Chapter.2.Polarization.Simulation

Chapter.2.Polarization.Simulation - Polarization of the...

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Polarization of the electric field 10-26-09 Eox 1 := ω 2 π 10 14 () := c 310 8 := Eoy 1 := i 1 1 2 := zo 10c ω := zo 4.775 10 6 × = k ω c := α 1 := ....................................................................................................................................... Ex z t x , Re Eox e ikz ⋅ω t −φ x + := Ey z t y , Re 1 Eoy e t y + := zz o 0 zo .01 , zo 2 .. := Polarization depends on: ** Relative phases of Ex and Ey ** Relative magnitudes of Ex and Ey
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Linarly polarized electric field n 2 := φ x 0 := φ y φ x π n + := 20 2 2 0 2 Ey z 0 y , () Ex z 0 x ,
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Circularly polarized electric field φ x 0 := φ y φ x π 2 + := 20 2 2 0 2 Ey z 0 y , () Ex z 0 x ,
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Eliptically polarized electric field φ x 0 := φ y φ x π 3 + := 20 2 2 0 2 Ey z 0 y , () Ex z 0 x ,
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Left handed circularly polarized [( ε -) negative helicity state] zz o 0 zo .01 , zo .2 .. := φ y φ x π 2 + := φ x 0 := 20 2 2 0 2 Ey z 0 y , () Ex z 0 x , Right handed circularly polarized [( ε+) positive helicity state] φ y φ x π 2 := 2 2 0 2 Ey z 0 y , Ex z 0 x ,
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Note that for the right handed circularly polarized light, at a fixed time, t, the tip of the electric field vector appears to be rotating counter clockwise.
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This note was uploaded on 12/19/2010 for the course PHYS 423 taught by Professor G. during the Spring '10 term at Missouri S&T.

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Chapter.2.Polarization.Simulation - Polarization of the...

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