The permittivity $\varepsilon_1$ is related to the complex refractive index $n_1$ by $\varepsilon_1 = n_1^2$. A lossless dielectric
has a real refractive index and so $\varepsilon_1$ is positive. Therefore, the condition $\eta > 0$ implies that
surface waves cannot exist at the interface between two dielectric media such as air and glass.
For surface waves to exist, one medium must be a dielectric with positive permittivity, say
medium 1 ($\varepsilon_1 > 0$), while the other medium must have a negative permittivity ($\varepsilon_2 < 0$). This is
the case of metals, which in the ideal case can be described by a purely imaginary index, $n_2 =
i\alpha_2$ with $\alpha_2$ real, which gives $\varepsilon_2 = n_2^2 = -\alpha_2^2 < 0$. In the following, we will assume such a
dielectric-metal interface with $\varepsilon_1 = n_1^2 > 0$ and $\varepsilon_2 = n_2^2 < 0$. By the way, the dielectric can be air
or even vacuum, which is why we call this wave a "surface" wave, because it is confined near
the surface of the metal with the dielectric side playing more of a spectator role.
8. Using the definition of $\kappa_i^2$ derived in question 2, show that the relation $\frac{\kappa_1}{\kappa_2} = -\frac{\varepsilon_1}{\varepsilon_2} = \eta$ leads to
$\frac{\beta^2}{k_0^2} = \frac{\varepsilon_1}{1-\eta}$. Deduce the upper bound on $\eta$ for a surface wave to exist.
9. Deduce from question 8 that the propagation constant $\beta$ can be expressed as $\beta = n_{eff}k_0$
where $n_{eff} = \sqrt{\frac{\varepsilon_1}{1-\eta}}$ is the effective refractive index of the surface wave.
10. Using the expressions derived in questions 2 and 8, show that $\kappa_1^2$ can be written as $\kappa_1^2 =
-\xi_1 k_0^2$ and give an expression for $\xi_1$ and $\xi_2$ in terms of $\varepsilon_1$, $\varepsilon_2$, and $\eta$.
