Past exam of the physics 2 paper of the University of Oxford / 2021 / 9 / d / Solution

The two optical frequencies have no persistent mutual optical coherence over the detector's averaging time, so their light intensities add. Assuming equal strengths and using $s=1.0\mathrm{mm}$, let $I_s$ be the one-slit light intensity at either wavelength. ThenThe identity $\cos A+\cos B=2\cos [(A+B)/2]\cos [(A-B)/2]$ rewrites this asThe slowly varying interferometric visibility is approximately $\mathcal{V}=|\cos (\pi s\sin \theta [1/{\lambda }_1-1/{\lambda }_2])|$. Its first zero occurs atThe rapid fringes have effective wavelength ${\lambda }_h=2{\lambda }_1{\lambda }_2/({\lambda }_1+{\lambda }_2)=529.81\mathrm{nm}$, so the first contrast minimum is at order $7.02\mu \mathrm{m}/{\lambda }_h=13.25$. Thus the conventional estimate is about 13 fringes on each side, before their first washout, plus the central maximum. “Clearly” is not a specified visibility threshold: the last of those fringes already has low contrast.

The fringes do not disappear permanently beyond that point. Their contrast revives; the first exact common maximum satisfies $27{\lambda }_1=26{\lambda }_2=14.04\mu \mathrm{m}$, at $|\theta |=0.804^{\circ }$. This is a concrete counterexample to interpreting 13 as a global bound on visible fringes. Unequal spectral strengths would leave nonzero residual contrast at the nominal washout; the plotted equal-strength case is the natural interpretation in the absence of specified strengths.