= Solution
The signal <phase velocity> in the coaxial cables is $v_c=c/2$, so the guided <wavelength> is $\lambda_c=v_c/f=12.5\,\mathrm{mm}$. The excess cable <length> is $\Delta\ell=3.125\,\mathrm{mm}=\lambda_c/4$, adding a relative delay of $20.83\,\mathrm{ps}$ and a relative <wave phase> of magnitude
$$
\delta_c=2\pi\frac{\Delta\ell}{\lambda_c}=\frac\pi2.
$$
Label the two inputs so that the first configuration has total <phase difference> $\phi-\pi/2$, where $\phi=2\pi b\sin\theta/\lambda$ is the geometric <phase difference>. Swapping the unequal cables reverses that added <wave phase>, giving $\phi+\pi/2$. For the same single-source received <wave intensity> $I_0$ in each arm,
$$
\boxed{I_A(\theta)=2I_0[1+\sin\phi(\theta)],\qquad
I_B(\theta)=2I_0[1-\sin\phi(\theta)].}
$$
The two <interference> patterns are shifted by half a fringe: maxima in one are nulls in the other. At broadside $\phi=0$ both give $2I_0$, so swapping does not change the strength of every possible source. For small <angles>, the maximum-to-maximum shift is $\lambda/(2b)\simeq\theta_R=2.056^\circ$ with the chosen baseline. Near broadside, the maxima of configuration A occur near $\theta=+\theta_R/2$ and those of B near $\theta=-\theta_R/2$; the signs are exchanged if the input labels are reversed. Independent satellites contribute separately according to these directional gains; their mutually uncorrelated signals should not be added as coherent <amplitudes>.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2021-resit-cable-swap-fringes.png]
{title=Calculated complementary two-dish signal strengths before and after swapping quarter-wavelength-delay cables}
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