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

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.