Johnny5 Posted May 12, 2005 Share Posted May 12, 2005 Unlike some of my recent threads, this one is actually a specific problem I need help with. I think the answer is snell's law, but its been years since I have solved it. You can get to Brewster's angle if you keep going. Here is the problem. You have an electromagnetic wave incident upon some surface. There is an incident wave, a transmitted wave, and a reflected wave. The permittivity constant of the one medium is e1 and the permittivity constant of the other medium is e2. Denote the angle of incidence by qi, and the transmitted angle by qt. Using boundary conditions on the electric field E, find a relationship between the angle of incidence, and the transmitted angle. Now I have been reading about 5 books on EM, and optics, as far as boundary conditions go. I remember doing it years ago, and again, none of the books are clear. I used Gauss' law at the boundary to get the boundary conditions on the E field. I think Ampere's law gets you the boundary conditions for the B field, but that doesn't matter in this problem, at least i dont think so. There is a boundary condition for the perpendicular components of the E field, and another for the parallel components. In the books they just say "electric field is continuous across the boundary." So here is what i think the answer is, but I'm not sure. e1 sin(qi ) = e2 sin(qt ) None of the books cover the vector relationships at the boundary very well. So any insight would be appreciated. Thank you PS: Also, I know that isn't exactly Snell's law, Snell's law is: n1 sin(qi ) = n2 sin(qt ) Where n1,n2 are the index of refractions of the mediums. But there is a relationship between them, the speed of light, and the permittivity. In general: n = (em/e0m0 )1/2 Where the numerator is the product of the permittivity and permeability of the medium, and the denominator is for free space. So equivalently: n = c(em)1/2 And letting v denote the speed of light in the medium we also have: n = c/v Link to comment Share on other sites More sharing options...
swansont Posted May 12, 2005 Share Posted May 12, 2005 Gauss's law tells you that the perpendicular component can only be discontinuos across a surface if there is a surface charge. The parallel component must be continuous. Link to comment Share on other sites More sharing options...
Johnny5 Posted May 12, 2005 Author Share Posted May 12, 2005 Gauss's law tells you that the perpendicular component can only be discontinuos across a surface if there is a surface charge. The parallel component must be continuous. What if the free charge is zero at the surface? I take it that this means that the perpendicular component must be continuous? And you are saying the parallel component will be continuous regardless of whether or not there is a surface charge? What I really want to do is figure out the relationship between the angle of incidence and the angle of reflection, in terms of the permittivities of both mediums. And I want to do that using the boundary conditions. I presume the correct answer is something close to Snell's law. Link to comment Share on other sites More sharing options...
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