= Solution
Place the two slits at transverse coordinates $+h/2$ and $-h/2$. Let the source point have coordinate $s$ and the observation point coordinate $x$, both measured in the same transverse direction. The total lower-slit path minus upper-slit path is
$$
\begin{aligned}
\Delta\ell(s,x)={}&\sqrt{R^2+(s+h/2)^2}-\sqrt{R^2+(s-h/2)^2}\\
&+\sqrt{D^2+(x+h/2)^2}-\sqrt{D^2+(x-h/2)^2}\\
\simeq{}&h\left(\frac sR+\frac xD\right).
\end{aligned}
$$
The last line uses the <paraxial approximation> at both source and screen. The source assumptions give $w,h\ll R$; the screen expression also assumes $h,|x|\ll D$. Reversing the source-coordinate orientation would change the sign of $s/R$, without changing the uniform-source result.
At the slits, the differential path delay from a source element is $hs/R$. Across the full filament its range is $hw/R$. Keeping all these phases nearly equal therefore requires
$$
\boxed{\frac{hw}{R}\ll\lambda.}
$$
This is a sufficient near-unit-<spatial coherence> condition, rather than a necessary condition for any observable <interference>. That distinction follows quantitatively below.
Different small elements of a thermal filament are mutually incoherent, so their <wave intensities> add. The two paths from one element are coherent. If that element supplies the same intensity $I_0B(s)\,ds$ through each slit, the <superposition principle> gives
$$
dI_x=\left|\sqrt{I_0B(s)\,ds}\,e^{i\varphi_+}+\sqrt{I_0B(s)\,ds}\,e^{i\varphi_-}\right|^2
=2I_0B(s)[1+\cos\delta(s,x)]\,ds,
$$
where
$$
\boxed{\delta(s,x)=\frac{2\pi h}{\lambda}\left(\frac xD+\frac sR\right).}
$$
Integrating intensities over the incoherent source gives
$$
\boxed{I_x=2I_0\int_{-w/2}^{w/2}B(s)\left[1+\cos\left(\frac{2\pi h x}{\lambda D}+\frac{2\pi h s}{\lambda R}\right)\right]ds.}
$$
Adding complex amplitudes across different source elements instead would incorrectly assume mutual <optical coherence> of the entire filament.
For $B(s)=1/w$, define $\alpha=2\pi h/(\lambda D)$ and $\beta=\pi h/(\lambda R)$. The source's signed quantity is
$$
V_{\rm printed}(x)=\frac1w\int_{-w/2}^{w/2}\cos(\alpha x+2\beta s)\,ds
=\cos(\alpha x)\frac{\sin(\beta w)}{\beta w}.
$$
Thus
$$
\boxed{\alpha=\frac{2\pi h}{\lambda D},\qquad \beta=\frac{\pi h}{\lambda R},\qquad I_x=2I_0\left[1+\cos(\alpha x)\frac{\sin(\beta w)}{\beta w}\right].}
$$
The continuous limit at $\beta w=0$ uses $\sin z/z\to1$.
\b[The printed $V$ is a signed local modulation, not the usual fringe visibility.] From the maximum and minimum of this scanned fringe pattern, the <interferometric visibility> is
$$
\boxed{\mathcal V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}=\left|\frac{\sin(\beta w)}{\beta w}\right|.}
$$
It is independent of the screen's fringe phase and is nonnegative. For a concrete distinction, take $\beta w=\pi/2$ and $\alpha x=\pi/2$: the printed $V$ is zero at that observation point, while the pattern has contrast $\mathcal V=2/\pi$. The same example has $hw/R=\lambda/2$, which is not much smaller than $\lambda$, yet substantial <interference> survives. When the signed coherence factor is negative, bright and dark fringe positions interchange; the contrast remains its absolute value. This is the distinction between <signed coherence and fringe visibility>.
\Image[/past-exam-of-the-electromagnetism-and-optics-course-of-the-university-of-oxford/2016-filament-coherence-and-fringes.png]
{title=Finite source width changes fringe contrast and can reverse fringe phase}
{description=Original uniform-filament calculation. The signed source factor and its absolute value differ; the latter is the standard <interferometric visibility>. The fringe traces illustrate high contrast, vanishing contrast, and phase reversal.}
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