# Why can't we duplicate the double slit experiment with two parrallel sources of light.. why must the light enter the two slits externally?

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.)

• 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. – Jon Custer Oct 14 '16 at 20:24
• 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. – rodrigo Oct 14 '16 at 20:24
• Even most cheap diode laser won't cut it, but if nothing else you can phase lock two lasers and then use them. – dmckee Oct 14 '16 at 21:00
• @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. – user130529 Oct 15 '16 at 7:22