= Solution
<Birefringence> means that two independent polarization eigenmodes have different <refractive indices> at the same <frequency> and propagation direction. The rotatory-power formula relevant here requires <circular birefringence>: the eigenmodes are opposite <circular polarizations>. General linear <birefringence> instead introduces <phase retardation> between linearly polarized components and need not rotate every incident <linear polarization> through one common angle.
Use the <complex amplitude> convention $\operatorname{Re}[\mathbf E e^{-i\omega t}]$ and transverse <unit vectors> $\mathbf e_1,\mathbf e_2$. Define
$$
\mathbf u_\pm=\frac{\mathbf e_1\mp i\mathbf e_2}{\sqrt2},\qquad
\mathbf e_1=\frac{\mathbf u_++\mathbf u_-}{\sqrt2}.
$$
Propagation through thickness $d$ adds <phases> $\phi_\pm=(2\pi/\lambda_0)n_\pm d$, where $\lambda_0$ is the <vacuum> <wavelength>. Recombining the two components gives
$$
\frac{e^{i\phi_+}\mathbf u_++e^{i\phi_-}\mathbf u_-}{\sqrt2}
=e^{i\bar\phi}(\mathbf e_1\cos\theta+\mathbf e_2\sin\theta),\qquad
\theta=\frac{\phi_+-\phi_-}{2}.
$$
Thus the rotation and the specific <rotatory power> per unit length are
$$
\boxed{\theta=\frac{\pi d}{\lambda_0}(n_+-n_-),\qquad
\rho\equiv\frac\theta d=\frac\pi{\lambda_0}(n_+-n_-)=\frac\omega{2c}(n_+-n_-).}
$$
The units of $\rho$ are $\mathrm{rad\,m^{-1}}$. The sign follows the specified circular basis and viewing convention. Equal attenuation of the two eigenmodes has been assumed; unequal attenuation also produces <elliptical polarization>.
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