= Solution
Use the symmetric lossless <beam splitter> convention with transmission amplitude $t=1/\sqrt2$ and reflection amplitude $r=i/\sqrt2$. The two mirrors contribute the same fixed <quantum phase>, which can be suppressed. At the rightward port $D_1$, the upper-arm route has one reflection at BS1 and one transmission at BS2, while the lower-arm route has one transmission at BS1 and one reflection at BS2. Their <probability amplitudes> therefore add as
$$
a_1=rt\,e^{i\theta}+tr=\frac{i}{2}(e^{i\theta}+1).
$$
By the <Born rule>,
$$
\boxed{P(D_1)=|a_1|^2=\frac{1+\cos\theta}{2}=\cos^2\!\left(\frac\theta2\right).}
$$
For the upward port, the two amplitudes are $r^2e^{i\theta}$ and $t^2$, giving $a_0=(1-e^{i\theta})/2$ and $P(D_0)=\sin^2(\theta/2)$. They sum to one, as required for an ideal lossless <Mach-Zehnder interferometer>. Thus $D_1$ is bright at zero relative <quantum phase> and dark at <quantum phase> $\pi$. A fixed unequal reflection <quantum phase> or a swapped port labeling would shift the fringe; the stated convention fixes the port and <quantum phase> calibration used below.
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