= Solution
Draw the <Fabry–Pérot etalon> with parallel mirrors and an incident beam making internal angle $\theta$ with their normal. Every emerging transmitted contribution has undergone one more internal round trip than its predecessor.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2-b6-etalon-rays.png]
{title=Successive transmitted paths and a common outgoing wavefront for a tilted beam in an air-gap etalon}
The extra internal path is $2d/\cos\theta$, but successive contributions leave at laterally separated points, a distance $2d\tan\theta$ apart. Comparing their phases on the same outgoing <wavefront> subtracts the path projection $2d\tan\theta\sin\theta$. Their net <optical path difference> in air is
$$
\Delta=\frac{2d}{\cos\theta}-2d\tan\theta\sin\theta=2d\cos\theta.
$$
Hence, in the reflection-phase convention implicit in the supplied transmission formula,
$$
\boxed{\delta=\frac{2\pi\Delta}{\lambda}=\frac{4\pi d\cos\theta}{\lambda}.}
$$
For a gap of <refractive index> $n$, replace $d$ by $nd$ and use the <vacuum> <wavelength>. Coating reflection phases can add a round-trip offset; the expression above assumes that offset is absorbed in the resonance order, or is zero modulo $2\pi$ as in the stated ideal model.
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