= Solution
The rigid bottom reflects the lower-layer wave. Its repeated returns to the interface create an <internal-wave cavity response> containing both upward and downward components in $-H<z<0$, while the upper layer contains the imposed incident wave and one net reflected wave.
Let the lower downward and upward particle-displacement coefficients be $D$ and $U$, multiplying $e^{im_2z}$ and $e^{-im_2z}$ respectively. Their vertical displacement has opposite polarization signs:
$$
\xi_z(z)=\cos\theta_2[-De^{im_2z}+Ue^{-im_2z}].
$$
Zero normal velocity at $z=-H$ gives $U=Dq$, where $q=e^{-2im_2H}$. Interface continuity consequently becomes
$$
\sin\theta_1(\eta_i+\eta_r)=\sin\theta_2D(1+q),\qquad
\cos\theta_1(\eta_i-\eta_r)=\cos\theta_2D(1-q).
$$
For the specified angles, $\sin\theta_2/\sin\theta_1=1/\sqrt3$ and $\cos\theta_2/\cos\theta_1=\sqrt3$. Thus
$$
\frac D{\eta_i}=\frac{\sqrt3}{2-e^{-2im_2H}},\qquad
\frac{\eta_r}{\eta_i}=\frac{-1+2e^{-2im_2H}}{2-e^{-2im_2H}}.
$$
The dependence on bottom depth is therefore
$$
\boxed{|D|=|U|=\frac{\sqrt3|\eta_i|}{\sqrt{1+8\sin^2(m_2H)}}.}
$$
In particular,
$$
\boxed{|D|_{\max}=\sqrt3|\eta_i|\quad\text{when }m_2H=n\pi.}
$$
Here $k=\pi/\lambda_i$ and $m_2=k/\sqrt3$, so the positive depths giving this maximum are
$$
\boxed{H=n\sqrt3\lambda_i,\qquad n=1,2,\ldots.}
$$
The lower vertical-displacement standing-wave envelope is $2\cos\theta_2|D|\,|\sin[m_2(z+H)]|$, while its horizontal-displacement envelope has the complementary cosine form. These distinguish the total standing disturbance from the amplitude of either traveling component. The traveling amplitude has minimum $|\eta_i|/\sqrt3$ at $m_2H=(n+1/2)\pi$.
Since $|q|=1$, $|-1+2q|=|2-q|$. Hence \b[the net upward-propagating wave in the upper layer has amplitude magnitude $|\eta_i|$ for every $H$], with the phase given above. At an enhancement maximum, $q=1$ and $\eta_r=\eta_i$. No energy escapes through the bottom, so complete eventual reflection is consistent with energy conservation.
Back to article page
