= Solution
On the $x$ axis the two geometric delays are identical:
$$
t_1=t_2=\frac{kx}{\sqrt2\omega}.
$$
Their field <wave amplitudes> therefore add as equal copies at zero relative delay. Stationarity removes dependence on the common delay, giving
$$
\boxed{I(x)=I_0+\operatorname{Re}\Gamma(0)=2I_0,}
$$
independent of $x$ in the ideal overlapping plane-wave region. The corresponding mean normal <energy> flux through a transverse plane is $\sqrt2I_0$. There is no increasing path difference along the bisector, so finite <temporal optical coherence> does not produce a longitudinal visibility envelope on this axis.
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