Confusion about the derivation of the first order minima of single slit diffraction I have a problem with the derivation of the first order minima of single slit diffraction. If the top ray cancels with the central ray which is $a/2$ below it, it means that the central ray should also be cancelled with the bottom ray which is $a/2$ below the central ray. Now there are three rays cancel with each other. 
Is it possible? Is my understanding wrong and where is the error in my understanding? 

 A: In fact, the top and bottom ray do not cancel with each other, as they are in phase (phase difference of $+\pi$ between the top and middle ray, and again $+\pi$ between the middle and bottom ray). So the field resulting from the top, middle ans bottom rays would be $E_0 + E_0 e^{i \pi} + E_0 e^{i2\pi} = E_0 - E_0 + E_0 = E_0$.
But maybe you missed the argument of finding the first minimum. You could sum all triplets of points in a weird way instead of summing over all pairs of points, and you would get the same result (it would just not be as easy to compute). The argument is that for each point in the top half of the slit, there is a corresponding point in the bottom half (at a distance of $a/2$ from the first point) for which the field at infinity is exactly opposite. So if you sum over every pairs of points that way, the full field would be $0 + 0 + 0 + \dots + 0 = 0$, corresponding to a minimum of intensity. The only caveat is that the middle point belongs both to the top and the bottom half, so the sommation breaks down here, but because it is a single point and you're summing infinitely many of them it does not make a difference.
Note that while this argument shows that for this particular direction, the intensity is minimum, it does not prove that this is the first minimum, as there could be a smaller angle for which summing all fields (possibly on a different way as here) would also give $0$. Such an angle does not exist in fact so this argument really gives the position of the first minimum.
