 Solution

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

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

Taking the squared modulus and using

cosh^{2} u = 1 + sinh^{2} u

gives

\frac{A _{T}}{A _{I}}^{2} = [1 + \frac{( k ^{2} + m ^{2} ) ^{2}}{4 k ^{2} m ^{2}} sinh^{2} 2 α]^{- 1} .

The incident wavevector makes an angle

ψ

with the downward vertical, so

sin 2 ψ = 2 km / (k^{2} + m^{2})

. Substitution gives the required formula.

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