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Reading about diffraction of EM radiation on edges, slits and multi slits as well as about electron diffraction behind a wire I came to the conclusion that the intensity distributions on an observers screen is the result of the interaction between radiation (electron beam) and a quantized field of surface electrons of the edges.

Going pregnant with such an idea, I started to read Physics.StackExchange. In this forum doesn't play a role that intensity pattern occurs behind every edge (in the contour of the geometrical shadow) and not only behind slits. Is it a possible point of view that behind a slit with a large enough distance between the edges of this slit occurs the same patterns like behind a single left edge and a single right edge - if we cut out the images of their intensity patterns and stick them together? I'm searching for arguments that this is not possible. This argumentation I need to contradict my idea about quantized field between photons and the surface electrons from one edge or the edges of single, double or multi slits.

That exists a well accepted and established theory about interference of EM radiation I know by myself, there is no need to talk about this fact. But I don't see a mistake in my explanation and need an argument to step away from my theory.

Beside the traditional point of view about interference of EM waves in the slits region even for single photons or electrons I found in the forum some evidence for the distribution pattern as a result of the quantized field of the interacting electrons from the edge(s): "... the electric field component perpendicular to the surface will be discontinuous, because there is electric field normal to the surface from bound surface charges...". This is an answer from @rob. Is this statement adoptable to my point of view?

I see no way to put this in different questions and hope it will be accepted as one question.

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HolgerFiedler
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    How then do you explain the fact that you get the same diffraction pattern whether the slits are made from a conductor or an insulator, even though light interacts with these different types of materials in different ways? – John Rennie Jan 07 '15 at 16:21
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    @John Rennie Thanks for replay at all. It's not known to me that there are experiments with different materials where the fringes distances where measured under equal circumstances for the distances from the source to the edges and to the screen. Moellenstedt experimented with different electrical potential and get not surprising different patterns. ... – HolgerFiedler Jan 07 '15 at 17:07
  • ... What I know is that there is a difference between the position of the first fringe from photons and from electrons. The first intensity fringe for photons lay half inside and half outside the geometrical shadow. And for electrons there is a repulsive potential and the fringe starts wider the geometrical shadow. – HolgerFiedler Jan 07 '15 at 17:08
  • And it is easy to explain single photon experiments. – HolgerFiedler Jan 07 '15 at 17:09
  • Here a asked about the slit's material. – HolgerFiedler Jan 07 '15 at 17:59
  • "In the forum doesn't play a role that intensity pattern occurs behind every edge and not only behind slits. Is it a possible point of view that a slit with the right distance between the edges gives the same patterns like a left and a right edge?" I do not understand these two sentences at all. Could you please rephrase them? – Ján Lalinský May 14 '15 at 22:35
  • What is a left edge? – Ján Lalinský May 14 '15 at 22:36
  • @JánLalinský I edited the question. A slit contain to edges and if the slit is positioned vertically, the sli has a left and a right edge. – HolgerFiedler May 15 '15 at 05:06
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    @JánLalinský Edges are geometric designations for a border area made from valence electrons and due to the sharpness of such edges in double slit experiments with a high electrostatic potential. – HolgerFiedler May 15 '15 at 05:15
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    Of course there is "an effect" in a absolutest sense, but the thing that optics people know that makes this a non-question is that it doesn't matter unless either the slit width or the wavelength is comparable to the depth of the boundary region. In other cases the final solution is dominated by the portion of wave passing far from the slit boundary. Certainly in the case of a plane wave passing an isolated edge the contribution from the half-plane completely overwhelm the edge effects. This follows from Huygens' principle. – dmckee --- ex-moderator kitten Nov 03 '15 at 19:18
  • @dmckee In which year a wavelength was measured first? How this wavelengths were measured, directly by comparing with an etalon or indirectly by what? – HolgerFiedler Nov 03 '15 at 19:44
  • IMO the problem is simple thinking of a S or D equation solutions, given potentials and boundary conditions. Due to the sinusoidal nature of the solutions there is no mystery to be attached to the way they come out when nature solves them. Just to understand that we probably did not take into account a potential or a boundary condition. – anna v Dec 27 '15 at 04:53
  • @annav Could you please extend your comment a little bit. For a not native speaker like me it is to cryptic. – HolgerFiedler Dec 27 '15 at 07:19
  • The underlying mathematics which describes very well quantum effects are solutions of the Shrodinger of Dirac or Klein Gordon equations. These are basically sines and cosines ( if you prefer complex exponentials). How the probability densities will materialize in an experiment depends on the potentials one puts in the equations and the boundary conditions. edge,slit etc. We know from the solutions of other easier to measure differential equations ( like water waves) that interferences appear when phases are kept and randomness rules when phases are lost. It is similar with – anna v Dec 27 '15 at 07:50
  • the probability function. It will display interferences if the phass are kept and this will depend on the specific boundary conditions. In my view nature is a great problem solver, it gives the correct probability distribution and it is up to us to understand what boundary condition or which potentials we did not include in our mathematics. – anna v Dec 27 '15 at 07:52
  • Holger, I have been very interested in this subject for a few years now. If I understand your question correctly I may have an answer for you. At the top of my page is a website. I have posted A paper that I wrote entitled "Single Edge Certainty" The paper shows you exactly how to derive any Fringe pattern for any edge or slit experiment. – Bill Alsept Dec 29 '15 at 22:40
  • @BillAlsept Let us discuss this in a chatroom – HolgerFiedler Dec 30 '15 at 19:01

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@Holger Fiedler Hi,

Thanks for your comment on my answer here. Here I would like to say that I am an experimental physicist and working on the ultrashort coherent XUV radiation. I have first hand experience with the interference of XUV radiation using double slits.

Your question is very genuine. Long ago I have thought over this question for quite a some time. The first comment made by @John Rennie is very pertinent. He is absolutely right in his saying that the slit material plays no role in the diffraction. If you can make double slit of same width, same separation, same thickness (this is just a practical consideration because with thicker slits alignment of the slit become tedious), same edge roughness you will get same diffraction pattern with any material (be it paper, copper, aluminum, steel or wood). Note that last two requirements of roughness and thickness are not that stringent.

To test this assumption I would like to suggest you to make a simulation. I presume that you must be familiar with some high level programming language like matlab or python.

Take a plane wave of wave vector $k$ (since you are dealing with time stationary diffraction you do not need $\omega t$ term) write the wave equation $exp\left(i kr\right)$, now construct two slits of width w and separation d and finally a screen at distance D. Take several points on the screen and on slits. Now trace the rays and calculate path from above equation and construct the intensity distribution on the plate by overlapping all the rays from double slit on a single point of screen.

The simulated interference pattern will be an exact replica of the observed diffraction pattern (If you can do such experiment). I have tested this myself, the matching between these two spectra is quantitative. Once you have made such program you can change your incident pulse to anything you like and so on or you can easily convert it into the simulation of knife edge.

The idea behind suggestion for making this simulation (where we did not consider any interaction between light and the electrons present in slit) is to illustrate that the wave nature of the light is responsible for the diffraction and not the interaction of the electrons at the edge of the slit. The interaction of the electrons in material is responsible for blocking of light.

I would like to mention that you can deduce all the above things (at least qualitatively) via very basic equations for plane wave but in my opinion by making such program you can see the effect of changing the parameters on the actual diffraction pattern and seeing is believing.

PS: If you like I can upload python code of my simulation.

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hsinghal
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  • The question is could one do a simulation of what I await about the electrons electric field on sharp edges. And this I think is it impossible. Since I'm an engineer I'm not in the position to do so. BTW has you read about the the distribution of magnetic dipole moments inside atoms? – HolgerFiedler Jul 21 '16 at 19:41
  • Thanks for your replay. Instead of a simulation I would prefer an experiment. Any idea an possibility? – HolgerFiedler Jul 21 '16 at 19:45
  • Hsinghal There is a historical overview of mine about electron diffraction. Unfortunately until now only in German https://ru.wikipedia.org/wiki/Участник:HolgerFiedler/Electron_diffraction . It is interesting that to the biprisma could be applied an electric potential and this changes the deflection and diffraction of the electrons path. More than this in the potentialfree case as well as with potential the intensity distribution on the observers screen is equally distributed. https://upload.wikimedia.org/wikipedia/commons/b/b6/Moellenstedt_biprisma_voltage_shadow.JPG – HolgerFiedler Jul 22 '16 at 04:36
  • Perhaps this is applicable to your work, XUV is used to mask microelectronic devices? – HolgerFiedler Jul 22 '16 at 04:37
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    One question - if it happens that there is a diffraction pattern from knife edge how this is explained by the wave theory? To have minimums and maximums you need at least two wave sources? Is that what you mean?

     – Baj Mile Feb 27 '17 at 23:41
  • @Baj Mile This is standard textbook question. if you put a knife edge in the path of wave the diffraction occur at edge but you can not see it in bright part only intensity modulation in dark part appear. – hsinghal Feb 28 '17 at 01:38
  • As an experimental physicist, you should be able to provide a refereed paper of an experiment that compared different materials. If there is no such paper, then it seems appropriate that you, an experimental physicist, should submit such a paper. – James Bowery Oct 09 '17 at 20:58
  • @James Bowery there might be some paper but that may be very old and may not be in English and not available online. But i think there is no paper because nothing was found. Material electrons have no role except to stop the light. – hsinghal Oct 10 '17 at 01:56
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    That's not the way science proceeds. A negative result to an experiment should be published if for no other reason than that the methodology can be critiqued by peers. Understandably, this is often relegated to post-graduate research assistants, but it is done. There is a crisis in some of the soft sciences due precisely to the failure to bother to even try to replicate. Science depends on experimental "redundancy". – James Bowery Oct 10 '17 at 03:20