Q.Light of wavelength $\lambda=6000\ \text{\AA}$ is incident normally on a single slit of width $a=0.2\ \text{mm}$, and the diffraction pattern is observed on a screen at a distance $D=1.5\ \text{m}$. Calculate the total width of the central maximum.

For a single slit of width a illuminated by a plane wavefront, dividing the slit into pairs of matching sub-sources shows that whenever the path difference between the two extreme edge rays, $a\sin\theta$, equals a whole wavelength $n\lambda$, the ENTIRE slit's light cancels in matching pairs, giving a diffraction MINIMUM at $a\sin\theta=n\lambda$. The (approximate) maxima fall roughly midway between successive minima, near $a\sin\theta=(n+\tfrac12)\lambda$. Using the small-angle approximation ($D\gg a$), this gives fringe spacing $W=\lambda D/a$ -- identical in FORM to Young's double-slit fringe width, but with slit WIDTH a replacing slit SEPARATION d -- with the crucial exception that the central bright fringe is TWICE as wide, $2\lambda D/a$, as every other fringe, since it spans between the first minima on either side. …