 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/39a/c/solution

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

Matching at

z = L

first expresses the middle solution in terms of

A_{T}

. Propagating its value and derivative across the layer of thickness

2 L

(cosh 2 α k sinh 2 α k^{- 1} sinh 2 α cosh 2 α), α = k L .

Matching this result to

A_{I} + A_{R}

and

- im A_{I} + im A_{R}

z = - L

, then eliminating

A_{R}

, gives

\frac{A _{T}}{A _{I}} = \frac{2 imk}{2 imk cosh 2 α + ( k ^{2} - m ^{2} ) sinh 2 α} .

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