 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 magnitudeLabel the two inputs so that the first configuration has total phase difference $\varphi -\pi /2$, where $\varphi =2\pi b\sin \theta /\lambda$ is the geometric phase difference. Swapping the unequal cables reverses that added wave phase, giving $\varphi +\pi /2$. For the same single-source received wave intensity $I_0$ in each arm,The two interference patterns are shifted by half a fringe: maxima in one are nulls in the other. At broadside $\varphi =0$ 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.