= Solution
\b[Spherical-particle <frequency> scaling.] For a <dielectric> sphere of radius $a\ll\lambda$, the quasistatic <electric polarizability> is
$$
\alpha_s=4\pi\epsilon_0a^3\frac{\epsilon_r-1}{\epsilon_r+2},\qquad\mathbf p(t)=\operatorname{Re}(\alpha_s\mathbf E_0e^{-i\omega t}).
$$
The <electric dipole radiation> formula and $I_0=\epsilon_0c|E_0|^2/2$ give
$$
\langle P\rangle=\frac{\omega^4|\alpha_s|^2|E_0|^2}{12\pi\epsilon_0c^3},\qquad\boxed{\sigma=\frac{\langle P\rangle}{I_0}=\frac{8\pi}{3}k^4a^6\left|\frac{\epsilon_r-1}{\epsilon_r+2}\right|^2.}
$$
Thus \b[<Rayleigh scattering> scales as $\omega^4$, or $\lambda^{-4}$], when the polarizability varies slowly with <frequency>. Material resonances and absorption can change this simple scaling. For the incident $z$ <polarization>, the dipole is along $z$: intensity is proportional to $\sin^2\chi$, where $\chi$ is the angle from $z$. There is no radiation on the dipole axis and a maximum in the $xy$ plane. In a direction $\hat{\mathbf n}$ the scattered <electric field> is along $\hat{\mathbf z}-(\hat{\mathbf z}\cdot\hat{\mathbf n})\hat{\mathbf n}$, hence linearly polarized wherever nonzero.<Rayleigh Scattering by Richard Fitzpatrick>[Small-particle dipole response and inverse-fourth-power wavelength dependence]{ref}
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2016/paper-1-rayleigh-pattern.png]
{title=A z-directed electric dipole radiates most strongly perpendicular to its axis and has axial zeros}
{source=paper-1-rayleigh-pattern.py}
\b[Needle model used in parts (a)–(d).] Let its axis be $\mathbf u=(u_x,u_y,u_z)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$. A very slender, high-permittivity needle has dominant longitudinal response, so use the <rank-one needle polarizability> $\boldsymbol\alpha\simeq\alpha_\parallel\mathbf u\mathbf u^{\mathsf T}$. Its transverse response is neglected at leading order. For observation along $y$, define $\mathbf v=u_x\hat{\mathbf x}+u_z\hat{\mathbf z}$, $s=|\mathbf v|^2=1-u_y^2$ and
$$
K=\frac{k^4|\alpha_\parallel|^2}{16\pi^2\epsilon_0^2R^2}.
$$
For a unit incident <Jones vector> $\mathbf e$, the common far-field relation is $\mathbf E_s\propto\mathbf v(\mathbf u\cdot\mathbf e)$, and $I_s=KI_0s|\mathbf u\cdot\mathbf e|^2$. Here $I_s$ is irradiance at distance $R$; multiplying by $R^2$ gives power per solid angle. No absolute value of $\alpha_\parallel$ is given or needed for the requested relative intensities.
\b[Diffuse random cloud, with the incident <polarization> of (a).] The single-particle amplitudes through $x$ and $z$ analysers are proportional to $u_xu_z$ and $u_z^2$. The <isotropic orientation fourth moment> gives the <needle-scattering analyzer moments>
$$
\langle u_x^2u_z^2\rangle=\frac1{15},\qquad\langle u_z^4\rangle=\frac15.
$$
For example, $\langle u_z^4\rangle=\tfrac12\int_{-1}^{1}\mu^4d\mu=1/5$; symmetry and $u_x^2+u_y^2+u_z^2=1$ give $2\langle u_x^2u_z^2\rangle=1/3-1/5$. Since the particles scatter mutually incoherently, intensities add. For $N$ illuminated particles,
$$
\boxed{I_x=\frac{NKI_0}{15},\qquad I_z=\frac{NKI_0}{5},\qquad I_x:I_z=1:3.}
$$
The cross correlation vanishes by reflection symmetry. The degree of <linear polarization> is $(I_z-I_x)/(I_z+I_x)=\boxed{1/2}$: random anisotropic scatterers produce partially depolarized light, even though every individual needle produces a single <linear polarization>.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2016/paper-1-needle-ensemble.png]
{title=Independent isotropically oriented needles give the analyser ratio one to three, while coherent mean amplitudes have no x component}
{source=paper-1-needle-ensemble.py}
\b[Continuous homogeneous <dielectric>.] One must sum amplitudes coherently and include their spatial phases, rather than retain the preceding sum of intensities. Random orientations give $\langle\mathbf u\mathbf u^{\mathsf T}\rangle=\mathbf I/3$, so the averaged response is isotropic and its induced <polarization> is along $z$. Its coherent radiation toward $y$, when allowed by the sample geometry, is therefore \b[purely $z$ polarized, with $I_x=0$]. The independent-needle prefactor is no longer a quantitative dense-medium model; local-field and collective effects determine the effective <permittivity>.
There is also <coherent continuum phase matching>: a uniform bulk driven by a <plane wave> has no arbitrary outgoing $90^\circ$ plane-wave component. In an ideal infinite homogeneous medium there is no diffuse side scattering at all; propagation, <refraction> and reflection are determined by the medium and its interfaces. A finite sample's coherent side signal depends on its shape and boundaries. Thus the old $1:3$ diffuse ratio is replaced by a coherent isotropic-medium response, not by a universal nonzero $N^2$ side-scattering intensity.
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