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I have been racking my mind over the double slit experiment. There are many excellent demonstrations of the double slit experiment, some with water, and some with light, available online. I am familiar with expressing a wave as:

$y_i(x,t)=A_isin(2πxλ_i)sin(2πf_it)$

and vartiants, so I have no problem envisioning and interference pattern. However, the analogy (if it is one) of light and water interference seems to break down to me because two sources of waves in the water can clearly create a diffraction patter with no slits. Two sources of light do not, in general,do the same thing, because I can turn on a few of my lamps and not get a diffraction pattern!

I know that the light sources are out of phase $\Delta\phi_0$, but could someone be so kind as to explain to me as to why they become in phase while passing through the slits?

Thank you kindly. (I have not seen two out of phase wave sources in water being made to pass through two slits.)

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    $\begingroup$ The two light sources have absolutely no phase with relation to each other - each photon has no relation with any other photon (unless, perhaps, they are mode-locked/phase-locked lasers which still takes a lot of effort). So, of course there is no visible interference - they don't interfere. In the normal double slit experiment, photons interfere with themselves (that is kind of the whole point), and each photon is in phase with itself. $\endgroup$ – Jon Custer Oct 14 '16 at 20:24
  • $\begingroup$ You can create diffraction patterns with different sources, but you need them to be exactly the same wavelength, such as two identical lasers. But with two ordinary lamps... it will not work mainly because they are not monochromatic. $\endgroup$ – rodrigo Oct 14 '16 at 20:24
  • $\begingroup$ Even most cheap diode laser won't cut it, but if nothing else you can phase lock two lasers and then use them. $\endgroup$ – dmckee Oct 14 '16 at 21:00
  • $\begingroup$ @Jon Custer As you say, in the normal double slit experiment, photons interfere with themselves, but the experiment is usually presented with an accumulation of photons which finally produces some visible interference, so how can that happen if the photons are note in phase? The Feynman lectures mention electrons (instead of photons) of same energy but seems to forget phase. This is a bit puzzling. $\endgroup$ – user130529 Oct 15 '16 at 7:22
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Where I have seen it, the two-slit are used to "divide" in two sources a common source, in order to have two coherent sources which interfere in different positions because of their different paths toward the given point.

Theoretically any pair of sources will interfere, but for the pattern to be invariable in time, the sources must be coherent.

As to why it does not happen as with water, well it does. When you observe a wavy sea surface, where peaky forms rather than waves are visible, yo are observing chaotic interference between many waves of different frequencies and amplitudes. And it probably happens with light too, but the variations are so fast, and the photons so many that no deviations of the mean brightness is apparent.

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  • $\begingroup$ mhleo, are two coherent sources really needed? Even behind a sharp edge a intensity distribution of light appears. Next, monochrome light is a good choice to get clear fringes Coherent light source means the phases of the light quanta are equal. This isn't needed for interference. The theory stated that the slits make the light in phase. Unanswered is the question how the slits make the incomming light quanta coherent. $\endgroup$ – HolgerFiedler Oct 15 '16 at 6:21
  • $\begingroup$ The classical explanation, which I refer to, does view light as a macroscopic wave, because one wave-front is large enough spatially to hit both slits. Therefore is an explanation incapable of addressing the points you mention. On the other hand QM explanations of the phenomenon (the ones I've seen) does not attempt to describe in terms of quanta, but in terms of a plane wave function with single momentum, so I think when there your questions remain unanswered. I think they are valid though. $\endgroup$ – rmhleo Oct 15 '16 at 6:33

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