= Solution
\b[Far-field transform.] For transverse aperture scale $b$ and screen distance $z$, the quadratic phase across the aperture is negligible when $kb^2/(2z)\ll1$, equivalently a small <Fresnel number> up to geometrical factors. Expanding the source-to-screen distance at transverse screen coordinate $X$ gives
$$
r\simeq z+\frac{X^2}{2z}-\frac{Xx}{z}+\frac{x^2}{2z}.
$$
Dropping the final aperture-dependent term and treating the prefactor as constant gives
$$
U(X)\propto\frac{e^{ikz+ikX^2/(2z)}}z\int h(x)e^{-i(kX/z)x}dx,
\qquad
\boxed{I(\theta)\propto\left|\int h(x)e^{-ikx\sin\theta}dx\right|^2}.
$$
Thus 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.
\b[<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 $s=qb/4$,
$$
H(q)=\int_0^{b/2}e^{-iqx}dx-\int_{-b/2}^0e^{-iqx}dx
=\frac{4\sin^2(qb/4)}{iq}.
$$
Therefore
$$
\boxed{I(\theta)\propto\frac{\sin^4(kb\sin\theta/4)}{\sin^2\theta}}.
$$
The apparent singularity at zero angle is removable: $H\to0$, so the centre is dark. Normalizing by the uncoated slit's on-axis intensity $I_0$ gives $I/I_0=\sin^4s/s^2$. Noncentral zeros have $s=j\pi$, or $\sin\theta=2j\lambda/b$. The two largest lobes are symmetric, at the first nonzero root of $\tan s=2s$, $s\simeq\pm1.166$, giving $\sin\theta\simeq\pm0.742\lambda/b$ when those directions propagate. Smaller lobes lie farther out under a decaying envelope.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2013/paper-2-phase-slit.png]
{title=Original calculated half-slit phase-reversal pattern and its change when the phase-layer thickness is doubled}
\b[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 pattern
$$
\boxed{I/I_0=\left[\frac{\sin(qb/2)}{qb/2}\right]^2}.
$$
It 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.
Back to article page
