Past exam of the physics A course of the University of Cambridge / 2013 / Paper 2 / Section B / B9 / Solution

Far-field transform. For transverse aperture scale $b$ and screen distance $z$, the quadratic phase across the aperture is negligible when $k{b}^2/(2z)\ll 1$, equivalently a small Fresnel number up to geometrical factors. Expanding the source-to-screen distance at transverse screen coordinate $X$ givesDropping the final aperture-dependent term and treating the prefactor as constant givesThus intensity is the modulus squared of the aperture Fourier transform. A lens focal plane supplies the same angular transform without a very distant screen. The infinitely extended vertical dimension is treated per unit vertical extent.

half-slit phase-reversal diffraction. Put $h=-1$ on the left half and $h=1$ on the right half. With $q=k\sin \theta$ and $s=qb/4$,ThereforeThe apparent singularity at zero angle is removable: $H\to 0$, so the centre is dark. Normalizing by the uncoated slit's on-axis intensity $I_0$ gives $I/I_0={\sin }^{4}s/s^2$. Noncentral zeros have $s=j\pi$, or $\sin \theta =2j\lambda /b$. The two largest lobes are symmetric, at the first nonzero root of $\tan s=2s$, $s\simeq \pm 1.166$, giving $\sin \theta \simeq \pm 0.742\lambda /b$ when those directions propagate. Smaller lobes lie farther out under a decaying envelope.

Doubled thickness. Neglecting absorption and interface reflection, the phase delay doubles from $\pi$ to $2\pi$. Both halves then transmit with the same phase, recovering the ordinary single-slit patternIt has a bright centre and first zeros at $\sin \theta =\pm \lambda /b$. The phase layer redistributes transmitted power rather than absorbing it in this idealization.