Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B2 / 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 $\varphi$ isConsequently the polarization overlap in interferometric visibility isFor equal returned intensities, two linear polarizations at angle $\theta$ have overlap $\cos \theta$ and thereforeChoose one linear direction as ${\mathbf{e}}_L=(1,0{)}^T$ and a circular polarization as ${\mathbf{e}}_C=(1,\pm i{)}^T/\sqrt{2}$. Their overlap has magnitude $1/\sqrt{2}$, givingA non-polarizing beam splitter does not by itself guarantee equal arm losses; if $I_1\ne I_2$, multiply either boxed polarization factor by $2\sqrt{I_1{I}_2}/(I_1+I_2)$. Perfect temporal coherence does not compensate for polarization or intensity imbalance.