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.
Best Answer
I boil everything down to Fourier transforms and I'll refer to the useful Wikipedia article several times. I won't care about pre-factors in the Fourier transforms for the sake of simplicity. You can look them up in the referred transforms.
First, consider the single slit with width
a, it has the functionrect(ax)and the Fourier transformsinc(x/a)(No. 201). Square it (Intensity) and you get the diffraction pattern.Second, the double slit. It is basically a convolution (I denote it by
**and multiplication by*) of two delta functionsδ(x-d/2)+δ(x+d/2)for the positions of the slits spaced bydand the box functionrect(ax)giving the single slit(s):f(x)=(δ(x-d/2)+δ(x+d/2))**rect(ax). Now the FT of the two delta functions is either directly the reverse of No. 304 or you can compute it as the sum of two shifted delta functions (shift theorem No. 102 and delta function No. 302). SoFT[(δ(x-d/2)+δ(x+d/2))]=cos(dx).By the convolution theorem (No 108) we get that
FT[g(x)**h(x)]=FT[g(x)]*FT[h(x)]and thus the resultingFT[f(x)]=cos(dx)*sinc(x/a). Since the pattern displays the intensity which is the square of the field we have to square everything and get the resulting interference pattern.Now, computing the minima is just maths and property of the
sinc²function resp. thecos². It turns out that thesinc²function has minima wheresin²has minima (i.e. wheresinis zero, with π periodicity) except for zero because of the definition of sinc that is 1 there. This explains the double width of the central maximum. The `cos², however, has equally spaced minima (π periodicity again)If you have infinitely small slits, you just get a uniform fourier transform and thus the double slit would be just the
cos².