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Solutions12

# Solutions12 - Physics 208 Assignment 12 Solutions RT...

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Physics 208, Assignment 12, Solutions, 4/20/09, RT Problem 1. Power P laser = 3 . 0 mW. λ = 0 . 67 μ m. Beam diameter D = 10 - 3 m. (a) I laser = 3 × 10 - 3 ( π/ 4)(10 - 3 ) 2 = 3 . 8 × 10 3 W / m 2 ; I sunlight = 1 . 0 × 10 3 W / m 2 ; I lightbulb = 10 2 4 π × (0 . 1) 2 = 0 . 80 × 10 3 W / m 2 . (b) Let N be number of photons per second. P laser = Nhf = N hc λ N = P laser λ hc = 3 × 10 - 3 × 670 1240 × (1 . 6 × 10 - 19 ) = 1 . 01 × 10 16 photons / s . (c) Rayleigh cone angle 2 × 1 . 22 λ/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 =620 km=390 miles. (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 =0.39 m. 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 0 E 2 E 1 0 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

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(b) λ laser = hc E 2 - E 1 = 1240 eV ./ nm (0 . 289 - 0 . 172) eV = 10 . 6 μ m , which is far infrared radiation. (c) λ pump = hc E 2 - E 0 = 1240 eV ./ nm (0 . 289) eV = 4 . 29 μ m , which is (less far) infrared. (d) Laboratory lasers have long slender cavities defined by mirrors at the ends. This collimates their light into a slender beam. The Mars
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Solutions12 - Physics 208 Assignment 12 Solutions RT...

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