= Solution
For each beam in <vacuum>, its <magnetic field> is $\mathbf B_j=\widehat{\mathbf k}_j\times\mathbf E_j/c$. Write the two complex scalar contributions, including the <beam splitter> factor, as $u_1=f(t-t_1)/\sqrt2$ and $u_2=f(t-t_2)/\sqrt2$. Since their <electric fields> are both along $z$,
$$
\mathbf E'=(u_1+u_2)\widehat{\mathbf z},\qquad
\mathbf B'=\frac1{c\sqrt2}\left[(u_1-u_2)\widehat{\mathbf x}-(u_1+u_2)\widehat{\mathbf y}\right].
$$
The optical-cycle and ensemble-averaged <Poynting vector> is
$$
\langle\mathbf S\rangle=\frac1{2\mu_0}\operatorname{Re}\langle\mathbf E'\times\mathbf B'^*\rangle
=\frac C{\sqrt2}\left[\langle|u_1+u_2|^2\rangle\widehat{\mathbf x}
+(\langle|u_1|^2\rangle-\langle|u_2|^2\rangle)\widehat{\mathbf y}\right].
$$
The mean individual <light intensities> are equal, so
$$
\boxed{\langle\mathbf S\rangle=\frac{I(y)}{\sqrt2}\widehat{\mathbf x}.}
$$
Thus <energy> flows along the bisector of the two beam directions, crossing a detector plane normal to $x$. Such a sensor can in principle record the variation across its $y$ coordinate. Its actual normal <energy> flux is $I(y)/\sqrt2$ in the normalization used above; this fixed factor does not change <interferometric visibility>.
There is a practical resolution condition in addition to the direction of <energy> flow. The <interference fringe spacing> $\lambda/\sqrt2$ is very small, so direct pixels would need spacing no larger than $\lambda/(2\sqrt2)$ as required by the <Nyquist–Shannon sampling theorem>, or optical magnification and sufficient angular aperture to preserve both incident directions. An ordinary <optical camera> with much larger effective pixels averages the <interference fringes>. The Poynting-vector argument establishes the allowed measurement geometry, not guaranteed resolution with any <optical camera>.
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