= Solution
Each mutually incoherent patch of an extended <star> produces a displaced <double-slit interference> pattern. The detector adds their <light intensities>. For two apertures separated by a projected <interferometric baseline> $D$, a source direction offset by a small <angle> $\vartheta_x$ changes their <phase difference> by $2\pi D\vartheta_x/\lambda$. The normalized sum over the source's angular brightness distribution is its <first-order optical coherence>, whose magnitude is the <interferometric visibility> for equally bright channels.
For a uniform circular stellar disk of full <angular diameter> $\alpha$, the normalized angular <Fourier transform> gives the <uniform-disk stellar visibility>
$$
\mathcal V(D)=\left|\frac{2J_1(x)}x\right|,\qquad
x=\frac{\pi D\alpha}{\lambda}.
$$
Here $J_1$ is the order-one <Bessel function of the first kind>. This follows by integrating $e^{i(2\pi D/\lambda)\vartheta_x}$ over the disk and dividing by its area. In particular, its <power series> gives $2J_1(x)/x=1-x^2/8+O(x^4)$.
For an ordinary slit separation $d=1\,\mathrm{mm}$, using the paper's $\alpha=22.6\times10^{-8}\,\mathrm{rad}$ and observing <wavelength> $\lambda=570\,\mathrm{nm}$,
$$
\frac{d\alpha}{\lambda}=3.965\times10^{-4},\qquad
x=1.2456\times10^{-3},\qquad
1-\mathcal V\simeq1.94\times10^{-7}.
$$
Even $d=10\,\mathrm{mm}$ gives a visibility loss of only $1.94\times10^{-5}$. Thus millimetre separation leaves the <interference fringes> almost indistinguishable from those of a point source; it gives very little sensitivity to the diameter.
The movable mirrors in the <Michelson stellar interferometer> collect light at a large adjustable separation $D$ and redirect it into closely spaced output beams. They increase the sampling <interferometric baseline> independently of the fixed output slit spacing $d$: $D$ sets sensitivity to source structure, whereas the output optics set the screen's <interference fringe spacing>. The first <Bessel function> zero is $x_1=3.83170597$, hence
$$
\boxed{D_{\rm null}=\frac{x_1}{\pi}\frac\lambda\alpha
=1.21967\frac{570\times10^{-9}\,\mathrm m}{22.6\times10^{-8}}
=3.076\,\mathrm m\simeq3.08\,\mathrm m.}
$$
The <interference fringes> disappear at this baseline even though each beam separately remains bright.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2022-q3-stellar-visibility.png]
{title=Original calculated <interferometric visibility> for the given stellar diameter and <wavelength>}
The value $3.08\,\mathrm m$ assumes a uniform disk. The original question supplies a diameter without a <surface brightness> law; real <limb darkening> changes the precise visibility zero. A one-dimensional uniformly bright strip of the same full angular width instead has a <sinc function> visibility and first zero $D=\lambda/\alpha=2.522\,\mathrm m$. That is a different source model, not the circular-disk result. The robust conclusion without a <surface brightness> model is that a baseline of a few metres, rather than millimetres, is required.
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