I'm assuming that the original light under consideration is
unpolarized. In this case, the maximum attenuation depends on the
order in which the polarizing plates act.
To see this, one needs to know that a circular polarizer is simply a
linear polarizer followed by a 1/4-wave plate with the fast direection
oriented at a 45 degree angle polarization direction of the preceeding
polarizer (see figure at
<http://www.fas.harvard.edu/~scdiroff/lds/LightOptics/CircularPolarization/CircularPolarization.html>
If unpolarized light is first passed through your linear polarizing
filter then the exiting light will be linearly polarized, let's say in
the horizontal plane and with half the intensity of the incoming
light. Now, if the circular polarizer is oriented such that its
linear polarizing element is oriented in the vertical plane, none of
the linearly polarized light from the first polarizer will be passed,
and the attenuation will be 100%. This is why circular polarizers
(properly oriented) can be used in photography to attenuate the
linearly polarized light produced by reflections.
If, instead, you pass unpolarized light through the circular
polarizer, it will emerge with half the intensity on the incident
light due to the effect of the linear polarizing element in the
circular polarizer. The next element, the 1/4-wave plate, has no
effect on the intensity, but it has the effect of "reapportioning" the
incident intensity into two, perpendicular, linearly polarized waves
that are 90 degrees out of phase with one another. Passing this light
through the linear polarizer will result in the loss of 1/2 of the
light incident on that polarizer. (It does not matter how this
polarizer is oriented.) So, overall, 1/4 of the original light will
be transmitted, and that light will be linearly polarized in the
direction of the final linear polarizer. In terms of attenuation, the
original unpolarized incident light will be 75% attenuated by this
arrangement.
If you know matrix algebra, all this can be easily proven using "Jones
matrices" (see <http://www.lunatechnologies.com/files/23jonesintro.pdf>,
or <http://optics.byu.edu/PrevText/BYUOptics04.pdf>, for example. The
fact that the order of the optical elements makes a difference to the
final result is simply a reflection of the fact that in matrix
algebra, multiplication of matrices is not commutative (A*B does not,
in general equal B*A). |
