2
$\begingroup$

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?


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?

$\endgroup$
2
  • 1
    $\begingroup$ It gets linear polarized behind the polarizer. Light vectrors can always analyzed to xyz constitutes. Of course light intensity will be less. $\endgroup$
    – Markoul11
    Commented Dec 22, 2022 at 15:26
  • $\begingroup$ Related by OP: physics.stackexchange.com/questions/742184/… $\endgroup$ Commented Dec 22, 2022 at 20:29

2 Answers 2

6
$\begingroup$

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.

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.

$\endgroup$
3
  • $\begingroup$ ProfRob, thank you. You say "Circularly polarised light can be decomposed into two waves..." By this, you mean an electric and a magnetic field? You also say "only one of the waves makes it through the polarizer..." By this, you mean that only either the electric or magnetic field would go through the linear polarizer? But light is an electromagnetic wave, and it needs both components to work. If either one were left behind, wouldn't the light wave simply cease to exist? $\endgroup$ Commented Dec 22, 2022 at 19:49
  • 3
    $\begingroup$ @CuriousExplorer No, neither of those things. Polarisation always refers to the electric field of an electromagnetic wave. $\endgroup$
    – ProfRob
    Commented Dec 22, 2022 at 20:10
  • $\begingroup$ @ProfRob Is it very tedious to prove mathematically that intensity becomes exactly half of original or is it very obvious since half part of the wave is now gone?? $\endgroup$
    – SHINU_MADE
    Commented Mar 24 at 16:02
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.