Interferometric visibility ($\mathcal{V}=(I_{\max }-I_{\min })/(I_{\max }+I_{\min })$)

Interferometric visibility measures the contrast of a scanned interference fringe. For the mutually coherent, equally polarized two-beam model in wave interference, with fixed individual wave intensities and $I_1+I_2>0$, the constructive interference intensity and destructive interference intensity giveEqual nonzero wave intensities give $\mathcal{V}=1$; $I_1=1,I_2=0.25$ give $\mathcal{V}=0.8$. Unequal wave intensities can therefore reduce interferometric visibility even with perfect mutual optical coherence. Random phase differences or distinguishable wave polarizations can reduce it further; zero contrast alone does not identify the cause. Both intensities zero leave the ratio undefined.

For normalized paths with complex amplitudes $\alpha ,\beta$ and identical internal states, the output probability is $1/2+\mathrm{Re}({\alpha }^{\ast }\beta e^{i\theta })$, so $\mathcal{V}=2|\alpha \beta |$. A photon polarization fringe alone determines transverse quantum coherence but not the sign of the population difference.

// Target: optics.bigb

Combining equal-intensity beams with complex degree of first-order coherence $g$ gives a fringe modulation proportional to $\mathrm{Re}(e^{i\phi }g)$. Maximizing and minimizing over observation phase gives visibility $V=|g|$. A symmetric source can make $g$ real, but it may still be negative. Its sign changes fringe phase by $\pi$; it does not represent negative physical contrast.