Fraunhofer Diffraction at a Single Slit

Consider a single slit of width a, with its width lying in the plane of the page and its (much longer) length running perpendicular to the page, illuminated by a plane wavefront $YY'$ (Fig. 7.15). Conceptually, the slit can be imagined as divided into an enormous number of extremely thin sub-slits (secondary sources), each emitting cylindrical wavelets, per Huygens' principle, the instant the plane wavefront reaches it; a cylindrical lens placed just beyond the slit converges these emerging beams onto a screen at its focal plane (distance D from the slit, effectively the lens's focal length F, since $D \gg a$ for all practical slit widths of $10^{-4}$ to $10^{-3}$ m against a typical D of order 10 m).

All the rays travelling parallel to the original (undeviated) axis, from every point across the slit from edge A to edge B, converge to the same CENTRAL point $P_0$ directly opposite the slit's midpoint O; since these rays are all parallel and therefore have equal optical path lengths to $P_0$, they arrive exactly in phase, producing the brightest possible CONSTRUCTIVE interference right at the centre.

Now consider a point P on the screen at angular position $\theta$ from the axis. Drawing a construction line AC from the top edge A, perpendicular to the direction of the rays heading towards P (so AC is normal to the direction $BP$), the segment $BC$ represents the PATH DIFFERENCE between the two extreme rays from the slit's edges A and B reaching angle $\theta$: $BC = a\sin\theta$. Suppose this path difference equals exactly one full wavelength, $a\sin\theta = \lambda$. Bisecting AC at its midpoint K conceptually splits the slit into two equal halves -- from A to the slit's centre O, and from O to B -- and any pair of points equally spaced from A and from O respectively (such as G above O and H below O, symmetric about O) then has a path difference of exactly $\lambda/2$ at point P, producing DESTRUCTIVE interference for that pair. Since this pairing exhausts the ENTIRE slit (every point in the upper half has a matching partner in the lower half, all with path difference $\lambda/2$), the light from the whole slit cancels out completely at P -- making P the location of the FIRST diffraction MINIMUM. The identical argument, extended to path differences of $2\lambda, 3\lambda, \ldots, n\lambda$ between the extreme rays, locates every successive minimum; the same reasoning applies symmetrically on the other side of the central maximum too. In general, the nth minimum occurs where $a\sin\theta = n\lambda$ ($n = \pm1, \pm2, \ldots$), and the (approximate, not exact) maxima fall roughly midway between successive minima, near $a\sin\theta = \left(n+\tfrac12\right)\lambda$.

Since $D \gg a$ makes $\theta$ a small angle, $\sin\theta \approx \tan\theta \approx y/D$ (y being the distance of the point on the screen from the centre), so the DISTANCES of the nth dark and (approximate) nth bright points from the centre work out to $y_{n,\text{dark}} = n\lambda D/a$ and $y_{n,\text{bright}} \approx \left(n+\tfrac12\right)\lambda D/a$. The spacing between consecutive minima (or, equivalently, between consecutive maxima away from the centre) is $W = \lambda D/a$ -- a formula that looks exactly like Young's double-slit fringe width, but with the slit WIDTH a in place of the slit SEPARATION d, and with one crucial exception: the CENTRAL bright fringe -- spanning between the first minimum on either side, at $\pm\lambda D/a$ -- is TWICE as wide, $W_c = 2\lambda D/a$, as every other bright fringe in the pattern. …