= Solution
At distance $x$ from its tip, the wedge thickness is $t_q=x\tan\theta_q$. Its <optical path difference> is therefore
$$
\Gamma_q=\Delta n_q x\tan\theta_q
=0.009(5\times10^{-3})\tan(0.5^\circ)
=3.927\times10^{-7}\,\mathrm m.
$$
The perpendicular slow directions make the two retardances subtract. A white-light black compensation fringe is the zero-order cancellation, so \b[the film retardance has magnitude approximately $393\,\mathrm{nm}$], with its slow direction along the rods. This interpretation assumes the observed fringe is the white-light compensation fringe; a single monochromatic dark fringe alone would have an integer-order ambiguity. Without the wedge, the <wavelength> dependence of $\sin^2(\pi\Gamma/\lambda)$ selectively transmits visible wavelengths and gives the film its <interference colour>.
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