A right-circularly polarized plane wave is normally incident from air onto a semi-infinite slab of plexiglas \(\left(\epsilon_{r}^{\prime}=3.45, \epsilon_{r}^{\prime \prime}=0\right) .\) Calculate the fractions of the incident power that are reflected and transmitted. Also, describe the polarizations of the reflected and transmitted waves.

Right-circularly polarized wave can be represented as a combination of two linearly polarized waves: one is s-polarized (parallel to the plane of incidence) and the other is p-polarized (perpendicular to the plane of incidence). This will allow us to evaluate the reflection and transmission coefficients for each individual polarization using Fresnel equations.

The Fresnel equations relate the amplitudes of reflected and transmitted electric fields to the incident electric field amplitude. They are given by: For s-polarized waves (TE waves): \(r_s = \frac{n_1 \cos{\theta_i} - n_2 \cos{\theta_t}}{n_1 \cos{\theta_i} + n_2 \cos{\theta_t}}\) For p-polarized waves (TM waves): \(r_p = \frac{n_2 \cos{\theta_i} - n_1 \cos{\theta_t}}{n_2 \cos{\theta_i} + n_1 \cos{\theta_t}}\) In this problem, the wave is normally incident, so \(\theta_i = 0\) and \(\theta_t = 0\). Also, since the incident medium is air, we have \(n_1=1\). Since the slab is plexiglass and \(\epsilon_{r}^{\prime}=3.45\), we can calculate the refractive index of the plexiglass as \(n_2=\sqrt{\epsilon_{r}^{\prime}}\).

By applying the Fresnel equations, we can find the reflection and transmission coefficients for both s- and p-polarized waves respectively: \(r_s = \frac{1- \sqrt{3.45}}{1 + \sqrt{3.45}}\) \(r_p = \frac{\sqrt{3.45} - 1}{\sqrt{3.45} + 1}\)

We can obtain the fractions of the incident power that are reflected and transmitted using the square of the reflection and transmission coefficients: \(R_s = |r_s|^2\) and \(R_p = |r_p|^2\). Since both s- and p-polarized components have equal power, we can average the two values to obtain the overall reflected and transmitted fractional power : \(R = \frac{R_s + R_p}{2}\), where R is the reflected power fraction \(T = 1 - R\), where T is the transmitted power fraction

The Fresnel coefficients and polarizations will now give us the polarization state of the reflected and transmitted waves. As both components have reverse signs of the reflection coefficient, the reflected wave will be left-circularly polarized. Since both components have the same signs for the transmission coefficient, the transmitted wave will remain right-circularly polarized.