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@Jahnavi I think you have this part figured out already, but to show it in detail: ## \\ ## ## E_{incident}(z,t)=E_o \hat{i} \cos(kz-\omega t) ##, and ## \\ ## ## E_{reflected}(z,t)=-E_o \hat{i} \cos(2ka-kz-\omega t) ##. ## \\ ## Evaluating at ## z=a ##: ## \\ ## ## E_{incident}(a,t)=E_o \hat{i} \cos(ka-\omega t) ##, and ## \\ ## ## E_{reflected}(a,t)=-E_o \hat{i} \cos(2ka-\omega t-ka)=-E_o \hat{i} \cos(ka-\omega t) ##. ## \\ ## Thereby, they cancel at ## z=a ##. ## \\ ## Also, in post 21, you asked about the vector ## \hat{i} ##. Yes, that means the electric field for this electromagnetic wave is transverse and points in the x-direction.Jahnavi said:At z = a , E0cos (ka- ωt) - E0cos (ka + ωt) = 2E0sin(ka)sin(ωt) ≠ 0
Can you show how they cancel ?
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