A \(10 \mathrm{MHz}\) uniform plane wave having an initial average power density of \(5 \mathrm{~W} / \mathrm{m}^{2}\) is normally incident from free space onto the surface of a lossy material in which \(\epsilon_{2}^{\prime \prime} / \epsilon_{2}^{\prime}=0.05, \epsilon_{r 2}^{\prime}=5\), and \(\mu_{2}=\mu_{0} .\) Calculate the distance into the lossy medium at which the transmitted wave power density is down by \(10 \mathrm{~dB}\) from the initial \(5 \mathrm{~W} / \mathrm{m}^{2}\).

First, we have to find the reflection coefficient at the interface between the two media. This can be calculated using the formula: \(R = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}\) where \(\eta_1\) is the intrinsic impedance of free space (approximately 120π Ω), \(\eta_2\) is the intrinsic impedance of the lossy material, and R is the reflection coefficient. Since the magnetic permeability of the lossy medium is equal to that of free space (\(\mu_2 = \mu_0\)), the relative permeability \(\mu_r = \frac{\mu_2}{\mu_0} = 1\). So, we can write: \(\eta_2 = \sqrt{\frac{\mu_2}{\epsilon_2'}} = \sqrt{\frac{\mu_0}{\epsilon_0 \epsilon_{r2}'}} = \frac{\eta_1}{\sqrt{\epsilon_{r2}'}} \) Now, we can find the reflection coefficient: \(R = \frac{\frac{\eta_1}{\sqrt{\epsilon_{r2}'}} - \eta_1}{\frac{\eta_1}{\sqrt{\epsilon_{r2}'}} + \eta_1}\)

Next, we need to calculate the transmitted power density at the surface of the lossy material. This can be found using the formula: \(T = S_1(1 - |R|^2)\) where \(S_1\) is the initial average power density (\(5 \mathrm{~W}/\mathrm{m}^2\)) and T is the transmitted power density.

We have to find the attenuation constant in the lossy medium. This can be calculated using the formula: \(\alpha = \omega \sqrt{\frac{\mu_2 \epsilon_2''}{2}} = 2\pi f \sqrt{\frac{\mu_0 \epsilon_0 \epsilon_{r2}''}{2}}\) We know \(\epsilon_{r2}'' = { 0.05 \times \epsilon_{r2}' }\)

Now we are ready to calculate the distance into the lossy medium at which the transmitted wave power density is down by 10 dB from the initial power density. The power reduction in dB can be expressed as: \(10 \mathrm{~dB} = 10 \log{\frac{T}{S_2}}\), where \(S_2\) is the transmitted power density at the required distance. Now we can rewrite the equation in terms of the distance in the lossy medium: \(x = \frac{1}{\alpha} \left[{10 \log{({\frac{T}{S_2}}) \frac{1}{10}} }\right]\) Now substitute the values of T and α in the equation to calculate the distance x.