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

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.