 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 $k_x=k{n}_g\sin \theta$. In water, the dispersion relation is $k_x^2+k_y^2=k^2{n}_w^2$, so above the critical angle,Choose the positive imaginary part because a field growing exponentially into the water would be unphysical. The evanescent field in total internal reflection is thereforeIt 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.