Solution (source code)

= 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>. Then
$$
\boxed{I(\theta)=4I_s\left[\cos^2\left(\frac{\pi s\sin\theta}{\lambda_1}\right)+\cos^2\left(\frac{\pi s\sin\theta}{\lambda_2}\right)\right],\quad \lambda_1=520\,\mathrm{nm},\quad\lambda_2=540\,\mathrm{nm}.}
$$
The identity $\cos A+\cos B=2\cos[(A+B)/2]\cos[(A-B)/2]$ rewrites this as
$$
I(\theta)=4I_s\left[1+\cos\left(\pi s\sin\theta\left(\frac1{\lambda_1}+\frac1{\lambda_2}\right)\right)\cos\left(\pi s\sin\theta\left(\frac1{\lambda_1}-\frac1{\lambda_2}\right)\right)\right].
$$
The slowly varying <interferometric visibility> is approximately $\mathcal V=|\cos(\pi s\sin\theta\,[1/\lambda_1-1/\lambda_2])|$. Its first zero occurs at
$$
s\sin\theta_{\rm zero}=\frac{\lambda_1\lambda_2}{2(\lambda_2-\lambda_1)}=7.02\,\mu\mathrm m,
\qquad \theta_{\rm zero}=0.402^\circ.
$$
The 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 \b[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.

\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2021-two-wavelength-fringes.png]
{title=Two-wavelength interference, its first visibility minima, and contrast revival over minus one to plus one degree}

The fringes do \b[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.