 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/40c/b/solution

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

Let

H = k β_{0} h

β_{0} = 1 - V^{2} / c_{0}^{2}

, and

β_{1} = V^{2} / c_{1}^{2} - 1

. Match pressure and normal velocity between an evanescent lower potential and an outgoing upper wave. The transmitted amplitude has

∣ A_{1} ∣^{2} = \frac{V ^{2} ϵ ^{2}}{β _{1}^{2} c o s h ^{2} H + ( ρ _{1} / ρ _{0} ) ^{2} β _{0}^{2} s i n h ^{2} H} .

Therefore

⟨ I_{y} ⟩ = \frac{ρ _{1} k ^{2} V ^{3} ϵ ^{2} β _{1}}{2 [ β _{1}^{2} c o s h ^{2} H + ( ρ _{1} / ρ _{0} ) ^{2} β _{0}^{2} s i n h ^{2} H ]} .

For

H ≫ 1

this is

\frac{2 ρ _{1} k ^{2} V ^{3} ϵ ^{2} β _{1} e ^{- 2 H}}{β _{1}^{2} + ( ρ _{1} / ρ _{0} ) ^{2} β _{0}^{2}},

showing tunnelling suppression. For

H ≪ 1

it tends to

ρ_{1} k^{2} V^{3} ϵ^{2} / (2 β_{1})

, as if the upper medium met the wall.

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