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Phase change of reflected wave - is there a polarization dependence?

  1. Jun 22, 2012 #1
    Does a light wave in air (n1 = 1) that is reflected off a glass surface (n2=1.5) experience a 180deg phase change? Looking at the Fresnel equations (theta = 0deg) I learn that:

    according to http://en.wikipedia.org/wiki/Fresnel_equations

    rp = (n2-n1)/(n1+n2) = 0.2 and rs = (n1-n2)/(n1+n2) = -0.2

    rp indicates no phase shift, while rs does indicate at 180 deg phase shift (rs is negative).

    Intuitively, the phase shift should not depend on polarization at normal incidence (theta = 0deg). How do I understand the above formulas? What if the polarization is circular?
  2. jcsd
  3. Jun 22, 2012 #2


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    For normal incidence, all incident waves have, effectively, the same (horizontal) polarisation because the field vectors are parallel with the plane of the reflecting surface. It's just a consequence of 'feeling' the geometry. Distinguishing between HP and VP is easy for large angles of incidence (glancing contact) but the parallel component of an incident VP wave gets larger as the angle of incidence approaches zero and the normal component goes to zero.

    After reflection for non-normal incidence, circularly polarised EM is no longer circular. At the Brewster angle, the polarisation becomes Plane.
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