Past exam of the physics A course of the University of Cambridge / 2010 / Paper 1 / Section B / B8

In Fraunhofer diffraction, the phase difference from aperture coordinate y to observation angle theta is $qy$, with $q=2\pi \sin \theta /\lambda$. Coherent superposition of the transmitted field elements therefore givesThe common propagation factor is omitted. This is the Fourier transform of the complex amplitude transmission, not of its intensity.

For a uniform slit of total width $W=0.2\mathrm{mm}$, integration gives ${\psi }_0=W\mathrm{sinc}(qW/2)$, where $\mathrm{sinc}s=\sin s/s$. Its normalized intensity is ${\mathrm{sinc}}^2(\pi W\sin \theta /\lambda )$, with a central maximum and first zeros at $\sin \theta =\pm \lambda /W=\pm 0.0025$.

For a half-covered slit choose $T=+1$ on the left half and $T=-1$ on the right. Put $h=W/2$ and $s=qh/2$. The two equal rectangles have centres separated by h, soThis half-slit phase-step diffraction factorization is the phase-reversed two-slit interference factor multiplied by the diffraction envelope of each half-slit. Here the half-slit width and centre separation both equal the earlier0.1mm separation. The central intensity is zero; the largest off-axis pair satisfies $\tan s=2s$, giving $|\sin \theta |\simeq 0.00186$. Other zeros are at $s=m\pi$. The plot includes the requested uniform-slit and half-covered patterns.

The slit transmissions have the same magnitude but opposite phase. An ideal narrow-slit representation isEquivalently draw two narrow rectangles at $y=\pm d/2$, one at+1 and the other at−1, and zero elsewhere. A finite rectangle width gives each slit its own diffraction envelope. The original aperture plot shows such rectangles with explicitly illustrative width; the negative value is a field phase reversal, not negative transmitted intensity.

For the narrow pair, the amplitude is $e^{-iqd/2}-e^{iqd/2}=-2i\sin (qd/2)$. Hence phase-reversed double-slit diffraction givesThere is a central dark fringe. Minima obey $d\sin \theta =m\lambda$, maxima $d\sin \theta =(m+1/2)\lambda$, and the near-axis angular fringe period is $\lambda /d=0.005\mathrm{rad}$. Finite slit widths multiply this interference profile by their single-slit envelope. The parent plot includes the narrow-pair profile.