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Unformatted text preview: 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: = E dxdydt = Idxdydt dxdy c n dt = n c I (b) See Gamma Toms course notes for the derivation dI dz = I 1 + I/I sat (c) See Gamma Toms course notes for the derivation ln I out I in + I out I in I sat = L (d) See Gamma Toms course notes for the derivation I out = I in e L (e) See Gamma Toms course notes for the derivation I out = I in + I sat L 2 2Level System Verdeyen 7.13 (a) dN 2 dt = I p h " N 2 g 2 g 1 N 1 # N 2 dN 1 dt = I p h " N 2 g 2 g 1 N 1 # + N 2 (b) First set dN 2 /dt = 0. Solving for N 2 we get: N 2 = g 2 g 1 N 1 N 2 h I p As I p , clearly N 2 ( g 2 /g 1 ) N 1 . 1 (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 = h 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, these2 due to high energy inelastic collisions is ignored....
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This note was uploaded on 02/28/2012 for the course ECE 455 taught by Professor Eden,j during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Eden,J

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