Message-Id: <3.0.32.19970330075958.006c644c@icefog.polarnet.com>
Date: Sun, 30 Mar 1997 08:01:00 -0900
To: pinhole@exploratorium.edu
From: "J. Lahr" <jlahr@polarnet.com>
Subject: Polarizing filters corrected
I woke up at 6AM this morning with the realization that I had made a
mistake in the previous Email on this subject.
Consider two ideal polarizing filters. Let the vector amplitude
of the polarized light that has passed through the first filter be
A. If this light strikes another filter that is rotated by the angle
theta, then the vector component A cos(theta) will pass through that filter
while the perpendicular component A sin(theta) will be blocked. The
intensity of light, however, is proportional to the vector amplitude squared,
so the following amplitudes and intensities would prevail:
LOCATION OF LIGHT VECTOR AMPLITUDE INTENSITY
After 1st polarizer A A*A
After 2nd polarizer A cos(theta) A*A cos(theta)*cos(theta)
Blocked by 2nd polarizer A sin(theta) A*A sin(theta)*sin(theta)
Note that since sin squared plus cos squared equals one, the intensity of the
light passed by the polarizing filter plus the intensity of the light blocked
by the polarizing filter equals the intensity of the incident light.
The following has been corrected from my previous Email.
Understanding the effect that the addition of polarizing filters between
two that are oriented at right angles has requires a an understanding of
vectors, as Paul D. pointed out. If a polarizer worked by totally blocking
all but the component of incident light that was perfectly aligned with
the filter, then only a negligible amount of unpolarized light would make
it through a single filter and no light would make it through two filters
unless they were perfectly aligned. Actually, if unpolarized light is
passed through a single polarizing filter, its intensity will be reduced
by one half. This is slowly sinking into my brain.
It's interesting to consider placing n filters between two with 90 degree
rotation. For example, for n = 2, each would be rotated 30 degrees from
the previous filter (90 divided by n+1). For "perfect" polarizers it
works out that the larger the value of n the more light exits the final
filter. The equation I derived is:
F = [cos(90/(n+1))] ** (2n+2) (The first term raised to the 2n+2 power)
where F is the fraction of light that makes it through the first filter
that also makes it through the final filter. A few values of F are:
n F
0 0.0
1 0.25
2 0.42
3 0.53
10 0.80
I would guess that real polarizing filters reduce the amount of light
with all polarizations by some amount, and that this added reduction
would eventually prevent more and more filters from actually increasing
the light output.
Hope I've got things right this time. If not, Paul D., please let me know
where I've gone off the track!
JCLahr
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