 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/36b/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

x y

-plane, so all three electric polarizations point along

z

. Write their signed scalar amplitudes as

E_{I}, E_{R}, E_{T}

. Tangential-

E

continuity gives

E_{I} + E_{R} = E_{T} .

H = \frac{1}{μ ω} k \times E .

Continuity of its tangential component gives

n_{-} cos θ_{I} (E_{I} - E_{R}) = n_{+} cos θ_{T} E_{T},

where the common proportionality between

∣ k ∣

and

n

\frac{E _{R}}{E _{I}} = \frac{n _{-} c o s θ _{I} - n _{+} c o s θ _{T}}{n _{-} c o s θ _{I} + n _{+} c o s θ _{T}} .

Using Snell's law to eliminate

θ_{T}

n_{+} cos θ_{T} = n_{+}^{2} - n_{-}^{2} sin^{2} θ_{I} .

Since

ϵ_{+} > ϵ_{-}

and

μ

is common,

n_{+} > n_{-}

. Therefore

(n_{+} cos θ_{T})^{2} - (n_{-} cos θ_{I})^{2} = n_{+}^{2} - n_{-}^{2} > 0.

The numerator is consequently strictly negative throughout

0 \leq θ_{I} \leq π /2

, and

\frac{∣ E _{R} ∣}{∣ E _{I} ∣} = \frac{n _{+}^{2} - n _{-}^{2} sin ^{2} θ _{I} - n _{-} cos θ _{I}}{n _{+}^{2} - n _{-}^{2} sin ^{2} θ _{I} + n _{-} cos θ _{I}} > 0 .

Thus transverse-electric polarization has no Brewster angle in this setting.

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