Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 79 / 2 / ii / Solution

The angles from part (i) have sinesBy Snell law for elastic and acoustic waves, $\sin {\theta }_f/{\alpha }_f=\sin {\theta }_P/\alpha$. Both angles lie in $[0,\pi /2)$, so ${\theta }_f={\theta }_z$. 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 satisfiesOne can verify this by squaring, using $D^2-s^2=d^2{R}^2$. ConsequentlyAdding the vertical fluid travel time from the interface to height $z$ givesThus 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.