Phase Shift of Reflected Wave in Conductive Material

E92M3
Messages
64
Reaction score
0

Homework Statement



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

Homework Equations



\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)}

\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)}

\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}}

The Attempt at a Solution



If I express \tilde{k} as \tilde{k}=Ke^{i\phi} where K is a constant, I have a phase difference \phi=\tan^{-1}\frac{\kappa}{k}. 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.
 
Physics news on Phys.org
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'.
ehild
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top