= Solution
At normal incidence between lossless media with real positive <acoustic impedances> $Z_1,Z_2$, continuity of pressure and normal particle <velocity> gives the pressure transmission amplitude $t_p=2Z_2/(Z_1+Z_2)$. Since the mean <acoustic intensity> is $|p|^2/(2Z)$, the fraction of incident <energy> transmitted is
$$
\mathcal T=\frac{Z_1}{Z_2}|t_p|^2=\frac{4Z_1Z_2}{(Z_1+Z_2)^2},\qquad
\mathcal R=\left(\frac{Z_2-Z_1}{Z_2+Z_1}\right)^2=1-\mathcal T.
$$
This is <acoustic reflection and transmission at an interface>. With the supplied <acoustic impedances>,
$$
\boxed{\mathcal T_{\rm air\to body}=9.81\times10^{-4}=0.0981\%},\qquad
\boxed{\mathcal T_{\rm water\to body}=0.99767=99.77\%}.
$$
A thin water-based gel layer excludes air gaps and approximately matches the body's <acoustic impedance>. About $99.90\%$ of incident <energy> would otherwise reflect at the air-body interface. The coupling is also needed on the return journey, so it strongly improves the detectable echo. These figures concern interface transmission, before absorption within tissue; the gel is approximated by the stated water value.
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