PHYS142 Team Problem 08 Measuring Polarization
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Physics 142 Team Problem 08: Measuring Polarization Recall polarizing unpo-
larized light reduces its intensity by 1/2, and sending polarized light through a filter with
a polarization direction of θ relative to the light’s polarization gives a mean intensity of S̄
where
S̄ = S̄0 cos2 θ (1)
1. You and your team are tasked with analyzing the results of an experiment designed to
measure how much a certain thin film rotates the polarization of the light that passes
through it. As the drawing shows, an unpolarized laser beam of average intensity 1000
W/m2 passes through a vertical polarizer. It then passes through the film. The light
then passes through another polarizer (called the analyzer). Imagine you can rotate the
analyzer freely. .
(a) What is the intensity of the beam after it passes through the first polarizer?
Solution: We know unpolarized light loses half of its intensity when we linearly
polarize it. So 500 W/m2 .
(b) As a test, you take out the magnetic film and rotate the analyzer to 7.50 degrees
clockwise from horizontal. What intensity of light should you expect?
Solution: If we say the vertical direction is 0◦ , the analyzer will be 97.50◦
degrees. We can use Malus’ Law
S̄ = S̄0 cos2 θ (2)
500 W/m2 goes in, but only 8.52 W/m2 come out, as the analyzer is very close
to 90◦ from the polarizer.
(c) You put the magnetic film back in. As you turn the analyzer through 180◦ , sketch
what you expect to see. Label the period of the function and the axes.
Solution: Students should draw a cosine-squared curve. The period should be
180◦ .
(d) The analyzer lets through zero light when turned to 49◦ counter-clockwise from
vertical. What is the direction of the polarization axis of the film as seen from the
laser? (Be clear – you need an angle and a direction from a reference point here.)
Solution: The analyzer is set to be perpendicular to the polarization axis of
the film. The analyzer is set to at 131◦ clockwise from vertical. That means the
film is polarized 41◦ clockwise from vertical.
(e) What is the maximum amount of light this set up will allow through?
Solution: We can tune the analyzer to 41◦ CW from vertical to allow through
all the light from the film, but the film will reduce the intensity to
S̄ = S̄0 cos2 θ (3)
or 285 W/m2 .
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