= Solution
The two <sodium> lines are distinct optical <frequencies>, so their rapidly oscillating mutual cross terms average to zero at the detector. Assuming equal line strengths and negligible individual linewidths, add their separate <interferograms>. Write $\bar k=(k_1+k_2)/2$ and $\Delta k=|k_1-k_2|$. With $I_0$ the total mean output intensity,
$$
I(x)=I_0\left[1+\frac12(\cos k_1x+\cos k_2x)\right]
=I_0[1+\cos(\bar kx)\cos(\Delta kx/2)].
$$
For mirror displacement $d=x/2$ this becomes
$$
\boxed{I(d)=I_0[1+\cos(2\bar kd)\cos(\Delta kd)]},\qquad
\mathcal V(d)=|\cos(\Delta kd)|.
$$
This is <spectral doublet from Michelson visibility revivals>. Since $\Delta\lambda\ll\bar\lambda$, $\Delta k\simeq2\pi\Delta\lambda/\bar\lambda^2$. Numerically,
$$
\boxed{\Delta d_{\rm fringe}\simeq\bar\lambda/2=294.65\,\mathrm{nm}},\qquad
\boxed{\Delta d_{\rm revival}=\pi/\Delta k\simeq0.2894\,\mathrm{mm}}.
$$
The first <interference visibility> zero occurs at $\boxed{d=0.1447\,\mathrm{mm}}$, with further zeros at odd multiples of that distance. Visibility is restored at integer multiples of $0.2894\,\mathrm{mm}$. The \i[signed] envelope has period $0.5788\,\mathrm{mm}$: alternate high-visibility packets have opposite fringe <phase>. Unequal line strengths leave a nonzero minimum visibility, and finite linewidths introduce a further decaying envelope; the wavelength data alone specify neither effect.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2-b8-interferograms.png]
{title=Monochromatic Michelson fringes and the fast-fringe and visibility scales of a sodium doublet}
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