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I know that when circularly polarized light passes through a linear polarizer, the resulting intensity is half the initial intensity. This is because we can decompose the components of $\mathbf{E}$ on the axis of the polarizer and the axis perpendicular to the polarizer, and on average the field on both axes will be equal.

Does the same effect occur with elliptical polarization? Can we say that $I=I_0/2$ after passage through a linear polarizer?

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You should again split up $\vec{E}$ in its components and work out the resulting field. You will notice that the resulting intensity, after passing through the polariser, is generally not half the original intensity.

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  • $\begingroup$ OK. Im not sure how to carry out this calculation. Any hints? $\endgroup$ – john melon Jun 6 '17 at 8:57
  • $\begingroup$ Decompose the field vector in the same way as you mentioned in the question. This way you will immediately see that one of the two components will have a greater amplitude. $\endgroup$ – NDewolf Jun 6 '17 at 9:08
  • $\begingroup$ This can be done with some basic trigonometry. $\endgroup$ – NDewolf Jun 6 '17 at 9:08
  • $\begingroup$ Could you please be a bit more specific? Thank you! I know the outcome should be $I(\Theta,\phi)$ with $\phi$ meaning the phasedifference and $\Theta$ meaning the scan angle of the linear polarizer. But I fail to get this kind of dependence... $\endgroup$ – Kay Jul 10 '17 at 21:38

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