A linearly polarized uniform plane wave, propagating in the forward \(z\) direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along \(y\left(\epsilon_{r y}\right)\) differs from that seen by waves polarized along \(x\left(\epsilon_{r x}\right) .\)

Since the wave is linearly polarized and propagating in the forward z direction, the electric field can be represented as: $$ \boldsymbol{E}_{inc} = E_0 \hat{\boldsymbol{x}} e^{-jkz}, $$ where \(E_0\) is the amplitude of the electric field, \(j\) is the imaginary unit, and \(k\) is the wave number. We are given that the polarization of this incident wave is along the x direction.

For a lossless anisotropic material, the wave equation can be represented as: $$ \nabla ^2 \boldsymbol{E} - k^2 (\epsilon_{r x} \hat{\boldsymbol{x}} \hat{\boldsymbol{x}} + \epsilon_{r y} \hat{\boldsymbol{y}} \hat{\boldsymbol{y}} + \hat{\boldsymbol{z}} \hat{\boldsymbol{z}}) \boldsymbol{E} = 0. $$ For the given plane wave, the Laplacian of the electric field will only have the \(\frac{\partial^2}{\partial z^2}\) term, and the resulting wave equation takes the form: $$ \frac{\partial^2}{\partial z^2} E_x - k^2 \epsilon_{r x} E_x = 0, $$ $$ \frac{\partial^2}{\partial z^2} E_y - k^2 \epsilon_{r y} E_y = 0. $$

The above two equations are essentially the same as the wave equations in two separate isotropic mediums, with their respective dielectric constants being \(\epsilon_{r x}\) and \(\epsilon_{r y}\). Therefore, we can find the propagation constants in each direction using the standard results in isotropic materials. The wave number in the x direction is given by: $$ k_x = k \sqrt{\epsilon_{r x}}. $$ And in the y direction, it is given by: $$ k_y = k \sqrt{\epsilon_{r y}}. $$

For the given linearly polarized incident wave with its polarization in the x direction, the power transfer will be mainly affected by the properties of the anisotropic medium along the x direction. The power transfer for the wave can be calculated using the Poynting vector: $$ \boldsymbol{S} = \frac{1}{2} \Re\lbrace \boldsymbol{E} \times \boldsymbol{H}^*\rbrace. $$ We know that the E-field and H-field are related: $$ \boldsymbol{H} = \sqrt{\frac{\epsilon_{r x}}{\mu_0}} \hat{\boldsymbol{y}} \times \boldsymbol{E}. $$ The Poynting vector for the incident wave becomes: $$ \boldsymbol{S} = \frac{1}{2} \Re\lbrace E_0^2 \sqrt{\frac{\epsilon_{r x}}{\mu_0}} e^{-j 2 k z} \rbrace \hat{z}. $$ Taking the real part, we have: $$ \boldsymbol{S} = \frac{1}{2} E_0^2 \sqrt{\frac{\epsilon_{r x}}{\mu_0}} \cos(2 k z) \hat{z}. $$ Finally, we can see that the power transfer through the anisotropic medium depends on the amplitude of the incident electric field, the dielectric constant along the x direction, and the position inside the medium along the propagation direction.