Physics Solution Manual for 1100 and 2101

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Unformatted text preview: is further reduced as the polarized light passes through the analyzer. Malus' law can be used in succession at each piece of polarizing material to determine the intensity that reaches the photocell, both with and without the presence of the analyzer. SOLUTION When the analyzer is present, the average intensity reaching the photocell is equal to the average intensity that leaves the analyzer. The average intensity leaving the analyzer is, according to Malus' law, Sinsert cos 2 φ , where S i nsert is the average intensity that leaves the insert and reaches the analyzer, and φ is the relative angle between the transmission axes of the analyzer and insert. From Figure 24.24a we see that φ= 90° − θ . The average intensity of the light leaving the insert is Sinsert = S0 cos 2 θ , according to Malus’ law. Therefore, when the analyzer is present as shown in Figure 24.24b, the average intensity leaving the analyzer and reaching the photocell is 1312 ELECTROMAGNETIC WAVES 2 = S= Sphotocell insert cos φ ( S0 cos2 θ ) cos2 (90° – θ ) This expression can be solved for S 0 to determine the average intensity leaving the polarizer: Sphotocell 110 W/m 2 == = 850 W/m 2 S0 (cos 2θ ) cos 2 (90° − θ ) (cos 2 23°)cos 2 (90° − 23°) If the analyzer were removed from the setup, everything else remaining the same, the intensity reaching the photocell would be equal to the intensity that leaves the insert. Therefore, if the analyzer were removed, the intensity reaching the photocell would be Sphotocell S= S0 cos 2 θ (850 W/m 2 ) cos 2 23°= 720 W/m 2 = insert = ______________________________________________________________________________ 59. SSM WWW REASONING AND SOLUTION The sun radiates sunlight (electromagnetic waves) uniformly in all directions, so the intensity at a distance r from the sun is given by Equation 16.9 as S = P /(4 π r 2 ) , where P is the power radiated by the sun. The power that strikes an area A⊥ oriented perpendicular to the direction in which the sunlight is radiated is P ′ = SA ⊥ , according to Equation 16.8. The 0.75- m 2 patch of flat land on the equator at point Q is not perpendicular to the direction of the sunlight, however. The figure at the right shows that A⊥ = (0.75 m 2 ) cos 27° Therefore, the power striking the patch of land is P = SA⊥ P′ = (0.75 m 2 ) cos 27° 2 4π r 3.9 ×1026 W = = 920 W (0.75 m 2 ) cos 27° 11 2 4π (1.5 ×10 m) ______________________________________________________________________________ Chapter 24 Problems 1313 60. REASONING According to Equation 24.5b, the average intensity S of the infrared 2 radiation is related to the rms value of the electric field Erms by S = cε 0 Erms . According to Equation 16.8, the average power P is equal to the average intensity times the area A to which the power is being delivered, the area being that of a circle or A = π r 2 . Thus, () = SA S π r 2 . The average power is given by Equation 6.10b as the energy Q absorbed P= by the leg divided by the time t, so that t = Q / P . The energy absorbed by the leg is related to the rise in temperature ∆T by Equation 12.4 as = cm ∆T , where c is the specific heat Q capacity and m is the mass. SOLUTION a. The average intensity of the infrared radiation is S = cε 0 Erms 2 (24.5b) ( ) ( ) 2 = 8 m/s 8.85 × 10−12 C2 / N ⋅ m 2 ( 2800 N/C ) =4 W/m 2 3.0 × 10 2.1 × 10 b. The average power delivered to the leg is ( ) = 10 ( 2.1 × P =S π r SA = 2 4 )( W/m π 4.0 × 10− m 2 2 ) 2 =10 W 1.1 × 2 (16.8) c. Combining the relations t = Q / P and = cm ∆T , the time required to raise the Q temperature by 2.0 C° is Q cm ∆T 3500 J/ ( kg ⋅ C° )( 0.28 kg ) ( 2.0 C° ) = = = 18 s P P 1.1 × 102 W ______________________________________________________________________________ = t 61. REASONING AND SOLUTION The intensity of the laser light is S = P/A = cu, where u is the energy density of the light. The energy in a section of length L of the cylindrical beam is U = uAL or PL ( 0.750 W )( 2.50 m ) = = 6.25 ×10 –9 J 8 c 3.00 × 10 m/s ______________________________________________________________________________ = U 1314 ELECTROMAGNETIC WAVES 62. REASONING AND SOLUTION The polarizer will transmit a maximum intensity of 1 S + S , when its axis is parallel to the polarization direction of the polarized component P 2U of the incident light. Then the light intensity at the photocell is Smax = ( 1 SU + SP ) cos2 θ 2 (1) The polarizer transmits minimum light intensity of 1 S U when its axis is perpendicular to 2 the polarization direction of the polarized incident light, so Smin = 1 S U cos 2 θ 2 (2) Solving Equation (2) for S U gives SU = 2 Smin cos 2 θ (3) Using Equation (3) in Equation (1) and solving give SP = Smax − Smin cos 2 θ (4) Using Equations (3) and (4), we find that the percent polarization is 100 ( Smax − Smin ) 100 ( Smax − Smin ) 100 SP cos 2 θ == SP + S U Smax − Smin + 2 Smin Smax + Smin cos 2 θ ______________________________________________...
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