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I set up a single slit and produced this interference pattern on a distant wall. enter image description here

I then placed another single slit in the deconstructive portion of the pattern. Here's a pic. The slit is between the two bright areas.

enter image description here

The dark fringe area produced no light after the second slit. Please consider my two questions.

1.If the dark areas are places where the waves are 180 out of phase then shouldn't the second slit disrupt the out of phase waves and create new wavelets that would the create a new interference pattern.

2.If the wave is not simply out of phase but destroyed. Where did the energy go? I put a beam splitter between the two slits and nothing returned to the source.

Here's a diagram of the setup.enter image description here

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  • $\begingroup$ It is not clear to me what we are seeing in the 2nd image, because you say that "the dark fringe area produced no light after the slit." So what is the light that we are seeing? $\endgroup$ – sammy gerbil Oct 20 '16 at 17:57
  • $\begingroup$ The second image shows the second slit positioned into a deconstructive zone between two constructive areas. The slit did not photograph very well. It is lost in the glare of the two constructive areas. $\endgroup$ – Lambda Oct 20 '16 at 18:27
  • $\begingroup$ Does the 2nd image show only the light emerging through the 2nd slit? The one which you said "produced no light after the slit"? Or is the light in this image light which doesn't pass through the 2nd slit? $\endgroup$ – sammy gerbil Oct 20 '16 at 18:41
  • $\begingroup$ It's the light that does not pass through slit 2. It's taken from the laser side of slit two. $\endgroup$ – Lambda Oct 20 '16 at 20:25
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The question puts forward 2 hypotheses for what causes the dark fringes in a single-slit experiment. In (1), the hypothesis is that it's due to the cancellation of 2 waves which are 180-degrees out of phase. Whereas in (2), the hypothesis is that it's due to the wave simply getting "destroyed".

Neither of these are quite right, but (1) is much closer to the truth. It's not really that 2 particular waves are physically canceling each other, it's more that if you add together all possible paths that light could take from the slit to the wall, you end up with a pattern which is zero at certain points (the nodes / dark fringes) and non-zero at most points. The cancellation is between an infinite number of paths, but it's true that since most of them will cancel out anyway, you only really need to consider the light rays that come from either edge of the slit.

The particular pattern of intensity you end up with is called a sinc function ($\frac{\sin x}{x}$), or more specifically ${\rm sinc}^2 x$.

The intensity is only exactly zero at single isolated points. So no matter how small you make the second slit, there is no way to avoid some light going through. But as long as it is narrow enough and centered on a node, you can reduce the intensity of the light passing through to very close to zero. So likely, there is some minimal amount of light getting through the 2nd slit you made, you are just not seeing it because it's far too dim.

You can do all of this using water waves instead of light waves if you prefer. Create a pool with a barrier which has a narrow slit near the middle. Then make some waves on one side of the pool, and see which waves end up getting through to the other side and crashing on the far wall. You can think of the waves as individual points in the pool moving up and down. Each time any point in the pool oscillates up and down, it send ripples out in all directions. Then all of the surrounding points are waving up and down, and those end up also sending out ripples in all directions. It doesn't make much of a difference whether there is a 2nd slit or not, if not many ripples are making it to a particular point on the wall, making a slit there is not going to change that. Not much light (or water waves) will get through a slit which you place near a dark fringe. Whereas a lot of light (or water waves) will get through a slit you place near a bright fringe.

At no point does anything like (2) happen where waves are simply destroyed and energy is lost.

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  • $\begingroup$ After reading your answer I spent some time investigating sinc functions. Your answer was helpful. It shifted my thinking from mechanical to probabilities. Given time and effort I will gain insight. For those like me trying to understand this here's a helpful video.youtube.com/watch?v=XaJnygMuT7M $\endgroup$ – Lambda Oct 20 '16 at 14:27
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I do not understand the thinking behind your 1st question. In what way do you think the 2nd slit "disrupts the out-of-phase waves and create new wavelets that would create a new interference pattern"? Perhaps as follows :

We can think of the illumination at slit 2 as either (a) individual light waves from different parts of slit 1 arriving with approximately the same amplitude but different phases, or (b) darkness resulting from the collective destructive interference of the waves in (a). In case (b) it is obvious that there will be darkness behind slit 2, but what about case (a)?

The different parts of the 1st slit can be thought of as coherent sources (Huygens' Construction). Each of these sources produces its own diffraction pattern behind the 2nd slit, similar to that in your 1st image. Each diffraction pattern has its own phase - or more accurately, each point in the pattern has a different phase depending on which source created it. But just as the sources at slit 1 are out of phase and interfere destructively at slit 2, so the diffraction patterns behind slit 2 are also out of phase, and interfere destructively at the screen.

So there is darkness behind the 2nd slit whichever way you think of the incident light.

Your 2nd question is answered in What happens to the energy when waves perfectly cancel each other? and many similar questions such as How does interference move energy from destructive to constructive regions? These answers explain that energy which is "lost" at places of destructive interference "reappears" at places of constructive interference. You cannot have destructive interference at one place without having constructive interference somewhere else. So there is always conservation of energy.

But these answers don't seem to explain why or how this happens. The explanation may be that waves are not objects themselves but descriptions of the transfer of energy or probability or something else. This distinction is explicit in quantum physics, between wave-functions and the particles they describe. So individual photons do not annihilate at points where there is destructive interference of the wave-function. But interference also occurs in Newtonian physics - eg sound in air, ripples on water. If we were to examine the collisions between elements of air or water, we would see the mechanics by which energy is transferred away from or towards certain regions at which we say destructive or constructive interference is occurring.

The wave picture is a shorthand method of describing some more complex underlying mechanism. Although it makes some things easier to understand (such as the appearance of interference patterns), it makes other things more difficult to understand (such as the disappearance of energy from one point and its reappearance somewhere else).

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  • $\begingroup$ Thanks for your response. My thinking behind my first question was that the light energy was still present in the dark fringes but cancelled. Kind of like two generators that are running in parallel, when their waves are in sync then the difference of potential between the two is zero, even though they may be producing high voltage when compared to ground. I was thinking the second slit would shift the phase, by causing the waves to travel different distances and expose the "voltage." Obviously, my experiment showed that my theory was incorrect. $\endgroup$ – Lambda Oct 21 '16 at 3:06

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