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When I create a double slit in a mirror by removing the paint and silver I get a beautiful interference pattern. What’s interesting to me is that I get one in the opposite direction too. What has me puzzled even more is that the “reflected” one never passes through the slits, yet it produces a beautiful interference pattern, equal in brightness and definition to the one that actually goes through the slits. Even when I turn the mirror around where the paint side is facing the laser I still get a clear pattern in both directions. When I simply use painted glass I get the same results. This has me so perplexed that I created a YouTube video showing what’s happening https://youtu.be/lELlRpmGOYE What’s going on? Wave simulator apps do not show congruent waves reflecting in both directions.

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    $\begingroup$ Part of the front waves is still artificially truncated from the reflection, so you get interferences patterns. $\endgroup$ – Exocytosis May 10 at 0:57
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With the back of the mirror nearest the laser the two exposed slits of glass reflect the laser light and act as two coherent sources.
It is a double slit/source arrangement and I would expect the fringe pattern to be less bright than the conventional straight though pattern because not that much light is reflected from an air/glass interface.

With the mirror reversed there are two effects ie the silvered side nearest the laser. The reflection of light from the part of the mirror which is still silvered and this is the narrow bright spot.
The reflection of light from the two glass/air interface slits which act as two coherent sources and produce a double slit interference pattern.
This fringe pattern can be seen because it spreads the light outside the narrow beam of light which has been reflected from the slivering on the mirror.

Update after some experimentation which confirms my original answer.

In the past to manufacture double slits I have used a microscope slide painted with aquadag (colloidal graphite but I must say that using a mirror is certainly easier and produces better slits.
To produce the slits I never moved the straight edge (metal ruler) and inclined the Stanley blade utility knife through a small angle between cuts as shown below.

enter image description here

The slits I used were about $0.28 \,\rm mm$ apart and each had a width of about $0.04 \,\rm mm$.
I used a laser pointer wavelength around $650\,\rm nm$ as the light source.

All the transmission and reflection fringes confirmed a double slit interference pattern modulated by a single slit diffraction envelope as shown below the photographs being produced using an oldish iPhone.

enter image description here

The photographs were not all taken from the same distance and that is why the fringe spacing looks different.
The first two are transmission with mirror back and mirror front towards the laser and the last two are reflection with mirror back and mirror front towards the laser.
The third photographs is reflection from the glass surface (and the back of the mirror) with the fringes much less bright than with transmission.
The fourth photograph also show the light from the laser being reflected from the rest of the mirror to produce a very bright spot.
Interference by reflecting slits is the basis of the reflection grating and is the reason for the colours seen when white light is reflected from a DVD.

Using a Canon camera enable greater control over the photographs.

enter image description here

When the laser light was being reflected from the front of the mirror as wll as the interference pattern due to the two slits and the bright reflection, another effect was observed as shown below.

enter image description here

I think that the circular fringes are due to the interference of light from reflection at the bottom surface of the mirror and reflection at the top surface of the mirror.

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  • $\begingroup$ Thanks for your answer. I think you’re right but a few things are still interesting to me: Mirror side facing the laser - why the light spreads out without diffraction taking place. Two coherent beams are reflected but they only “see” a slit - never pass through it. Why wouldn’t the edges reflect like the center????? $\endgroup$ – Lambda May 11 at 16:56
  • $\begingroup$ @Lambda I have looked at your video a number of times but I am still unsure exactly what the three patterns look like. How do the (a) reflected from the back of the mirror and (b) reflected from the front of the mirror patterns compare with the usual straight through pattern? Do they all have the close together, equally spaced, interferences fringes with the intensity modulated by a single slit diffraction pattern? $\endgroup$ – Farcher May 11 at 17:39
  • $\begingroup$ When the laser is directed toward the mirror side (silver facing the laser) the image that does not pass through the silt differs in two ways from the normal: 1. it has a bright spot in the center of the interference pattern. 2. It moves when the mirror is moved. If the mirror is moved say 45 degrees to the laser the image moves to 45 degree to the normal image. If the mirror is moved to almost 90 degrees to the laser a perfect circled is formed where both the reflected image and the regular image approach each other. Both interference patterns are visible on opposite sides of the circle. $\endgroup$ – Lambda May 13 at 3:44
  • $\begingroup$ When the paint side is facing the laser the bright center spot is not present, and the entire reflected image is not as bright as the normal image. The reflected image moves when the mirror is moved. $\endgroup$ – Lambda May 13 at 8:31
  • $\begingroup$ @Lambda I am going to have a go at reproducing your experiment. As you rotate the mirror does the pattern broaden? $\endgroup$ – Farcher May 13 at 8:47
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It looks like you have found yourself a case of Babinet's principle! It states that any diffraction pattern generated by some aperture $B$ is equal to that generated by its complimentary aperture $B'$ (meaning that if we combine the two we would block all the light).

If we look at what happens to the wavefront when it propagates through your mirror, we get the usual diffraction:

Transmission pattern

This happens because the edges of the slits are point sources as per Huygens's principle, so in the limit of small slits we simply get a point source. If the wave were to reflect back off of the surface like in your case, the edges would do exactly the same, that is:

Reflection pattern

(Here I haven't drawn the incident plane wave.)

Babinet delivered the proof that you can generalize this to any two apertures that are each other's complements.

Now to come back to the question in your title: the light does not necessarily need to pass through a slit, it merely needs some edge to diffract off of.

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  • $\begingroup$ I didn't know the name of the principle but I second this. $\endgroup$ – Exocytosis May 10 at 1:00
  • $\begingroup$ The application of Babinet's principle is right, but I don't see why you said "This happens because the edges of the slits start to behave as point sources, so in the limit of small slits we get such point sources." The person used a mirror as the double-slit material, so the backward reflection basically acts as aperture $B'$. Shouldn't this alone be sufficient to explain the reflected double-slit pattern? $\endgroup$ – SpiralRain May 10 at 2:30
  • $\begingroup$ @SpiralRain I was merely trying to refer to Huygens's principle to clear up why the light actually doesn't need to pass through the object in order to diffract. Hopefully this is clearer now. $\endgroup$ – Julius May 10 at 17:21
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An interference pattern in the far field is the same as a Fourier transform. For a perfect reflector, the reflection aperture is given by 1 - that of the transmission aperture (ignoring any phase shift). So the reflection transform is the Fourier transform of 1 minus the transmission transform.

Excepting a delta spike at zero (the transform of 1), the interference patterns will look (we don't see the phase but power spectrum) the same.

Note that due to the backwards zero angle delta spike, the patterns are visually different. Contrary to Julius' answer.

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