= Solution
In the <double-slit experiment>, each <photon> is detected as a localized event carrying energy $h\nu$, while repeated events build the <interference> pattern of coherent probability amplitudes from the two slits. Acquiring information that distinguishes the paths suppresses the <interference>; splitting the classical intensity into two incoherent alternatives would not reproduce the fringes.
Put one plate before each slit at normal incidence, with the same known <refractive index> $n$ and illumination of vacuum <wavelength> $\lambda$. Relative to the plate-free reference, their thickness difference $\Delta d=d_1-d_2$ produces an <optical path difference> $(n-1)\Delta d$ and a phase shift
$$
\Delta\phi=\frac{2\pi(n-1)\Delta d}{\lambda}.
$$
Observe the displacement of the full <interference fringe> pattern, or scan a controllable reference phase and estimate $\Delta\phi$. This compares the plates directly; a detectable phase displacement establishes unequal thickness under the common-material assumption. For fringe spacing $\lambda D/b$, the displacement is $-D(n-1)\Delta d/b$, where $b$ is slit separation and $D$ the screen distance, with sign set by the slit labeling.
A differential phase measurement can resolve a small difference sensitively and avoid determining two large absolute thicknesses. It does not guarantee a universally faster or more photon-efficient test: noise, coherence, alignment, and required accuracy matter. Moreover a null phase shift at one <wavelength> only implies
$$
\boxed{\Delta d=m\lambda/(n-1),\quad m\in\mathbb Z,}
$$
so equality requires a prior bound, a broadband or multiple-<wavelength> measurement, or continuously counting <interference fringes> as the plates are introduced. If the <refractive indices> differ, this apparatus measures optical thickness difference rather than geometrical thickness difference.
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