= Solution
Let the interface be $y=0$, with glass at $y<0$ and water at $y>0$, and choose $x$ parallel to the incident beam's tangential component. Define $k=\omega/c$ as the vacuum <wavenumber>, with glass and water <refractive indices> $n_g$ and $n_w$. Continuity of <phase> along the interface fixes $k_x=kn_g\sin\theta$. In water, the <dispersion relation> is $k_x^2+k_y^2=k^2n_w^2$, so above the <critical angle>,
$$
k_y^2=-k^2(n_g^2\sin^2\theta-n_w^2),\qquad k_y=i\kappa,\qquad \kappa=k\sqrt{n_g^2\sin^2\theta-n_w^2}>0.
$$
Choose the positive imaginary part because a field growing exponentially into the water would be unphysical. The <evanescent field in total internal reflection> is therefore
$$
\boxed{E(x,y,t)=E_0\exp[i(kn_g\sin\theta\,x-\omega t)]\exp[-k\sqrt{n_g^2\sin^2\theta-n_w^2}\,y].}
$$
It propagates in phase parallel to the interface but decays normally. Here $E_0$ is the transmitted field amplitude at the interface; relating it to the incident amplitude requires the appropriate polarization-dependent <Fresnel equations>. If the symbol $k$ instead denotes the incident glass <wavenumber>, replace every vacuum $k$ in the formula by $k/n_g$; equivalently $k_x=k\sin\theta$ and $\kappa=k\sqrt{\sin^2\theta-(n_w/n_g)^2}$. This states the wavenumber convention left implicit in the source.
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