Phase Shift of Reflected Wave in Conductive Material

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Homework Statement



An EM wave normally shed on a conductive material ([itex]\tilde{\epsilon}=\epsilon+i\epsilon_i,\mu_0[/itex]). Calculate the phase shift of the electric field of the reflected wave relative to that of the incident wave.

Homework Equations



[itex]\nabla^2\textbf{E}=\mu\epsilon\frac{\partial^2\textbf{E}}{\partial t^2}+\mu\sigma\frac{\partial\textbf{E}}{\partial t}\Rightarrow\tilde{\textbf{E}}=\tilde{\textbf{E}}_0e^{i(\tilde{k}z-\omega t)}[/itex]

[itex]\nabla^2\textbf{B}=\mu\epsilon\frac{\partial^2\textbf{B}}{\partial t^2}+\mu\sigma\frac{\partial\textbf{B}}{\partial t}\Rightarrow\tilde{\textbf{B}}=\tilde{\textbf{B}}_0e^{i(\tilde{k}z-\omega t)}[/itex]

[itex]\tilde{k}=k+i\kappa\qquad;\qquad k=\omega\sqrt\frac{\epsilon\mu}{2}\left[\sqrt{1+\left(\frac{\sigma}{\epsilon\mu}\right)^2}+1\right]^{\frac{1}{2}}\,,\qquad\kappa=\omega\sqrt\frac{\epsilon\mu}{2}\left[\sqrt{1+\left(\frac{\sigma}{\epsilon\mu}\right)^2}-1\right]^{\frac{1}{2}}[/itex]

The Attempt at a Solution



If I express [itex]\tilde{k}[/itex] as [itex]\tilde{k}=Ke^{i\phi}[/itex] where [itex]K[/itex] is a constant, I have a phase difference [itex]\phi=\tan^{-1}\frac{\kappa}{k}[/itex]. This however seem to be the phase difference between the E and B field and not between the incident and reflected wave. I am the grader of a course using the second half of Griffiths' EM textbook, but I don't seem to be able to find the solution to this question there.
 
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From Maxwell equations you can derive how the B and E waves are related. You also have to know that the tangential components of E and B/μ are continuous at an interface, and that can happen only if there is a reflected wave in addition to the incident wave. Applying the boundary conditions, you get the reflection coefficient at normal incidence in terms of the complex k as r=(ko-k)/(ko+k) (where ko is ω/c). Or you can write that k=ω/c N with N the complex refractive index, and then r= (No-N)/(No+N). You must find related material in Griffiths'.
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