Single slit diffraction: Minima

Why is there a minima when $$b\sin\phi$$ is an integral multiple of wavelength of the monochromatic light used for single slit diffraction?

I am convinced with the mathematical proof, but I'm looking for a non mathematical and physical explanation for why that should be the case (as in YDSE, it clearly isn't)

• It follows from HFP. – my2cts Dec 24 '19 at 9:21

Imagine that the slit is split into two halves of width $b/2$. A minima occurs when for each point on one half there is a corresponding point on the other half such that the path difference is half a wavelength. Such pairs cancel each other out.
Take the lowest point of the slit as $y = 0$. The pairs of points on the two halves which has a path difference of $\lambda/2$ is $(0,b/2), (\Delta y, b+\Delta y), (2\Delta y, b+2\Delta y), ... (n\Delta y, b)$
The path difference between the ends of the slit, i.e $y = 0$ and $y = b$ will be one wavelength, $\lambda$. Hence the condition $b\sin\phi = m\lambda$.