and (bottom) evanescent wave at an interface in red (reflected waves omitted). For example, consider
total internal reflection in two dimensions, with the interface between the media lying on the x axis, the
normal along y, and the
polarization along z. One might expect that for angles leading to total internal reflection, the solution would consist of an incident wave and a reflected wave, with no transmitted wave at all, but there is no such solution that obeys
Maxwell's equations. Maxwell's equations in a dielectric medium impose a boundary condition of continuity for the components of the fields
E||, H||, Dy, and
By. For the polarization considered in this example, the conditions on
E|| and
By are satisfied if the reflected wave has the same amplitude as the incident one, because these components of the incident and reflected waves superimpose destructively. Their
Hx components, however, superimpose constructively, so there can be no solution without a non-vanishing transmitted wave. The transmitted wave cannot, however, be a sinusoidal wave, since it would then transport energy away from the boundary, but since the incident and reflected waves have equal energy, this would violate
conservation of energy. We therefore conclude that the transmitted wave must be a non-vanishing solution to Maxwell's equations that is not a traveling wave, and the only such solutions in a dielectric are those that decay exponentially: evanescent waves. Mathematically, evanescent waves can be characterized by a
wave vector where one or more of the vector's components has an
imaginary value. Because the vector has imaginary components, it may have a magnitude that is less than its real components. For the plane of incidence as the xy plane at z = 0 and the interface of the two mediums as the xz plane at y=0 , the wave vector of the transmitted wave has the form : \mathbf{k_t} \ = \ k_y \hat{\mathbf{y}} + k_x \hat{\mathbf{x}} with k_x = k_t \sin \theta_t and k_y = k_t \cos \theta_t , where k_t is the magnitude of the wave vector of the transmitted wave (so the
wavenumber), \theta_t is the angle of refraction, and \hat{\mathbf{x}} and \hat{\mathbf{y}} are the unit vectors along the x axis direction and the y axis direction respectively. By using the
Snell's law n_i \sin \theta_i = n_t \sin \theta_t where n_i , n_t , and \theta_i are the refractive index of the medium where the incident wave and the reflected wave exist, the refractive index of the medium where the transmitted wave exists, and the angle of incidence respectively, : k_y =k_t \cos \theta_t = \pm k_t \left (1 - \frac{\sin^2 \theta_i}{n_{ti}^2}\right)^{1/2} . with n_{ti} = \frac{n_t}{n_i} . If a part of the condition of the
total internal reflection as \sin \theta_i > \sin \theta_c = n_{ti} , is satisfied, then : k_y = \pm i k_t \left( \frac{\sin^2 \theta_i}{n_{ti}^2} - 1\right)^{1/2} = \pm i \alpha . If the
polarization is perpendicular to the plane of incidence (along the z direction), then the electric field of any of the waves (incident, reflected, or transmitted) can be expressed as : \mathbf{E}(\mathbf{r},t) = \operatorname{Re} \left \{ E(\mathbf{r}) e^{ i \omega t } \right \} \mathbf{\hat{z}} where \mathbf{\hat{z}} is the
unit vector in the z axis direction. By assuming plane waves as E(\mathbf{r}) = E_0 e^{-i\mathbf{k}\cdot \mathbf{r}} , and substituting the transmitted wave vector \mathbf{k_t} into \mathbf{k} , we find for the transmitted wave: : E(\mathbf{r}) = E_o e^{-i ( -i \alpha y + \beta x ) } = E_o e^{-\alpha y - i \beta x } where \alpha = k_t \left( \frac{\sin^2 \theta_i}{n_{ti}^2} - 1\right)^{1/2} is the
attenuation constant, and \beta = k_x is the
phase constant. +i \alpha is ignored since it does not physically make sense (the wave amplification along
y the direction in this case). == Evanescent-wave coupling ==