= Polarization-dependent visibility
Let the two arms deliver intensities $I_1,I_2$ with unit <Jones vectors> $\mathbf e_1,\mathbf e_2$. For a perfectly coherent input, the recombined intensity under relative phase $\phi$ is
$$
I(\phi)=I_1+I_2+2\sqrt{I_1I_2}\operatorname{Re}[\mathbf e_1^\dagger\mathbf e_2e^{i\phi}].
$$
Consequently the <polarization overlap in interferometric visibility> is
$$
\mathcal V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}
=\frac{2\sqrt{I_1I_2}}{I_1+I_2}|\mathbf e_1^\dagger\mathbf e_2|.
$$
For equal returned intensities, two <linear polarizations> at angle $\theta$ have overlap $\cos\theta$ and therefore
$$
\boxed{\mathcal V_{\rm linear,linear}=|\cos\theta|.}
$$
Choose one linear direction as $\mathbf e_L=(1,0)^T$ and a <circular polarization> as $\mathbf e_C=(1,\pm i)^T/\sqrt2$. Their overlap has magnitude $1/\sqrt2$, giving
$$
\boxed{\mathcal V_{\rm linear,circular}=\frac1{\sqrt2}.}
$$
A non-polarizing <beam splitter> does not by itself guarantee equal arm losses; if $I_1\ne I_2$, multiply either boxed polarization factor by $2\sqrt{I_1I_2}/(I_1+I_2)$. Perfect temporal coherence does not compensate for polarization or intensity imbalance.
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