Past exam of the optics and electrodynamics course of the University of Cambridge / 2016 / Paper 1 / 4

Spherical-particle frequency scaling. For a dielectric sphere of radius $a\ll \lambda$, the quasistatic electric polarizability isThe electric dipole radiation formula and $I_0={ϵ}_{0}c|E_0{|}^2/2$ giveThus 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.^[ref]

Needle model used in parts (a)–(d). Let its axis be $\mathbf{u}=(u_x,u_y,u_z)=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. A very slender, high-permittivity needle has dominant longitudinal response, so use the rank-one needle polarizability $\mathbf{\alpha }\simeq {\alpha }_{\parallel }\mathbf{u}{\mathbf{u}}^{\mathsf{T}}$. Its transverse response is neglected at leading order. For observation along $y$, define $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, $s=|\mathbf{v}{|}^2=1-u_y^2$ andFor 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 $I_s=K{I}_{0}s|\mathbf{u}\cdot \mathbf{e}{|}^2$. 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.

Diffuse random cloud, with the incident polarization of (a). The single-particle amplitudes through $x$ and $z$ analysers are proportional to $u_x{u}_z$ and $u_z^2$. The isotropic orientation fourth moment gives the needle-scattering analyzer momentsFor example, $⟨u_z^4⟩=\frac{1}{2}{\int }_{-1}^1{\mu }^{4}d\mu =1/5$; symmetry and $u_x^2+u_y^2+u_z^2=1$ give $2⟨u_x^2{u}_z^2⟩=1/3-1/5$. Since the particles scatter mutually incoherently, intensities add. For $N$ illuminated particles,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.

Continuous homogeneous dielectric. One must sum amplitudes coherently and include their spatial phases, rather than retain the preceding sum of intensities. Random orientations give $⟨\mathbf{u}{\mathbf{u}}^{\mathsf{T}}⟩=\mathbf{I}/3$, 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 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.

Using the needle definitions in the question-level solution, $\mathbf{u}\cdot \mathbf{e}=u_z=\cos \theta$. The scattered field is parallel to $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, andThe light is linearly polarized along the needle's projection into the $xz$ observation plane. Its Jones vector in the $(x,z)$ basis is $(u_x,u_z{)}^{\mathsf{T}}/\sqrt{1-u_y^2}$ whenever the intensity is nonzero. A needle perpendicular to the incident electric field, or pointing directly at the observer, has zero leading-order signal.

Now $\mathbf{u}\cdot \mathbf{e}=u_y=\sin \theta \sin \varphi$, soWhenever nonzero, the scattered light has the same linear polarization along $\mathbf{v}$ as in (a). Although the incident field points toward the observer, the tilted anisotropic needle can induce a dipole component perpendicular to that direction. An isotropic sphere would instead have an axial radiation zero for this incident polarization.

Choose one circular convention as ${\mathbf{e}}_R=(\hat{\mathbf{z}}+i\hat{\mathbf{y}})/\sqrt{2}$ with time dependence $e^{-i\omega t}$. Reversing the handedness convention changes the sign of $i$ but not the result here. The scattered complex amplitude isThe scalar factor contributes a common phase to both output components. Consequently the scattered wave is linearly polarized along $\mathbf{v}$, rather than circular or elliptical, withThe dominant longitudinal needle response has rank one, so both incident circular components emerge in the same real output direction.

Unpolarized incident light has equal, mutually incoherent $y$- and $z$-polarized intensities. Both components scatter into the same direction $\mathbf{v}$, so their coherency matrices add toThis matrix has rank one when the signal is nonzero. The light is therefore fully linearly polarized along $\mathbf{v}$, withA fixed highly anisotropic needle can act as a polarizing scatterer even for unpolarized input. The equality of total intensities with (c) does not mean the incident states were identical. These conclusions use the leading longitudinal-polarizability approximation; finite transverse response can change the polarization purity.