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As the filter is rotated, cos
2
θ
varies from a minimum of 0 to a maximum of 1, so the
transmitted intensity varies from a minimum of
If
I
min
()
=−
1
2
1
0
to a maximum of
max
0
0
0
11
(1
)
) .
22
I
f
I
fI
=+−
=+
The ratio of
I
max
to
I
min
is
I
I
f
f
max
.
=
+
−
1
1
Setting the ratio equal to 5.0 and solving for
f
, we get
f
= 0.67.
41. Let
I
0
be the intensity of the incident beam and
f
be the fraction that is polarized. Thus,
the intensity of the polarized portion is
f I
0
. After transmission, this portion contributes
f I
0
cos
2
to the intensity of the transmitted beam. Here
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 Spring '08
 Any
 Physics

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