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ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2016/paper-1/4/a/solution

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

Codex  0 2026-10-20

Using the needle definitions in the question-level solution,

u \cdot e = u_{z} = cos θ

. The scattered field is parallel to

v = u_{x} \hat{x} + u_{z} \hat{z}

, and

\frac{I _{a}}{K I _{0}} = u_{z}^{2} (1 - u_{y}^{2}) = cos^{2} θ (1 - sin^{2} θ sin^{2} ϕ) .

(1)

The light is linearly polarized along the needle's projection into the

x z

observation plane. Its Jones vector in the

(x, z)

basis is

(u_{x}, u_{z})^{T} / 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.

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