Q.In the diffraction pattern produced by a single slit, explain why the central maximum is very much brighter than any of the secondary maxima on either side of it.

In single-slit diffraction, the resultant amplitude at any point on the screen is found by adding up the contributions of secondary wavelets from every part of the slit width $a$, each with its own phase depending on its position across the slit and the angle $\theta$ under consideration.

At the centre ($\theta=0$), every wavelet from every part of the slit travels the same path length to reach the screen, so all of them add up exactly in phase -- the full width of the slit contributes constructively, giving the largest possible resultant amplitude and hence the brightest (central) maximum.

Away from the centre, at the angular position of a secondary maximum, the phases of wavelets from across the slit are no longer all aligned; only a small effective fraction of the slit's width contributes wavelets that are still roughly in phase with each other at that particular angle, while the wavelets from the rest of the slit largely cancel each other out in pairs. Since only a small fraction of the slit effectively contributes, the resultant amplitude -- and hence the intensity, which goes as amplitude squared -- at a secondary maximum is very much smaller than at the central maximum, and it becomes smaller still for the higher-order secondary maxima farther from the centre, where an even smaller effective fraction of the slit remains roughly in phase.

Compare how much of the slit's width contributes constructively at theta = 0 (all of it) versus at a secondary maximum (only a small fraction).

• Thinking the secondary maxima are dim simply because they are 'farther from the centre' -- the real reason is the reduced effective coherent contribution across the slit, not distance by itself.
• Confusing this single-slit brightness pattern with the double-slit case, where all the bright fringes ARE roughly equally bright.