Q.Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.

[!TLDR] Pairing points across a single slit's width shows the entire slit cancels out (in matching $\lambda/2$-path-difference pairs) whenever the extreme-ray path difference $a\sin\theta$ is a whole multiple of $\lambda$, giving minima at $a\sin\theta=n\lambda$; the maxima fall roughly midway, near $a\sin\theta=(n+\t …

Consider a single slit of width a, illuminated by a plane wavefront, with its width lying in the plane of the diagram and length running perpendicular to it. Conceptually divide the slit into an enormous number of infinitesimally thin sub-slits (secondary Huygens sources), all in phase with each other the instant the plane wavefront arrives. Rays travelling parallel to the original (undeviated) axis, from every point across the slit, converge to the central point $P_0$ directly opposite the slit's midpoint O; since these are all parallel with equal optical path lengths to $P_0$, they arrive exactly in phase, giving the brightest possible constructive interference right at the centre.

Now consider a screen point P at angular position $\theta$. Drawing a construction line AC from the top edge A, perpendicular to the direction of the rays heading towards P, the segment BC (from the bottom edge B) represents the PATH DIFFERENCE between the two extreme rays reaching angle $\theta$: $BC=a\sin\theta$. Suppose this equals exactly one wavelength: $a\sin\theta=\lambda$. Bisect AC at its midpoint K -- this conceptually splits the slit into two equal halves, from A to the centre O, and from O to B. Take any pair of points symmetric about O, one in each half (say G, above O, and H, below O, at equal distances from O) -- since the total path difference across the FULL slit width a is $\lambda$, the path difference across HALF the slit width (i.e. between any such matched pair G, H) is exactly $\lambda/2$, which is precisely the condition for DESTRUCTIVE interference between that pair. Since this same pairing argument applies to EVERY point in the upper half matched with its partner in the lower half -- the pairing exhausts the entire slit -- the contributions from the whole slit cancel out completely at P, making it a point of zero intensity: the FIRST diffraction MINIMUM.

The identical argument extends directly to path differences of $2\lambda, 3\lambda, \ldots, n\lambda$ between the extreme rays (each such case can similarly be divided into $2n$ equal matched-pair segments, each contributing zero), locating every successive minimum, with the same reasoning applying symmetrically on either side of the central maximum. In general, the CONDITION FOR MINIMA is: $a\sin\theta=n\lambda$, for $n=\pm1,\pm2,\pm3,\ldots$ (n = 0 is excluded, since that is the central bright maximum itself, not a minimum). …

Divide the slit into matched pairs of points separated by half the slit width, show that a path difference of $\lambda$ between the extreme rays makes every such pair cancel destructively (giving the entire slit zero net contribution), and extend the same pairw …

Applying the double-slit interference conditions ($a\sin\theta=n\lambda$ for BRIGHT fringes) directly to single-slit diffraction, instead of correctly recognising that for diffraction it is $a\sin\theta=n\lambda$ that gives the DARK fringes (minima), with bright fringes instead falling near …