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We want to understand with Electromagnetism what happens at a surface. From Maxwell's equations we can understand what happens to the components of the $\stackrel{\u20d7}{E}$ and $\stackrel{\u20d7}{B}$ fields: First lets look at the $\stackrel{\u20d7}{E}$ field using Gauss' law. Recall $$\oint \epsilon \stackrel{\u20d7}{E}\cdot d\stackrel{\u20d7}{s}=\int \rho dV$$
Consider the diagram, the field on the incident side is $\stackrel{\u20d7}{{E}_{i}}+\stackrel{\u20d7}{{E}_{r}}$ . On the transmission side, the field is $\stackrel{\u20d7}{{E}_{t}}$ . We can collapse the cylinder down so that it is a pancake with an infinitelysmall height. When we do this there are no field lines through the side of the cylinder. Thus there is only a flux through the top and the bottom of thecylinder and we have; $$\oint \epsilon \stackrel{\u20d7}{E}\cdot d\stackrel{\u20d7}{s}=\oint \left[{\epsilon}_{i}\left({E}_{i\perp}+{E}_{r\perp}\right)-{\epsilon}_{t}{E}_{t\perp}\right]ds=0.$$ I have set $\int \rho dV=0$ since we will only consider cases without free charges. So we have $${\epsilon}_{i}{E}_{i\perp}+{\epsilon}_{i}{E}_{r\perp}={\epsilon}_{t}{E}_{t\perp}$$ if ${\widehat{u}}_{n}$ is a unit vector normal to the surface this can be written $${\epsilon}_{i}{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{E}}_{i}+{\epsilon}_{i}{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{E}}_{r}={\epsilon}_{t}{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{E}}_{t}$$
Similarly Gauss' law of Magnetism $$\oint \stackrel{\u20d7}{B}\cdot d\stackrel{\u20d7}{s}=0$$ gives $${B}_{i\perp}+{B}_{r\perp}={B}_{t\perp}$$ or $${\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{B}}_{i}+{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{B}}_{r}={\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{B}}_{t}$$
Amperes law can also be applied to an interface.Then
$$\int \frac{\stackrel{\u20d7}{B}}{\mu}\cdot d\stackrel{\u20d7}{l}=\int \stackrel{\u20d7}{j}\cdot d\stackrel{\u20d7}{s}+\frac{d}{dt}\int \epsilon \stackrel{\u20d7}{E}\cdot d\stackrel{\u20d7}{s}$$ (note that in this case $d\stackrel{\u20d7}{s}$ is perpendicular to the page)Now we will not consider cases with surface currents. Also we can shrink the vertical ends of the loop so that the area of the box is 0 so that $\int \epsilon \stackrel{\u20d7}{E}\cdot d\stackrel{\u20d7}{s}=0$ . Thus we get at asurface $$\frac{{B}_{i\parallel}+{B}_{r\parallel}}{{\mu}_{i}}=\frac{{B}_{t\parallel}}{{\mu}_{t}}$$ or $$\frac{{\widehat{u}}_{n}\times {\stackrel{\u20d7}{B}}_{i}}{{\mu}_{i}}+\frac{{\widehat{u}}_{n}\times {\stackrel{\u20d7}{B}}_{r}}{{\mu}_{r}}=\frac{{\widehat{u}}_{n}\times {\stackrel{\u20d7}{B}}_{t}}{{\mu}_{t}}$$
Similarly we can use Faraday's law $$\int \stackrel{\u20d7}{E}\cdot d\stackrel{\u20d7}{l}=-\frac{d}{dt}\int \stackrel{\u20d7}{B}\cdot d\stackrel{\u20d7}{s}$$ and play the same game with the edges to get $${E}_{i\parallel}+{E}_{r\parallel}={E}_{t\parallel}$$ or $${\widehat{u}}_{n}\times {\stackrel{\u20d7}{E}}_{i}+{\widehat{u}}_{n}\times {\stackrel{\u20d7}{E}}_{r}={\widehat{u}}_{n}\times {\stackrel{\u20d7}{E}}_{t}$$ (notice $\epsilon $ does not appear)
In summary we have derived what happens to the $\stackrel{\u20d7}{E}$ and $\stackrel{\u20d7}{B}$ fields at the interface between two media: $${\epsilon}_{i}{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{E}}_{i}+{\epsilon}_{i}{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{E}}_{r}={\epsilon}_{t}{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{E}}_{t}$$
$${\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{B}}_{i}+{\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{B}}_{r}={\widehat{u}}_{n}\cdot {\stackrel{\u20d7}{B}}_{t}$$
$$\frac{{\widehat{u}}_{n}\times {\stackrel{\u20d7}{B}}_{i}}{{\mu}_{i}}+\frac{{\widehat{u}}_{n}\times {\stackrel{\u20d7}{B}}_{r}}{{\mu}_{i}}=\frac{{\widehat{u}}_{n}\times {\stackrel{\u20d7}{B}}_{t}}{{\mu}_{t}}$$
$${\widehat{u}}_{n}\times {\stackrel{\u20d7}{E}}_{i}+{\widehat{u}}_{n}\times {\stackrel{\u20d7}{E}}_{r}={\widehat{u}}_{n}\times {\stackrel{\u20d7}{E}}_{t}$$
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