= Solution
A material interface obeys $\eta_t+\phi_{j,x}\eta_x=\phi_{j,y}$ at its actual height. Keeping only first-order terms and evaluating at the undisturbed interface gives $\phi_{1,y}=\phi_{2,y}=\eta_t$ at $y=0$. Linearized unsteady Bernoulli gives the pressure perturbation $-\rho_j(\phi_{j,t}+g\eta)$, after absorbing time-only constants into the potentials. Pressure <continuity> therefore gives
$$
\boxed{\rho_1\phi_{1,t}-\rho_2\phi_{2,t}+g(\rho_1-\rho_2)\eta=0\quad\text{at }y=0}.
$$
For $\vartheta=kx-\omega t$, choose potentials decaying away from the interface:
$$
\phi_1=\frac{A\omega}{k}e^{ky}\sin\vartheta\quad(y<0),\qquad
\phi_2=-\frac{A\omega}{k}e^{-ky}\sin\vartheta\quad(y>0).
$$
They are harmonic and their normal derivatives equal $A\omega\sin\vartheta=\eta_t$ at zero. The dynamic condition reduces to
$$
\boxed{\omega^2=gk\frac{\rho_1-\rho_2}{\rho_1+\rho_2},\qquad c=\frac\omega k=\sqrt{\frac{g(\rho_1-\rho_2)}{k(\rho_1+\rho_2)}}}.
$$
If the denser fluid is above, the squared frequency is negative. The corresponding perturbations grow or decay exponentially instead of oscillating: this is <Rayleigh-Taylor instability>, with growing rate $\sqrt{gk(\rho_2-\rho_1)/(\rho_1+\rho_2)}$ in the ideal model without surface tension.
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