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# HW6_soln - ECE 455 Optical Electronics Homework 5 Solutions...

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ECE 455: Optical Electronics Homework 5 Solutions Due: Oct. 15, 2009 1 Derivations (a) Consider the flux of energy accross the dx - dy plane in a time dt . This energy is: E = Idxdydt In the time dt , the light has moved a distance dz = c n dt . Therefore the energy density is: ρ = E dxdydt = Idxdydt dxdy c n dt = n c I (b) See Gamma Tom’s course notes for the derivation dI dz = γ 0 I 1 + I/I sat (c) See Gamma Tom’s course notes for the derivation ln I out I in + I out - I in I sat = γ 0 L (d) See Gamma Tom’s course notes for the derivation I out = I in e γ 0 L (e) See Gamma Tom’s course notes for the derivation I out = I in + γ 0 I sat L 2 2-Level System Verdeyen 7.13 (a) dN 2 dt = - σI p " N 2 - g 2 g 1 N 1 # - N 2 τ dN 1 dt = σI p " N 2 - g 2 g 1 N 1 # + N 2 τ (b) First set dN 2 /dt = 0. Solving for N 2 we get: N 2 = g 2 g 1 N 1 - N 2 I p στ As I p → ∞ , clearly N 2 ( g 2 /g 1 ) N 1 . 1

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(c) Again assume a steady state distribution. Set N 1 = 2 N 2 in the equation we found in part A. Then we can solve for I p . I p = τσ 2 g 2 g 1 - 1 ! - 1 = 7 . 78 W/cm 2 (d) In writing the rate equations, there are many terms that are neglected becauase they are so small compared to the optical effects. For example, the rate of excitation from state 1 to state 2 due to high energy inelastic collisions is ignored. When the pump beam is turned off, these effects are all the only mechanisms that exchange particles between populations 1 and 2. The
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