Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section B / B9 / Propagation and reflection phases

Let the first ray reflect at the upper surface and the second traverse the film, reflect below, and emerge at a laterally displaced point. Its internal round-trip geometric length is $2d/\cos \theta$, but the external path to a common outgoing wavefront must also be included. The lateral separation is $2d\tan \theta$, whose projection along the external ray is $2d\tan \theta \sin {\theta }_i$. The net optical path difference is thereforeusing Snell's law $\sin {\theta }_i=n\sin \theta$. Multiplying by the vacuum wavenumber givesFor the usual index ordering $1<n<n_2$, including the stated later value $n=1.2$, both reflections are towards higher refractive index and both reverse the electric-field sign. Their relative reflection phase is therefore zero; $\delta$ is also their total relative phase modulo $2\pi$. A different index ordering or polarization-dependent reflection phase would require that additional term.