= Solution
The normal to the section is $[010]$, so its in-plane principal vibration directions are $[100]$ and $[001]$. Their refractive-index difference is
$$
\boxed{\Delta n=1.45-1.32=0.13.}
$$
The slow direction is $[001]$. The optical path retardation is
$$
\Gamma=t\Delta n=(4\ {\rm \mu m})(0.13)=0.52\ {\rm \mu m}.
$$
At an angle $\alpha$ between a principal direction and the <polarizer>, ideal crossed-polar transmission has the form
$$
I(\lambda)=I_P\sin^2(2\alpha)
\sin^2\left(\frac{\pi\Gamma}{\lambda}\right),
$$
where $I_P$ is the intensity after the first <polarizer>. The two components recombine parallel to the incident <polarization> when $\Gamma=m\lambda$, and the crossed analyser removes them. Consequently
$$
\boxed{\lambda=\Gamma/m,\qquad
\lambda_{\rm visible}=520\ {\rm nm}\quad(m=1).}
$$
The next order is260nm, outside ordinary visible illumination. At exact principal-axis alignment, $\sin2\alpha=0$ and every wavelength is extinguished, explaining the black appearance rather than a special single-wavelength cancellation.
\Image[/past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2015/paper-1-crossed-polars.png]
{title=Original crossed-polar transmission calculation for the given retardation, including the visible extinction wavelength and azimuthal brightness factor}
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