Solution

To estimate the penetration depth of an electromagnetic wave, the free electron plasma model is used.

At optical frequencies, the permittivity of the conductor is

$\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2}, \qquad \omega = 2\pi\nu$,

where the plasma frequency in SI is

$\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$.

With e = $1.6\times10^{-19}\,\text{C}, \varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}, m_e = 9.1\times10^{-31}\,\text{kg}$:

$\omega_p^2 = \frac{10^{28}(1.6\times10^{-19})^2}{8.85\times10^{-12}\times9.1\times10^{-31}} \approx 3.2\times10^{31}\,\text{rad}^2/\text{s}^2, \quad \omega_p \approx 5.6\times10^{15}\,\text{rad/s}$

$Since \omega = 2\pi\times10^{15} \approx 6.3\times10^{15}\,\text{rad/s} > \omega_p$
the wave propagates with attenuation. The penetration depth \delta in a conductor is given by the damping length of the field intensity:

$\delta = \frac{c}{\sqrt{\omega_p^2 - \omega^2}} \approx \frac{c}{\omega_p} \quad (\text{since } \omega \sim \omega_p)$

Numerically,

$\delta \approx \frac{3\times10^{10}\,\text{cm/s}}{5.6\times10^{15}\,\text{s}^{-1}} \approx 5.4\times10^{-6}\,\text{cm}$

$\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$.