Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section B / B9 / Thickness from angular fringe spacing

Eliminate the internal angle using Snell's law:Neighbouring bright fringes differ in phase by $2\pi$. The local angular fringe spacing of a transparent film therefore givesWith $\Delta {\theta }_i=0.4\pi /180=6.981\times 10^{-3}\mathrm{rad}$, ${\theta }_i\simeq 10^{\circ }$, $n=1.2$ and $\lambda =500\mathrm{nm}$,The data specify a local spacing near $10^{\circ }$, so differentiating is appropriate; choosing particular neighbouring fringe angles slightly shifts this approximate result. The phase decreases as incidence angle increases.