= Solution
The angles from part (i) have sines
$$
\sin\theta_P=\frac{\alpha R}{\sqrt{s^2+\alpha^2R^2}},\qquad
\sin\theta_z=\frac{\alpha_fR}{\sqrt{s^2+\alpha^2R^2}}.
$$
By <Snell law for elastic and acoustic waves>, $\sin\theta_f/\alpha_f=\sin\theta_P/\alpha$. Both angles lie in $[0,\pi/2)$, so \b[$\theta_f=\theta_z$]. The outgoing fluid ray tilts away from the inward-tilted normal by exactly its tilt from vertical; it therefore travels vertically upwards. This is <hyperbolic interface collimation of P-waves>.
The source-to-interface distance satisfies
$$
r(R)=\sqrt{R^2+Z(R)^2}=\frac{\alpha D+\alpha_fs}{d^2}.
$$
One can verify this by squaring, using $D^2-s^2=d^2R^2$. Consequently
$$
\frac r\alpha-\frac Z{\alpha_f}=-\frac{s}{\alpha\alpha_f}
=\frac{z_0}{\alpha}-\frac{z_0}{\alpha_f}.
$$
Adding the vertical fluid travel time from the interface to height $z$ gives
$$
\boxed{\tau(R,z)=\frac{r(R)}\alpha+\frac{z-Z(R)}{\alpha_f}
=\frac{z_0}\alpha+\frac{z-z_0}{\alpha_f}.}
$$
Thus surfaces of equal arrival time in the fluid are horizontal planes. Their leading <wave amplitude> may vary with $R$: a planar <wavefront> does not imply a spatially uniform plane-wave solution at every order.
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