= Solution
Compare a directly transmitted ray emerging at $A$ with a ray that makes one extra round trip inside the <Fabry–Pérot interferometer> and emerges at $C$. Both emerge parallel at angle $\theta$ to the plate normal. The extra internal distance is $2d/\cos\theta$, and their emergence points differ laterally by $AC=2d\tan\theta$. The first ray has already travelled $AC\sin\theta$ towards a common outgoing <wavefront> when the second reaches $C$. The <air-gap optical path difference in a Fabry–Pérot interferometer> follows because <optical path length> equals geometric path length in the air-gap approximation:
$$
\Delta=\frac{2d}{\cos\theta}-2d\tan\theta\sin\theta
=\frac{2d(1-\sin^2\theta)}{\cos\theta}
=\boxed{2d\cos\theta}.
$$
The subtraction is essential: comparing distances only inside the plates would compare different outgoing <wavefronts>. The corresponding propagation <phase difference> is $2\pi\Delta/\lambda$. Coating reflection phases can add a separate phase offset; they do not change this geometric <optical path difference>.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-2-fabry-perot.png]
{title=One extra round trip and the common outgoing wavefront in an air-spaced Fabry–Pérot interferometer}
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