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Understanding Brewster's Angle

  1. Sep 3, 2013 #1

    I'm struggling to conceptually understand why Brewster's angle occurs. I know Fresnel's equations and can see that at some angle, the reflected parallel component goes to zero. What I don't understand is why this is conceptually happening. I understand that the material is made up of 'dipoles' and free charges which can absorb and deflect the wave, but I cannot see why the perpendicular component is not treated the same way as the parallel component. I don't see how they appear any differently to the material. If someone could provide some insight into this, that would be great.

    Thank you.
  2. jcsd
  3. Sep 3, 2013 #2


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    The Fresnel formulas come from Maxwell's equations and continuity requirements at the boundary between the two materials. The boundary conditions are different for the field components in tangential and perpendicular direction of the medium. See, e.g., Jackson Classical Electrodynamics.
  4. Sep 3, 2013 #3
    Thank you for the reply. I understand the derivation of these equations (at least I think I do), and I understand that they are based upon the boundary conditions (tangential fields equal on either side of the boundary and perpendicular components not necessarily equal, and a charge distribution forms on the boundary to oppose it).

    My question is more a conceptual one. I don't 'get' it. I see the maths, the maths works nicely. I get the equations and everything, but I don't understand conceptually what is going on. I don't see how changing the angle of the field will affect anything if initially, the incoming field is unpolarised. As the angle increase, why are the parallel and perpendicular components interacting differently with the boundary?
  5. Sep 3, 2013 #4


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    The reflected beam can be considered as radiation of the exited dipoles in the material. It can be shown that a dipole does not radiate in its own direction. The dipoles in the material are perpendicular to the refracted beam. At the Brewster Angle, the refracted and reflected beams are perpendicular, so the dipoles are oriented parallel to the reflected beam in case of in-plane polarization (when the electric field is parallel to the plane of incidence) . At perpendicular polarization, the polarization is out-of plane, perpendicular to the reflected ray.


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    Last edited: Sep 3, 2013
  6. Sep 6, 2013 #5
    An oscillating dipole cannot radiate along its axis (you can convince yourself of this by drawing a charge bobbing up and down with the electric field lines attached at different points in time). For the one polarization, the axis of the induced dipoles lines up with the reflected ray at Brewster's angle, thus no reflected ray can be generated. For the other polarization, the axis of the dipoles is parallel to the material's surface and never lines up with reflected ray direction, even at Brewster's angle. The axis of the induced dipoles in the material is lined up with the electric field vector of the transmitted wave.

    Take a look at my lecture notes, page 17: http://faculty.uml.edu/cbaird/95.658(2013)/Lecture1.pdf
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