 Solution

For a monochromatic photon of vacuum wavelength $\lambda$, a plate of thickness $d$ replaces air by glass and adds phase $\varphi =2\pi (n-1)d/\lambda$. 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 $D_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$. The successive quantum states, omitting common propagation phases, areThe first output component feeds detector 2 and the second feeds detector 1, givingThe imbalance gives $p_1-p_2=\cos \varphi$. Thus the measured probabilities determine only $\varphi$ up to $\varphi \mapsto \pm \varphi +2\pi m$, and hence only a corresponding family of thicknesses. There is no unique absolute thickness from this one measurement. The zero-plate limit recovers $p_1=1$, $p_2=0$.