In Young's double slit experiment, when you see the diffraction pattern, why does the intensity of the light fade out as you move from the central maximum? I think it has something to do with the wave-particle duality but I'm not sure and my teacher has no idea. See images below to see what I'm talking about. Where does that energy go? I'm not talking about the actual interference pattern, but the intensity of the light.
(source: cyberphysics.co.uk)
Best Answer
The answer I gave before was wrong, as was kindly pointed out by CuriousOne. The fading of the intensity isn't because of the $1/r^2$ fall-off of light, since the diffraction formulas are only for small angles anyway.
First of all, it's clear from the second figure in the question that the only relevant thing is the diffraction and not the interference; that is, we might as well consider a single slit. Also, since this is only a wave phenomenon, the wave-particle duality only comes into play if you think light is made of particles (which it is, with a certain definition of "particle"), but we don't need that here; the classical wave behavior of light can explain this perfectly well.
The main idea is that as we get farther away from the center of the diffraction pattern, the phase difference between all the rays coming from different parts of the slit gets bigger and bigger. When we are very near the center, the rays are almost in phase, and if we move a little bit, they are still almost in phase. But if we keep moving, they get a bit out of phase, and the interference between them all makes the intensity go down. As we get farther and farther away from the center, the phase difference gets bigger; and since the dependence of the phase with the distance from the center (or the angle) is different for all the rays, they won't get in phase again (as would happen if we had two very narrow slits). Simply put, the reason that away from the center the intensity goes down is that the waves coming from the slit are all out of phase with each other.