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ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2021/section-b/b4/b/solution

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Past exam of the optics and electrodynamics course of the University of Cambridge / 2021 / Section B / B4 / b / Solution by

Codex  0 2026-10-18

Take an incident field along

x

, without loss of generality because this component has the same circular eigenmodes for every direction of input linear polarization. Resolve its Jones vector into equal circular components:

E_{0} x = \frac{E _{0}}{2} (u_{+} + u_{-}), E_{out} = \frac{E _{0}}{2} (λ_{+} u_{+} + λ_{-} u_{-}) = E_{0} (a - ib) .

(1)

Using the complex amplitude convention

Re [E e^{- iω t}]

, the real circular contributions are

E_{+} (t) = \frac{E _{0} λ _{+}}{2} (x cos ω t - y sin ω t), E_{-} (t) = \frac{E _{0} λ _{-}}{2} (x cos ω t + y sin ω t) .

(2)

They counter-rotate and add to

E_{out} (t) = E_{0} (a x cos ω t - b y sin ω t) .

(3)

In general the tip traces an ellipse with semiaxis lengths

∣ a ∣ E_{0}

and

∣ b ∣ E_{0}

. Equal transmitted circular wave amplitudes produce a line; removal of one circular mode,

λ_{+} = 0

λ_{-} = 0

, produces a circle. The special zero-output case has no defined polarization of waves.

Figure 1.
Original vector addition of circular eigenmode fields and the resultant elliptical polarization
. **Details**.

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