InterviewSolution
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InterviewSolution
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A linearly polarised electromagnetic wave given as `E=E_(0)haticos(kz-omegat)` is incident normally on a perfectly reflecting wall `z=a`. Assuming that the material of the optically inactive, the reflected wave will be give asA. `vec(E)_(r)=-E_(0)hat(i)cos(kx-omegat)`B. `vec(E)_(r)=E_(0)hat(i)cos(kz+omegat)`C. `vec(E)_(r)=-E_(0)hat(i)cos(kz+omegat)`D. `vec(E)_(r)=E_(0) hat(i)sin(kz-omegat)` |
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Answer» Correct Answer - B As the well is perfectly reflecting, there is no change in amplitude `E_(0)`. Also the wall is optically inactive, so, there is no phase change. After reflection, the wave travels along `-ve` z direction, `therefore" "vecE_(r)=E_(0)hati cos (-kz-omegat)` `=E_(0)haticos(kz+omegat)" "(because cos(-theta)=costheta)` |
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