= Solution
Now take $\widehat{\mathbf n}=(0,\sin\psi,\cos\psi)$, with the same angle to the incident $z$ direction. The perpendicular direction is $\widehat{\mathbf x}$ and the in-plane transverse direction is $(0,\cos\psi,-\sin\psi)$. The roles of the two <electric dipole> populations interchange: $x$ <electric dipoles> radiate at full strength, whereas the observed <wave amplitude> from a $y$ <electric dipole> is reduced by $\cos\psi$.
Thus $I_\perp=CN_1$ and $I_\parallel=CN_2\cos^2\psi$, giving
$$
\boxed{\mathcal P_{yz}(\psi)=\frac{|N_1-N_2\cos^2\psi|}{N_1+N_2\cos^2\psi}.}
$$
This is also obtained by the symmetry $x\leftrightarrow y$, $N_1\leftrightarrow N_2$ in part (a). For $N_1=N_2$, both planes give the familiar <Rayleigh scattering> <degree of polarization> $\sin^2\psi/(1+\cos^2\psi)$: the diffuse signal is unpolarized forward and fully linearly polarized at right angles. At a dark direction with zero denominator, no <degree of polarization> can be measured.
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