= Solution
The outward normal from the semicircular barrier is $\mathbf e_r=(-\cos\phi,\sin\phi)$. The leading radial flux in the thick region is
$$
q_s\simeq h^3(-\cos\phi-h_s).
$$
For
$$
h=H(\phi)\left(1-\frac s{\Delta(\phi)}\right),
$$
one has $h_s=-H/\Delta$. Negligible radial flux therefore requires
$$
\boxed{H=\Delta\cos\phi}.
$$
This expresses local hydrostatic blocking: the free-surface gradient opposes the downslope gravitational flux. Streamlines arriving from upslope divide at $\phi=0$ and run around the two sides of the barrier toward $\phi=\pm\pi/2$.
At leading order the azimuthal flux density is $q_\phi\simeq h^3\sin\phi$. Its integral across the thick region is
$$
Q_\phi=\int_0^\Delta q_\phi\,ds
\simeq\frac14H^3\Delta\sin\phi.
$$
This must equal the upstream unit flux intercepted between the symmetry axis and that polar angle, namely $R\sin\phi$. Hence
$$
H^3\Delta=4R.
$$
Combining this with $H=\Delta\cos\phi$ gives
$$
\boxed{H=(4R\cos\phi)^{1/4}},
\qquad
\boxed{\Delta=\frac{(4R\cos\phi)^{1/4}}{\cos\phi}}.
$$
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