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Visualizing the double slit experiment, there are light lines and dark lines. The dark lines I understand are caused by the interference cancelling waves. What I don't understand is where the energy of the cancelled waves goes. What happens to the energy of those waves?

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An intuitive way to see it: Intensity of light in constructive interference: $I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2$. Intensity of light in destructive interference: $I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2$. One slit has intensity $I_1$ and another slit has intensity $I_2$. And the "total intensity" has $I = I_1 + I_2$. As you can see, it is a redistribution.

A more clear: Assume $I_1 = I_2 = I_0$. Therefore, both slits radiates the same light intensity. Then, by conservation of energy, the "total" intensity is: $I = I_1 + I_2 = 2I_0$. But then: $$ \mbox{Constructive Interference: } I = I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = 4I_0 $$ $$ \mbox{Destructive Interference: } I = I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = 0 $$

So, what happened here? Simple! We took the "$2I_0$" and redistributed it in such way that now, there are places with intensity $4I_0$ and places with intensity $0$.

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  • $\begingroup$ So there will always be an even number of light and dark spots? $\endgroup$ – Kontakt Dec 1 '14 at 19:36
  • $\begingroup$ The angular space between dark and light spots is $\sin\theta = \lambda/d$, for $d$ is the distance between the slits. $\endgroup$ – Physicist137 Dec 2 '14 at 14:01
  • $\begingroup$ I'm confused as to the mechanism of the energy transfer. If the waves do not interact anywhere before the lines, how canthe energy move? $\endgroup$ – Kontakt Dec 3 '14 at 18:39
  • $\begingroup$ @Kontakt This waves are electromagnetic waves. So, $E = v(x)e^{i\omega t}$. Therefore, electromagnetic waves comes up from both slits. They interact with each other before the lines by summing it's electric and magnetic fields. But you can only see this interaction when the light hit the plate and you have the lines. A proof they interact before it, if you move the plate to closer the slit, the interference pattern won't be destroyed. If you still have questions about it, you can ask another question (different from this one of course). =). $\endgroup$ – Physicist137 Dec 3 '14 at 19:07
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The interference pattern has parts were the wave cancel (destructive interference) and parts where they are added (constructive interference), so the energy gets redistributed, there are no losess.

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It's not quite that simple. If you fire one photon at a time, then one photon always hits the wall/detector. Energy is always conserved (you never loose a photon).

What happens is that over time the photons land with more probability on the bright bands than the dark bands.

Have a look at the Wikipedia page on Double Slit Experiment. The images in particular highlight what I'm saying.

An important version of this experiment involves single particles (or waves—for consistency, they are called particles here). Sending particles through a double-slit apparatus one at a time results in single particles appearing on the screen, as expected. Remarkably, however, an interference pattern emerges when these particles are allowed to build up one by one (see the image to the right). This demonstrates the wave-particle duality, which states that all matter exhibits both wave and particle properties: the particle is measured as a single pulse at a single position, while the wave describes the probability of absorbing the particle at a specific place of the detector. This phenomenon has been shown to occur with photons, electrons, atoms and even some molecules, including buckyballs. So experiments with electrons add confirmatory evidence to the view that electrons, protons, neutrons, and even larger entities that are ordinarily called particles nevertheless have their own wave nature and even their own specific frequencies.

The best way to visualise this (intuitively) is to imagine a wave guiding the photon and the photon follows the peaks of interference. See this video for a good visualisation of this with water.

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Interesting question. If explore water waves one can see that the medium oscillates. The oszillation and not the medium moves from the point of disturbance away. Exploring waves from two points of disturbance one can see that at every point at the surface of the medium the level of the surface is the sum from the two waves. The minimum und the maximum of the two disturbances with equal amplitudes is zero. If produce not circular but plane waves with parallel disturbance lines and the same amplitudes there is a line in the middle between the two disturbance lines there the waves eliminate each over. The kinetic energy of the two waves changes to heat.

If explore the waves from two points of disturbance (and equal amplitudes) there are only points there minimum meets maximum and the energy gets dislocated into heat. AND always this points are moving. See the difference to the fringes of the double slit experiments? They are not moving. They are not the result of the disturbance of a medium. To conclude from the frozen picture of the interference of water waves to the static fringes of photons or electrons behind a edge, a slit or a double slit is strange.

It would be more plausible to take in account that every surface and then more every edge is surrounded by an electric field from the electrons on the surface. This field interacts with the photons (or electrons). The common electromagnetic field is quantized and this is the reason for the fringes. This explanation works for single particle experiments and for edges instead of slits too.

That means the photons get focused from continuous distribution in front of the slit to fringes behind the slit. What we see is the structure of the EM field of edges. And the space between the fringes is really empty (approximately) and the fringes contain the double (approximately) amount of particles.

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