= Solution
In <thin-film interference>, coherent beams reflected at the two faces of a transparent film overlap. Let the film have <refractive index> $n_f$, thickness $t$, and internal ray angle $\theta_f$, determined by <Snell's law>. For vacuum <wavelength> $\lambda_0$, the propagation part of the relative <phase> is
$$
\boxed{\delta_{\rm prop}=\frac{4\pi n_ft\cos\theta_f}{\lambda_0}}.
$$
Although the internal round-trip ray length is $2t/\cos\theta_f$, the two emergent rays must be compared on the same outgoing <wavefront>. Subtracting their external lateral path contribution gives the effective <optical path difference> $2n_ft\cos\theta_f$, explaining the cosine rather than its reciprocal.
Reflection also changes <phase>. At normal incidence in lossless positive-index media, the electric-field <Fresnel equations> give $r_{ij}=(n_i-n_j)/(n_i+n_j)$: reflection towards higher index reverses sign, whereas reflection towards lower index does not. If the two interfering reflections differ by one such <phase> reversal, their total relative <phase> is $\delta_{\rm prop}+\pi$ modulo $2\pi$. Then reflected maxima and minima obey
$$
\boxed{2n_ft\cos\theta_f=(m+\tfrac12)\lambda_0\quad\text{maxima}},\qquad
\boxed{2n_ft\cos\theta_f=m\lambda_0\quad\text{minima}}.
$$
If there are zero or two reversals, the conditions interchange. These bright/dark conditions assume overlapping beams with adequate <temporal optical coherence>; complete cancellation further requires equal amplitudes. In transmitted light, a lossless complete film has complementary intensity variations because reflected plus transmitted <power> is conserved.
For multiple internal reflections at normal incidence, summing the geometric series gives the exact film reflection amplitude
$$
r_{\rm film}=\frac{r_{01}+r_{12}e^{2i\beta}}{1+r_{01}r_{12}e^{2i\beta}},\qquad
\beta=\frac{2\pi n_ft}{\lambda_0},\qquad \mathcal R=|r_{\rm film}|^2.
$$
This includes both the amplitude mismatch and repeated-beam interference. A normal-incidence antireflection coating on substrate index $n_s>n_0$ can cancel reflection at one wavelength when $n_f=\sqrt{n_0n_s}$ and $t=\lambda_0/(4n_f)$: both interface reflections reverse sign and the film's round-trip propagation adds $\pi$, with equal interface amplitudes. A varying thickness or angle selects different maxima for different wavelengths, producing the colours of transparent films in white light. Broad bandwidth and large incidence angles generally prevent perfect cancellation across all wavelengths.
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