= Solution
A <Michelson interferometer> splits a collimated beam at a partially reflecting <beam splitter>. Two mutually perpendicular arms reflect from mirrors and recombine at the same <beam splitter>; the superposed returning beams produce <interference fringes> at the detector. Translate one mirror along its arm while holding the other fixed.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2-a3-michelson.png]
{title=Original Michelson geometry for measuring wavelength by counting fringes while one mirror moves}
If that mirror moves a measured distance $\Delta x$, the round-trip <optical path difference> changes by $2\Delta x$ in air, approximated as <vacuum>. Every change of one <wavelength> moves one complete bright-dark-bright fringe cycle past a fixed detector point. Count $N$ such cycles to obtain
$$
\Delta\phi=\frac{2\pi}{\lambda}(2\Delta x)=2\pi N,\qquad
\boxed{\lambda=\frac{2\Delta x}{N}.}
$$
Use a <spectral line> selected with a suitable <optical filter>, and count many <interference fringes> to reduce the fractional uncertainty of the distance and count. Keep the two arms within the source's <optical coherence length> and maintain overlap and alignment. If the <vacuum> <wavelength> is required for an arm filled with air of known <refractive index> $n$, the corrected relation is $\lambda_0=2n\Delta x/N$ at stable $n$.
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