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For a wave characterized by Eox cos(w.t-k-z) V E(z,t) = Egy cos (w.t-k-z) m 0 a) plot E₁ (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave? b) plot E₁ (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave?

For a wave characterized by Eox cos(w.t-k-z) V E(z,t) = Egy cos (w.t-k-z) m 0 a) plot E₁ (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave? b) plot E₁ (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave?

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For a wave characterized by Eox cos(w.t-k-z) ● E(z, t) = Eoy cos(w•t_k•z) 0 a) plot E, (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave? V m b) plot E, (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave? = 108 W:= rad S Ex=5 Ex:=5 k:=0.5 rad m Egy:=5 Eoy:=-4
Transcribed Image Text:For a wave characterized by Eox cos(w.t-k-z) ● E(z, t) = Eoy cos(w•t_k•z) 0 a) plot E, (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave? V m b) plot E, (z, t) vs Ex(z, t) for one period at z=0. What type of polarization is this wave? = 108 W:= rad S Ex=5 Ex:=5 k:=0.5 rad m Egy:=5 Eoy:=-4
Expert Solution
Step 1: Determine the given data:

straight E subscript straight x left parenthesis straight z comma straight t right parenthesis equals straight E subscript 0 straight x end subscript cos left parenthesis straight omega straight t minus kz right parenthesis straight E subscript straight y left parenthesis straight z comma straight t right parenthesis equals straight E subscript 0 straight y end subscript cos left parenthesis straight omega straight t minus kz right parenthesis straight E subscript ox equals 5 space comma space straight E subscript 0 straight y end subscript equals 5 straight E subscript ox equals 5 space comma space straight E subscript 0 straight y end subscript equals negative 4 straight omega equals 10 to the power of 8 space rad divided by straight s comma space straight k equals 0.5 space rad divided by straight m

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