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Elliptical polarization

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Circularpolarization example

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As a result, only one of the waves makes it through the polariser and the transmitted light will be linearly polarised and of half the intensity of the original beam. This will be true, irrespective of how you rotate the polariser.

Circularly polarizedlightequation

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What does circularly polarizedlightlook like

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The reason I’m asking is because I’ve heard that in photography linear polarizers can cut through smog but not through fog, which generates circularly polarized light. This seems strange, because one would assume that the linear polarizer would absorb all of the circularly polarized light and would thus cut through the fog. Why is this not so?

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Circularly polarized light is composed of 50% linearly horizontally polarized light and 50% linearly vertically polarized light. The two linearly polarized components are 90 degrees out of phase. If perfectly circularly polarized light passes through a perfect linear polarizer that transmits horizontal (vertical) polarized light then the output will be perfectly linearly horizontal (vertical) polarized light with 50% of the power of the incident circular polarized light. The answer is the same whether the incident light is right- or left-hand circularly polarized.

Right-circularly polarizedlight

Circular polarised lightmeaning

I have seen a lot of examples of what happens when circularly polarized light passes through a circular polarizer composed of a quarter-wave plate and a linear polarizer, but what would happen to the circularly polarized light if it passed through only the linear polarizer without a quarter-wave plate?

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Circularly polarised light can be decomposed into two electromagnetic waves, with their respective electric fields linearly polarised at right angles to each other, of equal amplitude but 90 degrees out of phase. One of the polarisations can be chosen to line up with the polariser and the other at right angles to it. e.g. $${\bf E} = E_0 \sin(\omega t - kz)\ {\bf i} + E_o \sin(\omega t -kz -\pi/2)\ {\bf j}\ .$$ Note that you can choose any pair of orthogonal unit vectors that are in a plane at right angles to the wave motion. So you can choose to have one of those unit vectors be along the axis of the linear polariser.

Elliptically polarizedlight

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