= Solution
The geometric path difference for a direction at zenith angle $\theta+\alpha$ is $b\sin(\theta+\alpha)$. Equal cable lengths leave that difference unchanged; the inserted electronic delay cancels $b\sin\theta$ for the tracking center. The residual delay is therefore
$$
\tau=\frac bc[\sin(\theta+\alpha)-\sin\theta]\simeq\frac{b\alpha\cos\theta}{c}.
$$
The <complex visibility> of the unresolved source has phase $2\pi\nu\tau$, giving $\exp(ikb\alpha\cos\theta)$ in the chosen baseline ordering. Reversing the signal order complex-conjugates it without changing the measured visibility magnitude. Tracking the <antennas> geometrically does not itself provide this <fringe stopping>.
\Image[/past-exam-of-the-frontiers-of-observational-astrophysics-course-of-the-university-of-cambridge/2012/paper-1-interferometer-delay.png]
{title=Tracking delay in a two-antenna interferometer}
{description=Original functional schematic. The nearer antenna receives the tracked wave earlier and its electronic delay compensates that geometric lead. Angles are exaggerated and cable paths are schematic.}
For a top-hat band centered at $\nu_0$, let $f=\nu-\nu_0$. Averaging the residual fringe gives
$$
\frac1{\Delta\nu}\int_{-\Delta\nu/2}^{\Delta\nu/2}e^{2\pi i(\nu_0+f)\tau}\,df
=e^{2\pi i\nu_0\tau}\operatorname{sinc}(\pi\Delta\nu\tau),
\qquad\operatorname{sinc}x=\frac{\sin x}{x}.
$$
After removing the central fringe phase, its envelope is exactly
$$
\boxed{\operatorname{sinc}\left(\frac{\pi\Delta\nu b\alpha\cos\theta}{c}\right).}
$$
The full complex average retains the displayed central-frequency phase; the bare sinc in the question is the phase-referenced envelope or an average written in frequency-offset coordinates. This <bandwidth decorrelation of a point source> is cancellation between different frequency-dependent phases, not absorption of source radiation.
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