I want to reproduce this experiment by myself. What I need for this. What parameters of slits and laser/another light source it needs? Is it possible to make DIY-detector?

  • $\begingroup$ The detector could be a whole separate question, especially if you are thinking of building a photomultiplier or something. Honestly, a successful experiment will show itself qualitatively upon visual inspection. If you want a more quantitative analysis, perhaps taking a RAW picture and looking at pixel values will suffice. $\endgroup$
    – user10851
    Feb 15, 2013 at 19:24
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    $\begingroup$ You might be interested in quantum eraser experiment, that can also be reproduced at home, youtube.com/watch?v=R-6St1rDbzo $\endgroup$
    – raindrop
    Feb 16, 2013 at 18:18
  • $\begingroup$ duplicate or near-duplicate of physics.stackexchange.com/q/38440 $\endgroup$
    – user4552
    May 22, 2013 at 4:42
  • $\begingroup$ I forgot about this possibility making the double-slit: use MS Paint or Inkscape and print them on overhead transparency. $\endgroup$
    – Noah
    Sep 23, 2013 at 0:12

3 Answers 3


It's actually quite easy to perform the experiment in the comfort of your own home. The simplest setup I have seen (as depicted in this, and other youtube videos) is to use a laser pointer and pencil lead, but you can certainly be more systematic and cut slits in some opaque material as well.

I would encourage you to experiment to answer the question of how far apart the slits need to be etc., but some basic math behind this is as follows: If the slits are a distance $d$ apart, if the light has wavelength $\lambda$, and if the distance between the slits and the screen is $L$, then the spacing $\Delta y$ between successive fringes on the wall will approximately be $$ \Delta y \approx \frac{\lambda L}{d} $$ So let's say the laser is red so that $\lambda\approx 700 \mathrm{nm}$, the slits are $1\,\mathrm{mm}$ apart, and the screen is $1.5\,\mathrm m$ away from the slits, then we have $$ \Delta y \approx \frac{(700\,\mathrm{nm})(1.5\,\mathrm{m})}{1\,\mathrm{mm}} = 1.05\,\mathrm{mm} $$ So you can actually try this and see if your results agree! (I might actually try this myself come to think of it; thanks for the question!)


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    $\begingroup$ You can also use human hairs to produce at least single-slit diffraction patterns. Two hairs/graphite leads can in principle also do it, through Babinet's principle. $\endgroup$ Feb 15, 2013 at 17:07
  • $\begingroup$ What's about detector near one slit? $\endgroup$
    – Robotex
    Feb 15, 2013 at 18:06
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    $\begingroup$ I would add from my own experience that cutting slits in an opaque material often produces slits that are not entirely negligible in width (compared to $d$), so there may be a broad single-slit pattern modulating the double-slit pattern that is sought. $\endgroup$
    – user10851
    Feb 15, 2013 at 19:21
  • $\begingroup$ What's about depth of slits? Does it have sense? $\endgroup$
    – Robotex
    Feb 16, 2013 at 14:07

Absolutely, though the result does depend somewhat upon your definition of "at home".

Simply seeing the interference pattern is as simple as a laser pointer and a few narrow apertures (see the other answer(s))

People have successfully even done single-photon interference at home!


Laser pointer, nit comb, bit of cardboard from a cereal box to control the number of slits. Works perfectly!


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