= Thickness from angular fringe spacing
Eliminate the internal angle using <Snell's law>:
$$
\delta(\theta_i)=\frac{4\pi d}{\lambda}\sqrt{n^2-\sin^2\theta_i},\qquad
\frac{d\delta}{d\theta_i}=-\frac{4\pi d}{\lambda}\frac{\sin\theta_i\cos\theta_i}{\sqrt{n^2-\sin^2\theta_i}}.
$$
Neighbouring bright fringes differ in <phase> by $2\pi$. The local <angular fringe spacing of a transparent film> therefore gives
$$
d\simeq\frac{\lambda\sqrt{n^2-\sin^2\theta_i}}{2\sin\theta_i\cos\theta_i\,\Delta\theta_i}.
$$
With $\Delta\theta_i=0.4\pi/180=6.981\times10^{-3}\,\mathrm{rad}$, $\theta_i\simeq10^\circ$, $n=1.2$ and $\lambda=500\,\mathrm{nm}$,
$$
\boxed{d\simeq2.49\times10^{-4}\,\mathrm m\simeq0.25\,\mathrm{mm}}.
$$
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.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2-b9-film-fringes.png]
{title=Two-beam reflected intensity versus phase and angular fringes for the inferred film thickness}
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