 Solution

At distance $x$ from its tip, the wedge thickness is $t_q=x\tan {\theta }_q$. Its optical path difference is thereforeThe perpendicular slow directions make the two retardances subtract. A white-light black compensation fringe is the zero-order cancellation, so 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.