 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 shiftObserve the displacement of the full interference fringe pattern, or scan a controllable reference phase and estimate $\Delta \varphi$. 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 impliesso 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.