Physics Solution Manual for 1100 and 2101

7 b letters and words held up to a mirror are reversed

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Unformatted text preview: the 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 one-half 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 one-half 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.

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