Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 39A (Waves)

Substitution of the monochromatic ansatz givesThus the outer solutions oscillate with $m=k\sqrt{N_0^2/{\omega }^2-1}$ and the middle solutions are combinations of $\cosh kz$ and $\sinh kz$. The incident wave has phase $kx-mz-\omega t$; its group velocity is upward, whereas the reflected sign has downward group velocity. The transmitted wave must again use the upward-energy sign. This gives exactly the three forms stated.

Continuity of normal velocity and pressure at each density interface requires continuity of $\hat{w}$ and $d\hat{w}/dz$ at $z=\pm L$. These four conditions determine the four unknown amplitudes.

Matching at $z=L$ first expresses the middle solution in terms of $A_T$. Propagating its value and derivative across the layer of thickness $2L$ uses the transfer matrixMatching this result to $A_I+A_R$ and $-im{A}_I+im{A}_R$ at $z=-L$, then eliminating $A_R$, gives

Taking the squared modulus and using ${\cosh }^{2}u=1+{\sinh }^{2}u$ givesThe incident wavevector makes an angle $\psi$ with the downward vertical, so $\sin 2\psi =2km/(k^2+m^2)$. Substitution gives the required formula.