= Solution
Take equal arm lengths, no plate, and coherent illumination. At a coated <beam splitter>, choose the real-amplitude convention where transmission has amplitude $1/\sqrt2$ and reflection has amplitude $+1/\sqrt2$ from one side and $-1/\sqrt2$ from the other; which reflection carries the minus sign is a basis convention. The two paths interfere constructively at detector 1 and destructively at detector 2 in the diagram's labeling. Thus
$$
\boxed{I_1/I_{\rm in}=1,\qquad I_2/I_{\rm in}=0.}
$$
For example, writing the <beam splitter> as $B=2^{-1/2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ and the two equal-reflection <mirrors> as $M=-\begin{pmatrix}0&1\\1&0\end{pmatrix}$, one obtains $BMB(1,0)^T=(0,-1)^T$. The first output component propagates downwards to detector 2; the nonzero second component propagates upwards to detector 1. The overall minus sign is an unobservable global phase. The redistribution is <interference> of amplitudes, rather than two successive independent coin tosses.
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