Physics Solution Manual for 1100 and 2101

6 reasoning and solution a the height of the shortest

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Unformatted text preview: ses is 2 = S1 cos 2 θ = S2 A = S1 cos 2 θ = S2 B = S1 cos 2 θ S2 = C cos 0° ( 24 W/m ) = 24 W/m = ( 24 W/m ) cos 90° 0 W/m 30.0° ) ( 24 W/m ) cos ( 60.0° −= 2 2 2 2 2 2 2 2 18 W/m 2 ______________________________________________________________________________ 45. REASONING In experiment 1, the light falling on the polarizer is unpolarized, so the average intensity S transmitted by it is S = 1 S ′ , where S ′ is the intensity of the incident 2 light. The intensity of the light passing through the analyzer and reaching the photocell can be found by using Malus’ law. In the second experiment, all of the polarized light passes through the polarizer, so the average intensity reaching the analyzer is S ′ . The intensity of the light passing through the analyzer and reaching the photocell can again be found by using Malus’ law. By setting the intensities of the light reaching the photocells in experiments 1 and 2 equal, we can determine the number of additional degrees that the analyzer in experiment 2 must be rotated relative to that in experiment 1. 1304 ELECTROMAGNETIC WAVES SOLUTION In experiment 1 the light intensity incident on the analyzer is S0 = 1 S ′ . 2 Malus' law (Equation 24.7) gives the average intensity S1 of the light reaching the photocell as = S0 cos 2 60.0° 1 S ′ cos 2 60.0° S1 =2 In experiment 2 the incident light is polarized along the axis of the polarizer, so all the light is transmitted by the polarizer. Thus, the light intensity incident on the analyzer is S0 = S ′ . Malus' law again gives the average intensity S2 of the light reaching the photocell as = S0 cos 2 θ S ′ cos 2 θ S2 = Setting S1 = S2 , we have 1 2 S ′ cos 2 60.0° = S ′ cos 2 θ Algebraically eliminating S ′ from this equation, we have that 1 cos 2 60.0° = 2 cos 2 θ or = cos −1 θ ) ( 1 cos 2 60.0= ° 2 69.3° The number of additional degrees that the analyzer must be rotated is 69.3° − 60.0° = 9.3° . The angle θ is increased by the additional rotation. _____________________________________________________________________________________________ 46. REASONING The rms value Erms of the electric field is related to the average intensity S 2 of the light by S = cε o Erms (Equation 24.5b). The average intensity S of the light transmitted by the polarizer is related to the incident intensity S0 by Malus’ law, S = S0 cos 2 θ , where θ is the angle between the transmission axis and the direction of polarization. These two relations will allow us to determine the rms value of the electric field. SOLUTION Combining the two equations given above and solving for the rms value of the electric field, we have = Erms S0 cos 2 θ = cε 0 (15 W/m ) cos m/s ) 8.85 × 10 2 2 25° = 68 N/C 2 2 C / N⋅m ______________________________________________________________________________ (3.0 × 10 8 −12 ( ) Chapter 24 Problems 47. 1305 SSM WWW REASONING The average intensity of light leaving each analyzer is given by Malus' Law (Equation 24.7). Thus, intensity of the light transmitted through the first analyzer is S1 = S0 cos 2 27° Similarly, the intensity of the light transmitted through the second analyzer is S 2 = S1 cos 2 27° = S0 cos 4 27° And the intensity of the light transmitted through the third analyzer is S 3 = S2 cos 2 27° = S 0 cos 6 27° If we generalize for the Nth analyzer, we deduce that S N = S N –1 cos 2 27° = S 0 cos 2 N 27° Since we want the light reaching the photocell to have an intensity that is reduced by a least a factor of one hundred relative to the first analyzer, we want S N / S 0 = 0.010 . Therefore, we need to find N such that cos 2 N 27° = 0.010 . This expression can be solved for N. SOLUTION Taking the common logarithm of both sides of the last expression gives log 0.010 = 20 2 log (cos 27 °) ______________________________________________________________________________ 2 N log(cos 27 °) = log 0.010 or N= 48. REASONING No light intensity will pass through two adjacent polarizers that are in a crossed configuration, that is, whose transmission axes are oriented perpendicular to one another. With this in mind, let’s remove the polarizers one by one (see the following drawing). When A is removed, no two of the remaining adjacent polarizers are crossed. When B is removed, A and C are left in a crossed configuration. When C is removed, B and D are left in a crossed configuration. When D is removed, no two of the remaining adjacent polarizers are crossed. Thus, when sheet B or C is removed, the intensity transmitted on the right is zero, and when sheet A or D is removed, the intensity transmitted on the right is greater than zero. 1306 ELECTROMAGNETIC WAVES horizontal vertical 30.0º 60.0º Light D C B A We can anticipate that the greater intensity is transmitted on the right when sheet D is removed. To begin with, we note that the polarization directions of the light striking B and C are the same, no matter whether A or D is removed, with the result that B and C absorb the same fraction of t...
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