Ashish Posted February 5, 2009 Share Posted February 5, 2009 In order to show unpolarized light we use the can you tell me please about what this arrow on the unpolarized represent whether it represent electric field or what? Link to comment Share on other sites More sharing options...
insane_alien Posted February 5, 2009 Share Posted February 5, 2009 either electric or magnetic field, it doesn't really matter. Link to comment Share on other sites More sharing options...
Ashish Posted February 5, 2009 Author Share Posted February 5, 2009 can you tell me why? I've read its electric field Link to comment Share on other sites More sharing options...
Klaynos Posted February 5, 2009 Share Posted February 5, 2009 The direction of the electric field vector is the definition of polarisation. Link to comment Share on other sites More sharing options...
swansont Posted February 5, 2009 Share Posted February 5, 2009 By convention it's the electric field; the magnetic field will always be at 90º so there's no need to depict it. Edit: Gah! One minute late! Kurse you, Klaynos! Link to comment Share on other sites More sharing options...
timo Posted February 5, 2009 Share Posted February 5, 2009 I have a follow-up question on polarization of light. Assume I want to polarize microwaves with a metal grate. Assume the grate is oriented such that the bars of the grate are vertically aligned. Staying in the definition above: What will be the polarization of the beam of microwaves after the grate - and why? Link to comment Share on other sites More sharing options...
Klaynos Posted February 5, 2009 Share Posted February 5, 2009 I think... Perpendicular to the grating, the E field induced a charge in the grating along it's axis, and gets absorbed through non-radiative damping. I wonder though about the diffraction through these slits, or if they're below the diffraction limit, the surface plasmon polaritons you would create and their associated fields... I will check with a microwave physicist in the morning Link to comment Share on other sites More sharing options...
swansont Posted February 6, 2009 Share Posted February 6, 2009 Apply the boundary conditions. The perpendicular component of the field has to go to zero for a conductor where there is no net charge, and the parallel component suffers no discontinuity. This assumes that the perpendicular direction can't support the wave, i.e. it's smaller than the wavelength. Link to comment Share on other sites More sharing options...
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