= Solution
For a monochromatic <photon> of vacuum <wavelength> $\lambda$, a plate of thickness $d$ replaces air by glass and adds phase $\phi=2\pi(n-1)d/\lambda$. Neglect losses and reflections, or assume antireflection coatings. Between the first <beam splitter> and mirror 1, the affected arm is the second basis component, so $D_\phi=\operatorname{diag}(1,e^{i\phi})$. The successive <quantum states>, omitting common propagation phases, are
$$
\begin{aligned}
\binom10&\xrightarrow{B}\frac1{\sqrt2}\binom11
\xrightarrow{D_\phi}\frac1{\sqrt2}\binom1{e^{i\phi}}
\xrightarrow{M}-\frac1{\sqrt2}\binom{e^{i\phi}}1\\
&\xrightarrow{B}-\frac12\binom{e^{i\phi}-1}{e^{i\phi}+1}.
\end{aligned}
$$
The first output component feeds detector 2 and the second feeds detector 1, giving
$$
\boxed{p_2=\sin^2(\phi/2),\qquad p_1=\cos^2(\phi/2).}
$$
The imbalance gives $p_1-p_2=\cos\phi$. Thus the measured <probabilities> determine only $\phi$ up to $\phi\mapsto\pm\phi+2\pi m$, and hence only a corresponding family of thicknesses. There is no unique absolute thickness from this one measurement. The zero-plate limit recovers $p_1=1$, $p_2=0$.
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