 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/15d/b/ii/solution

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

A e^{i (k x - Et /ℏ)}

J = \frac{ℏ k}{m} ∣ A ∣^{2} .

The incident amplitude is one and the transmitted amplitude is

T

, so

\frac{J _{tr}}{J _{in}} = ∣ T ∣^{2} = sech^{2} (2 ka) .

\partial ρ / \partial t = 0

, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

\frac{∣ J _{refl} ∣}{J _{in}} = 1 - sech^{2} (2 ka) = tanh^{2} (2 ka) .

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