Solutions12 - Physics 208, Assignment 12, Solutions,...

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Unformatted text preview: Physics 208, Assignment 12, Solutions, 4/20/09, RT Problem 1. Power P laser = 3 . 0 mW. = 0 . 67 m. Beam diameter D = 10- 3 m. (a) I laser = 3 10- 3 ( / 4)(10- 3 ) 2 = 3 . 8 10 3 W / m 2 ; I sunlight = 1 . 10 3 W / m 2 ; I lightbulb = 10 2 4 (0 . 1) 2 = 0 . 80 10 3 W / m 2 . (b) Let N be number of photons per second. P laser = Nhf = N hc N = P laser hc = 3 10- 3 670 1240 (1 . 6 10- 19 ) = 1 . 01 10 16 photons / s . (c) Rayleigh cone angle 2 1 . 22 /D =0.0016. (d) The distance to moon is 3 . 8 10 8 m. At that distance the beam diameter is 0 . 0016 3 . 8 10 8 =620km=390miles. (e) To reduce the beam diameter by a factor of 390 it would be necessary to increase the laser diameter (proportionally) to 10- 3 390 =0.39m. Problem 2. The energy level diagram for CO 2 is shown. Presumably there are strong absorption processes populating excited levels 1 and 2, and strong spontaneous emission from level 1 back to ground state 0. But spontaneous emission from level 2 is weak. This makes it possible for there to be more molecules in level 2 than in level 1. E E 2 E 1 0.172 eV 0.289 eV (a) Only in the presence of such a population inversion are there molecules in a higher level, in this case level 2, waiting a long enough time for there to be appreciable probability of radiation of energy hf = E 2- E 1 coming along to stimulate de-excitation transi- tion to the lower level. The competing inverse absorption rate from level 1 to level 2 is small because of the population inversion. 1 (b) laser = hc E 2- E 1 = 1240 eV ./ nm (0 . 289- . 172) eV = 10 . 6 m , which is far infrared radiation....
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Solutions12 - Physics 208, Assignment 12, Solutions,...

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