Solutions to Problems : Chapter 25

Solutions to Problems : Chapter 25 Problems appeared on the end of chapter 25 of the Textbook (Problem 16, 30, 42, 44, 58, 60, 66, 72) 16. Picture the Problem: Radio signals travel from Earth to a distant spacecraft. Strategy: Divide the distance by the speed of light to calculate the time for the signal to reach the craft. Δt = Solution: Calculate the time: 4.5 × 1012 m d = = 1.5 × 10 4 s c 3.00 × 108 m/s Insight: This time delay is 4 hours and 10 minutes. When NASA sends a signal to the craft it takes 8 hours and 20 minutes for NASA to receive a confirmation from the satellite. 30. Picture the Problem: The radiation emitted by humans has a wavelength of about 9.0 µm. Strategy: Solve equation 25-4 to calculate the frequency. Then compare the frequencies to the ranges given in section 25-3 of the text. f = Solution: 1. (a) Calculate the frequency: c λ = 3.00 × 108 m/s = 3.3 × 1013 Hz 9.0 × 10−6 m 2. (b) This frequency falls in the infrared range (1012 Hz to 4.3 × 1014 Hz). 42. Picture the Problem: A sinusoidal electric field has a maximum value of 65 V/m. Strategy: Divide the peak electric field by the square root of two to calculate the rms magnitude of the electric field. Solution: Calculate the rms electric field: Erms = Emax 2 = 65 V/m 2 = 46 V/m Insight: The rms magnetic field for this wave is 1.5×10−7 T. 44. Picture the Problem: A given electromagnetic wave has a maximum intensity of 5.00 W/m2. Strategy: Solve equation 25-10 for the maximum electric field. Solution: Calculate Emax : 2 I max = cε 0 Emax Emax = I max = cε 0 5.00 W/m 2 ( 3.00 ×10 m/s )(8.85×10 8 -12 C2 / N ⋅ m 2 ) = 43.4 V/m Insight: Verify for yourself that the maximum magnetic field for this wave is 0.145 µT. 58. Picture the Problem: A 75.0-W lightbulb emits electromagnetic waves uniformly in all directions. Strategy: Use equation 14-7 to calculate the intensity of the light 3.5 m from the source. Insert the intensity into equation 25-10 to calculate the rms electric field, and then solve equation 25-9 for the magnetic field. P 75 W Solution: 1. Divide the power by I av = av = = 0.4872 W/m 2 2 A 4π ( 3.50 m ) area: 2. Calculate the electric field: 2 I av = cε 0 Erms I av = cε 0 Erms = 3. Find the magnetic field: Brms = 0.4872 W/m 2 ( 3.00 ×10 m/s )(8.85 ×10 8 −12 C2 / Nm 2 ) = 13.5 V/m Erms 13.55 V/m = = 45.2 nT c 3.00 × 108 m/s Insight: The magnetic field could also have been calculated using I av = c μ0 2 Brms (equation 25-10). 60. Picture the Problem: A 2.8-mW laser beam has a diameter of 2.4 mm. Strategy: Write the intensity as the average power divided by the area of the beam. Write the intensity in terms of the rms electric field using equation 25-10 and solve for the electric field. Solution: 1. Write I av in terms of Erms : I av = 2. Solve for the electric field: Erms = Pav 2 = cε 0 Erms A = Pav Acε 0 2.8 × 10−3 W π (1.2 × 10−3 m) ( 3.00 × 108 m/s )( 8.85 × 10−12 C 2 /N ⋅ m 2 ) 2 Erms = 0.48 kV/m Insight: Note that the electric field is inversely proportional to the beam diameter. If the diameter is doubled to 4.8 mm, the electric field will drop to 240 V/m. 66. Picture the Problem: The image shows unpolarized light incident upon two polarizers, the transmission axes of which are oriented at some angle with respect to each other. Strategy: Set the intensity after the first polarizer equal to half the intensity before (equation 25-14). Use Malus’ Law (equation 25-13) to calculate the intensity after the second polarizer. Divide the result by the initial intensity to determine the relative intensity. Solution: 1. Calculate the intensity after the first polarizer: I1 = 2. Calculate the intensity after the second polarizer: I 2 = I1 cos 2 θ = 3. Divide the final intensity by the initial: I2 1 = cos 2 30.0° = 0.375 I0 2 1 I0 2 1 I 0 cos 2 θ 2 Insight: The exact orientation of the two polarizers is not important, only the relative orientation of their transmission axes. 72. Picture the Problem: The image shows unpolarized laser light passing through three polarizers. Strategy: Use equation 25-14 to calculate the intensity after the first polarizer. Then use Malus’s Law (equation 25-13) to calculate the intensity as the light passes through each of the other polarizers. 1 I0 2 Solution: 1. (a) Use equation 25-14 to calculate I at point A: I= 2. (b) Use Malus’s Law to calculate the intensity at point B: ⎛1 ⎞ I = ⎜ I 0 ⎟ cos 2 30.0° = 0.375 I 0 ⎝2 ⎠ 3. (c) Use Malus’s Law to calculate the intensity at point C: I = ( 0.375I 0 ) cos 2 ( 90.0° − 30.0° ) = 0.0938I 0 4. (d) Use Malus’s Law to calculate the intensity at point C, with the second polarizer removed: ⎛1 ⎞ I = ⎜ I 0 ⎟ cos 2 90.0° = 0 ⎝2 ⎠ Insight: The second filter rotates the polarization so that some light can pass through the third filter.