 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/40c/c/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Eliminating

T

from the two boundary conditions gives

\frac{1 + R}{1 - R} = \frac{λ ^{'} k ^{'} c o s θ}{λk c o s θ ^{'}} .

\frac{k ^{'}}{k} = \frac{s i n θ}{s i n θ ^{'}} .

R = \frac{λ ^{'} sin 2 θ - λ sin 2 θ ^{'}}{λ ^{'} sin 2 θ + λ sin 2 θ ^{'}} .

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