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transmission axis are initially parallel to each other, and the maximum amount of light is
transmitted. As the polarizing material is rotated, the intensity of the transmitted light
decreases in accord with Malus’ law. Chapter 24 Problems 1301 If the incident light is polarized along the y axis, the direction of polarization and the
transmission axis are initially perpendicular to each other, and no light is transmitted. As the
polarizing material is rotated, the average intensity of the transmitted light increases.
SOLUTION
a. Since the incident light is unpolarized, the average intensity of the transmitted light is
onehalf the average intensity of the incident light. Therefore, for both α = 0° and 35°, we
have
=
S =
S0 1
2 1
2 =
( 7.0 W/m )
2 3.5 W/m 2 b. When the incident light is polarized along the z axis, the direction of polarization and the
transmission axis are initially parallel to each other. Therefore, the angle α is the same as
the angle θ between the transmission axis of the polarizer and the direction of the
polarization. According to Malus’ law (Equation 24.7), the average intensity of the
transmitted light is given by = S=
S
cos 2 θ
0 cos 0°
( 7.0 W/m )= = S=
S
cos 2 θ
0 =
( 7.0 W/m ) cos 35° 2 2 2 2 7.0 W/m 2
4.7 W/m 2 c. When the incident light is polarized along the y axis, the direction of polarization and the
transmission axis are initially perpendicular to each other. The angle θ in Malus’ law is the
angle between the direction of polarization (along the y axis) and the transmission axis
(measured relative to the z axis). It is related to the angle α according to θ = 90.0° – α.
The average intensity of the transmitted light is, therefore, S
cos θ
= S=
0 ( 7.0 W/m ) cos 2 =
( 90.0° − 0° ) S
cos 2 θ
= S=
0 ( 7.0 W/m ) cos 2 =
( 90.0° − 35° ) 2 2 2 0 W/m 2 2.3 W/m 2 The table below summarizes the results:
Average Intensity of Transmitted Light
α = 0°
α = 35° Incident Light
(a) Unpolarized
3.5 W/m2
3.5 W/m2
(b) Polarized parallel to
7.0 W/m2
4.7 W/m2
z axis
(c) Polarized parallel to
0 W/m2
2.3 W/m2
y axis
______________________________________________________________________________ 1302 ELECTROMAGNETIC WAVES 41. SSM REASONING AND SOLUTION The average intensity of light leaving each polarizer is given by Malus' Law : S = S0 cos 2 θ (Equation 24.7). Solving for the angle θ
and noting that S = 0.100 S0 (since 90.0% of the intensity is absorbed), we have θ
= cos − 1 0.100 S0
=
S0 71.6° ____________________________________________________________________________________________ 42. REASONING Malus’ law, S = S0 cos 2 θ (Equation 24.7), relates the average intensity S0 of polarized light incident on the polarizing sheet to the average intensity S of light
transmitted by the sheet, where θ is the angle between the polarization axis of the incident
light and the transmission axis of the polarizing sheet. The incident light is horizontally
polarized, so the angle θ is measured from the horizontal, and is, therefore, the angle we
seek.
SOLUTION Solving S = S0 cos 2 θ (Equation 24.7) for θ, we obtain
=
cos 2 θ S
=
or
cos θ
S0 S
S
= cos −1 θ
or S
S0 0 (2) Substituting the given values of the average incident and transmitted intensities yields 0.764 W/m 2 θ=
= cos −1 0.883 W/m 2 21.5 43. REASONING If the intensity of the unpolarized light is I0, the intensity of the polarized
light leaving the polarizer is
insert is 1I
20 1I .
20 By Malus’ law, the intensity of the light leaving the cos 2θ . From the results of Conceptual Example 8, the intensity of light leaving the analyzer is 1I
20 cos 2θ sin 2θ . SOLUTION The intensity I of light that reaches the photocell is
=
I 1=
I cos 2θ sin 2 θ
2o 1 (150
2 =
W/m 2 )cos 2 30.0° sin 2 30.0° 14 W/m 2 ______________________________________________________________________________ Chapter 24 Problems 1303 44. REASONING
Drawing A The transmission axes of the polarizer and analyzer are parallel to each other,
so all the light transmitted by the polarizer is completely transmitted by the analyzer.
Drawing B The transmission axes of the polarizer and analyzer are perpendicular to each
other, so no light is transmitted through the analyzer.
Drawing C The transmission axes of the polarizer and analyzer make an angle of 30.0°
with respect to each other. Thus, some of the light transmitted by the polarizer, but not all,
is transmitted through the analyzer.
Therefore, we expect the transmitted intensities to be in the following decreasing order
(largest first): A, C, B.
SOLUTION Since the incident light is unpolarized, the average intensity S1 of the light
transmitted by the polarizer is onehalf the average intensity S0 of the incident light, or =
S1 =
S0 1
2 1
2 =
( 48 W/m )
2 24 W/m 2 . The average intensity S 2 of the light transmitted by the analyzer is given by Malus’ law, Equation 24.7, as S 2 = S1 cos θ , where θ is the angle
between the direction of polarization and the transmission axis. The average intensity of the
transmitted beams for each of the three ca...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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