Understanding the Formula for Dark Fringe in Single-Slit Diffraction

[KFC]

Hi guys,
I have a doublt about deducing the formula of dark fringe for single-slit diffraction. In the text, it reads "Consider two narrowstrips, one just below the top edge of the drawing of the slit and one at its center. The difference in path length to point P is $$(a/2)\sin\theta$$, where a is the slit width. Suppose this path difference happens to be equal to $$\lambda/2$$; then light from the two strips arrives at point P with a half-cycle phase difference, and cancellation occurs."

Why it said "happens to be ..." ? Why it must be the top point and the center of the slit? It seems not that convincible! What about I pick the top point of the slit and a point apart the top edge of the slit by a/2.25 and said "Suppose the path difference (from these two points) happens to be $$\lambda/2$$ ..." then I will also get a similar formula but with different $$\theta$$! So the text choose that specific points make sense?

 

[Born2bwire]

A rough approximation with a wide slit is that light passing through the center of the slit does so unperturbed (think of it as in wavelets). However, the light that strikes the edge of the slit will be diffracted. This diffracted light at the edges will travel outward back towards the center of the slit and interfere with the light that originally passed through unperturbed.

 

[Dadface]

I think the book is referring to a simplified method of finding the positions of the first minima.The wavefront is divided in half and the book is considering two corresponding points on the two half wavefronts and calculating the angle such that the waves from these two points interfere destructively.At this angle there will also be destructive interference between the waves from every other set of corresponding points on the two half wavefronts,in other words, by dividing in half the contribution from every single point on the incident wavefront can be accounted for.With your method KFC you have not divided the wavefront in half and although from some sets of points there will be destructive interference you cannot extend this method to work out the contribution from the remaining points.