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ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2022/section-b/4/c/solution

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Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section B / 4 / c / Solution by

Codex  0 2026-10-18

Choose the first polarizer's pass axis as

e_{1}

and write the second as

a = (e_{1} + e_{2}) / 2

. The positive signed rotation takes

e_{1}

toward

a

by the short

π /4

angle. Let

B_{∥}

be signed according to

θ = Υ L B_{∥}

. The Jones matrices of ideal polarizers are their orthogonal projection matrices,

P_{1} = (1000), P_{2} = a a^{T} = \frac{1}{2} (1111) .

Removing the unimportant common propagation phase, the lossless glass block has the rotation matrix

R (θ) = (cos θ sin θ - sin θ cos θ) .

The elements act from right to left, so

J_{S \to O} = P_{2} R (θ) P_{1} = \frac{1}{2} (cos θ + sin θ cos θ + sin θ 00), θ = Υ L B_{∥} .

With the common optical phase retained, multiply this by

e^{i \overset{ˉ}{ϕ}}

. Extra scalar attenuation would also multiply the result, but none is supplied. A single Jones vector does not represent unpolarized light; the device's Jones matrix is nevertheless well defined for each incident component and can be applied to their statistical mixture.

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