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As it turns out, descriptions of this are already all over the internet:

I don't think describing the procedure is necessary, just click on any of the links.

I tried it both with my fingers and with cards a lot and it seems to always work. I only tried artificial light sources, not sun light, for safety reasons.

However, isn't the light of incandescent light bulbs (I – among other kinds of light sources – tried it with several different ones of those) black body radiation which isn't coherent whatsoever? Why does it still work?

I'm not using any optical instruments (glasses, contact lenses, prisms, etc.), btw. Just a light source, my eyes, my fingers, and cards.

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UTF-8
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  • Not sure there's a true question here that hasn't been answered multiple times in the linked posts. – Jake Watrous Feb 08 '17 at 02:30
  • @JakeWatrous The first linked page doesn't even mention light requirements. The second linked page has an answer to the question on that page, however, mine is different: Why does it work even if specifically non-coherent light is used. The third linked page does discuss coherence and the post I linked to specifically says: "For interference to be strongly visible, one or both of two types of coherence is required." But when choosing black body radiation from a light source of fairly big size compared to how far I'm away from it, shouldn't there be neither temporal nor spacial coherence? – UTF-8 Feb 08 '17 at 02:37
  • Seems like the Reddit post makes a convincing case for it being optical diffraction and/or a trick of our eyes and not black body radiation, though. What convinced you black body radiation is the soul possible cause? – Jake Watrous Feb 08 '17 at 02:54

2 Answers2

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Black body radiation and any large in spatial extent source radiation is incoherent , true.

Point sources by their construction create coherence.

space coherence

This illustrates space coherence

Incoherent light going through a small hole creates coherence, and that is how Young got coherence for the double slit experiment.

When light is scattered off an obstacle, as the edge of a paper, or the slit between fingers, the reflecting points on edge act like a point source creating spatial coherence in the plane perpendicular to the edge.

So coherence does not come from the black body radiation of the source, but from the geometry at the level of scattered light and the dimensions involved

anna v
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  • Nice answer. A follow up. Apart from spatial coherence, which makes sense, there also has to be a frequent coherency. Isn't most of the visible spectrum coming out of that point source and out of the slots? And if so how can it work? (I've seen Young's experiment so many times that question never occurred to me before!) – Bob Bee Feb 08 '17 at 05:42
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    @BobBee well, in single slit there is no frequency coherence , one gets the rainbow colors in the interference pattern. One would have to put a filter of color to get a limit on frequency spread. Young had colors and used them to estimate differences in wavelength https://en.wikipedia.org/wiki/Young%27s_interference_experiment#Young.27s_work_on_wave_theory – anna v Feb 08 '17 at 06:28
  • Yes, that makes sense. I've done array interferometers in microwaves and we had filters. I just wanted to see if I was missing something else obvious. Interesting thong o microwaves, even if no filter in front, but behind, i.e. Even broadband, as long as a single source we got the multiple receivers to show interference, basically relative phase coherence, of course different for each freq filter behind the multiple receivers. I know a different arrangement, but relative phase coherence was the driver. Thanks much – Bob Bee Feb 08 '17 at 19:29
  • @annav I only see a rainbow right where the incandescent bulb is. Everywhere else on the slit between the cards, there only is the light of the usual color with dark bars in-between. Given that my incandescent light bulb doesn't produce monochromatic light, shouldn't there either be rainbows everywhere in the silt or at least not straight dark bars going all the way along the slit parallel to the cards? Shouldn't light of different wave lengths (provided the wave lengths aren't multiples of very few "prime wave lengths") have a really hard time interfering destructively in regular patterns? – UTF-8 Feb 08 '17 at 22:53
  • It is all a matter of wavelengths and the way your retina records what you are seeing. Dark will be dark, so you see the dark lines, but your retina will mix again the bright part with colors into the previous light. You need a film or large distances to see the separate colors . at around 6 minutes https://www.youtube.com/watch?v=Iuv6hY6zsd0 . – anna v Feb 09 '17 at 06:29
  • @annav How light from a thermal source gets coherent behind a small hole – HolgerFiedler Feb 11 '17 at 07:52
  • @HolgerFiedler have a look at this question and the answers therein http://physics.stackexchange.com/questions/76692/is-coherent-light-required-for-interference-in-youngs-double-slit-experiment . I would say that a point source creates a spherifal wave that can be described well with sinusoidals and phases with the classical maxwell equations, and lo , the data shows it is so. and it is not behind , what goes through the small hole is coherent by construction – anna v Feb 11 '17 at 12:34
  • @anna Do you mean this answer?: http://physics.stackexchange.com/a/244669/46708 – HolgerFiedler Feb 12 '17 at 08:37
  • @actually I meant Numrok's answer. I also have an answer bellow – anna v Feb 12 '17 at 08:44
  • @anna It's clear to me that this will be an extended discussion which is possible to do only from face to face. To do such discussion in writing is only possible if to touch only single point step by step and for each point have an question at the and which will be answered by the opponent "with yes or no because..." . Bill Dixon wrote "I quote the great physicist Paul Dirac (The Principles of Quantum Mechanics, Oxford Science Publications, Fourth Edition, p 9) “If the two components are now made to interfere, we should require a photon in one component to be able to interfere with one in ... – HolgerFiedler Feb 12 '17 at 13:01
  • ?.. the other. Sometimes these two photons would have to annihilate one another and sometimes they would have to produce four photons. This would contradict the conservation of energy. The new theory, which connects the wave function with the probabilities for one photon, gets over the difficulty by making each photon go partly into each of the two components. Each photon then interferes only with itself. Interference between two different photons never occurs.” Do you agree? – HolgerFiedler Feb 12 '17 at 13:02
  • @HolgerFiedler No. Annihilation is interaction . What happens is that the wavefunctions of the two photons overlap (psi1+psi2). to get a measurement,( i.e. see the interference )one has to take the complex conjugate square which will have a term with the interference pattern. Superposition of wave functions is not interaction. – anna v Feb 12 '17 at 13:14
  • there exist interfence patterns from two laser beams, pure photons . Single photon double slit interference has to do with the fields produced by the slits and the superposition of the photon wave function with that. – anna v Feb 12 '17 at 13:24
  • Ok, let the superposition not be an interaction. What about electrons bended by single edges into intensity distribution. How to explain the superposition behind single edges? See page 8 of my excerpt http://s3.amazonaws.com/academia.edu.documents/48285665/2016-07-27_Electron_deflection_rev2016-08-09.pdf?AWSAccessKeyId=AKIAIWOWYYGZ2Y53UL3A&Expires=1486909798&Signature=qtfkgQhy4Fxr8WY6XWlHTR0t2HU%3D&response-content-disposition=attachment%3B%20filename%3DDeflection_of_electron_beams_at_edges.pdf – HolgerFiedler Feb 12 '17 at 13:32
  • The electron on an edge is no longer a plane wave ; the quantum mechanical problem "electron +edge" gives a different wavefunction which when complex conjugate squared at the detection level shows interferences. Alternatively with field theoretical tools the electron wave packet is distorted by the fields near the edge , again with the same result. The link does not work for me – anna v Feb 12 '17 at 15:01
  • https://www.academia.edu/27983554/Deflection_of_electron_beams_at_edges – HolgerFiedler Feb 12 '17 at 15:17
  • The only difference in your pov in the conclusion with mine is that you talk of an "electron wave" whereas I see it as an"electron probability amplitude wave" – anna v Feb 12 '17 at 16:06
  • My point of view is 4.2 quantized electric potential between the edge and the electron? – HolgerFiedler Feb 12 '17 at 16:32
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Visible light is only a portion of the black body radiation spectrum and these frequencies are a small enough subset that the modes of wave propagation are visible through the slit. (I am avoiding the word interference as the phenomenon is really a result of the photon as a wave function.)

PhysicsDave
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