= Solution
The normal-incidence <acoustic impedance> is $Z=\rho V_p$, with units $\mathrm{kg\,m^{-2}\,s^{-1}}$. Thus
$$
\boxed{Z_1=1.50\times10^6,\qquad Z_2=2.25\times10^6,\qquad Z_3=5.00\times10^6.}
$$
Define the <pressure-amplitude reflection coefficient> and <pressure-amplitude transmission coefficient> for incidence downwards from layer $i$ into layer $j$ by
$$
r_p=\frac{p_r}{p_i}=\frac{Z_j-Z_i}{Z_j+Z_i},\qquad t_p=\frac{p_t}{p_i}=\frac{2Z_j}{Z_j+Z_i}=1+r_p.
$$
Here $p$ denotes compressional normal-stress amplitude; in the <water> it is the acoustic <pressure>. Continuity of normal <traction> and normal particle <velocity> gives these expressions. They describe each interface separately, before any layer <layer reverberation>.
|| Interface, downward incidence
|| $r_p$
|| $t_p$
|| Reflected <energy> fraction $R_E$
|| Transmitted <energy> fraction $T_E$
| Water to sediment
| $1/5=0.2000$
| $6/5=1.2000$
| $1/25=0.0400$
| $24/25=0.9600$
| Sediment to basement
| $11/29=0.37931$
| $40/29=1.37931$
| $121/841=0.14388$
| $720/841=0.85612$
For a lossless normal-incidence <plane wave>, intensity is proportional to $p^2/Z$, so
$$
R_E=r_p^2,\qquad T_E=\frac{Z_i}{Z_j}t_p^2=\frac{4Z_iZ_j}{(Z_i+Z_j)^2},\qquad R_E+T_E=1.
$$
A transmitted pressure amplitude exceeding unity is therefore compatible with <conservation of energy>. If instead the coefficient is defined for particle <velocity> measured along the same downward coordinate, $r_v=-r_p$ and $t_v=2Z_i/(Z_i+Z_j)=1-r_p$: the two transmitted velocity coefficients are $0.8000$ and $18/29=0.62069$. Upward incidence reverses the sign of $r_p$, interchanges the transmitted pressure and velocity coefficients, and leaves $R_E,T_E$ unchanged. At normal incidence between these ideal <isotropic> layers there is no P-to-S <seismic mode conversion>.
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