InterviewSolution
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InterviewSolution
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What kind of polarizations has a plane electromagetic wave if the projections of the vector `E` on the `x` and `y` axes are perpendicular to the propagation direction and are defind by the following equations: (a) `E_(x) = Ecos (omegat - kz), E_(y) = E sin (omegat - kz)`, (b) `E_(x) = E cos(omegat - kz), E_(y) = E cos (omegat - kz + pi//4)` (c) `E_(x) = E cos (omega t - kz), E_(y) = E cos (omega t - kz + pi)`? |
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Answer» The wave is moving in the direction of `z`-axis (a) Here `E_(x) = E cos (omegat - kz), E_(y) = E sin(omegat - kz)` `(E_(x)^(2))/(E^(2)) + (E_(y)^(2))/(E^(2)) = 1` so the up of the electric vector moves along a circle. For the right handed coordinate system this represents circular anticlockwise polarization when observed towards the incoming wave. (b) `E_(x) = E cos (omegat - kz), E_(y) = Ecos (omegat - kz+(pi)/(4))` so `(E_(y))/(E) = (1)/(sqrt(2)) co s(omegat - kz) - (1)/(sqrt(2)) sin (omegat - kz)` or `((E_(y))/(E) - (1)/(sqrt(2))(E_(x))/(E))^(2) = (1)/(2)(1-(E_(x)^(2))/(E^(2)))` or `(E_(y)^(2))/(E^(2)) + (E_(x)^(2))/(E^(2)) - sqrt(2) (E_(y)E_(x))/(E^(2)) = (1)/(2)` This is clearly an ellipse. By comparing with the pervious case (compare the phase of `E_(y)` in the two cases ) we see this represents elliptical clockwise polarization when viewed towards the incoming wave. We write the equations as `E_(x) + E_(y) = 2Ecos(omegat - kz+(pi)/(8)) cos((pi)/(8))` `E_(x) - E_(y) = +2E sin (omegat - kz + (pi)/(8)) sin ((pi)/(8))` Thus `((E_(x) - E_(y))/(2Ecos((pi)/(8))))^(2) + ((E_(x) - E_(y))/(2esin((pi)/(8))))^(2) = 1` Since `cos((pi)/(8)) gt sin((pi)/(8))`, the major axis is the direction of the stright line `y = x`. (c) `E_(x) = Ecos (omegat - kz)` `E_(y) = Ecos(omegat - kz + pi) =- E cos (omegat - kz)` Thus the top of the electric vector traces the curve `E_(y) =- E_(x)` which is a stright line `~(y =- x)`. It corresponds to plane polarization. |
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