= Solution
Reflections from the two film surfaces interfere with propagation phase $4\pi nd\cos\theta_t/\lambda_0$ plus the relative reflection phase. Reflection from a lower to a higher <refractive index> contributes a $\pi$ phase reversal. For a higher-index film surrounded by air there is one relative reversal, hence
$$
\boxed{\text{reflection maxima: }2nd\cos\theta_t=(m+\tfrac12)\lambda_0,\quad
\text{minima: }2nd\cos\theta_t=m\lambda_0}.
$$
Thickness or angle variation therefore produces fringes or colours under broadband illumination. An <antireflection coating> between indices $n_0,n_s$ instead uses, at <normal incidence>, $n_f=\sqrt{n_0n_s}$ and $n_fd=\lambda_0/4$: the two reflected <amplitudes> then cancel at the design <wavelength>. Absorption, <polarization>, unequal <amplitudes> and multiple reflections determine the full contrast.
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