= Solution
From the second-edge <amplitude> in part (c),
$$
\boxed{\frac FA=i\frac{k_0}{k_1}=i\sqrt{\frac{k_0}{k_2}}.}
$$
This uses $\psi_{\mathrm R}=Fe^{ik_2(x-a)}$. If the transmitted wave is instead written $Te^{ik_2x}$, its coefficient is $\boxed{T/A=i\sqrt{k_0/k_2}\,e^{-ik_2a}}$; the extra factor only changes the <quantum phase> convention.
Using the <probability current> of a <plane wave>,
$$
J_{\mathrm{in}}=\frac{\hbar k_0}{m}|A|^2,\qquad J_{\mathrm{trans}}=\frac{\hbar k_2}{m}|F|^2=\frac{\hbar k_2}{m}\frac{k_0}{k_2}|A|^2=\boxed{J_{\mathrm{in}}.}
$$
Thus the flux <transmission probability> is one. The larger transmitted density $|F|^2>|A|^2$ compensates for the smaller transmitted <velocity> $\hbar k_2/m$; equal currents need not have equal wave amplitudes.
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