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Direction of reflected light

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data
    i cant understand why the reflected light has the direction of oscillating inside and outside of the page while the direction of refracted wave is upwards and downwards (perpendicuar to the plane of wave propagation. ) Can the direction of reflected light vibrate perpendicuar to the plane of wave propagation.? why ? i cant find the reason in my book and the internet . please help!


    2. Relevant equations



    3. The attempt at a solution
     

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  2. jcsd
  3. Sep 7, 2014 #2

    ehild

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    The figure is rather misleading. Both the incident and refracted light are unpolarized, they have components both oscillating in the plane and perpendicularly to the plane of incidence. At the Brewster angle, the reflected light has only the perpendicularly oscillating component.

    ehild
     
  4. Sep 7, 2014 #3

    At the Brewster angle, the reflected light has only the perpendicularly oscillating component. what do you mean by it? Is it oscillate inside and oustide of the page or up and down at the plane?
    i read a lot of online notes. they give the refelected wave oscillate in and out of the page... Why is it so? can it oscillate up and down at the plane
     
  5. Sep 7, 2014 #4
    here's the better note
     

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  6. Sep 7, 2014 #5

    ehild

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    At the Brewster angle, there is no light which oscillates in the plane. (You say, up and down). Only that component can be reflected which oscillates in and out of the plane.

    It is the electric field, that oscillates. We speak about two modes of waves, with parallel and perpendicular polarization. The polarization means the direction of the electric field, which can be perpendicular to the plane of incidence (oscillates in and out of the plane) while parallel polarization means that the electric field oscillates in the plane of incidence. The two modes reflect differently, and the parallel wave does not reflect from the boundary when the angle of incidence is equal to the Brewster angle. If the incident light had also perpendicular component, it is reflected. At the Brewster angle, the reflected light has only perpendicular component (oscillating in and out of the plane)
    When the incident light is parallel polarized, and the boundary is illuminated at the Brewster angle, there is no reflected light from the surface.

    The last figure is misleading, the refracted light can contain both parallel and perpendicular components.


    ehild
     
  7. Sep 7, 2014 #6
    What do you mean by the light is parallel polarized? Is it parallel to the glass surface ? Or parallel to the plane of incidence? (circlrd part)?
     
  8. Sep 7, 2014 #7
    Please refer to the picture...
     

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  9. Sep 7, 2014 #8

    ehild

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    Last edited: Sep 7, 2014
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