= Solution
The strong plume's outgoing flux equals the total lower-layer supply, $Q_2(h_2)=Q$. Apply the additional closure in this part and the lower-layer balance from part (b):
$$
C(B_1+B_2)^{1/3}h_2^{5/3}
=C(B_1^{1/3}+B_2^{1/3})h_1^{5/3}.
$$
Dividing by $CB_2^{1/3}h_1^{5/3}$ gives
$$
(1+\psi)^{1/3}\left(\frac{h_2}{h_1}\right)^{5/3}=1+\psi^{1/3},
$$
so
$$
\boxed{\frac{h_2}{h_1}=\left[\frac{1+\psi^{1/3}}{(1+\psi)^{1/3}}\right]^{3/5}.}
$$
For $0<\psi<1$ this puts the second matching level above the first. It tends to one as $\psi\to0$ and to $2^{2/5}$ as $\psi\to1$. At the latter endpoint the two warm layers already have identical density, so $h_2$ is only a formal matching label, not a distinct observable interface. The expression follows from the stipulated closure; modeling the actual plume in the changed ambient requires the matched continuation discussed in part (d).
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