. In diffraction they are not perfectly dark. Width of the ...


1

Diffraction occurs when a wave encounters an obstacle or a slit these characteristic behaviors are exhibited when a wave encounters an obstacle or a slit that is comparable in size to its wavelength, whereas Interference is the phenomenon where waves meet each other and combine additively or substractively to form composite waves. In a sense there are ...


2

Assuming your screen is "far away", then the pattern you see on the screen is the Fourier transform of the aperture function. So if you take the (inverse) Fourier transform of the image on the screen, you get back to the aperture function. Small complication: the intensity image no longer contains the phase information which you need to fully reconstruct ...



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