Problem 4.1 (Verdeyen Problem #8.7) (a) From (7.4.7), simulated emission cross section is defined as following.

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1 Problem 4.1 (Verdeyen Problem #8.7) (a) From (7.4.7), simulated emission cross section is defined as following. σσ(νν) = AA 21 λλ 2 8ππnn 2 gg(νν) AA 21 = ssssss 1 From the figure, the emission is state 2 to state 1, which has a transition energy of 5.5eV 3.2eV = 2.3eV. (b) λ = 1.24ss EEgg(ss) = 12.4 = 0.539μμμμ = 539nnμμ 2.3 gg(νν) iiss tthss lliinnsssshaaaass ffffnnssttiiffnn, wwhiissh iiss LLffLLssnnttLLiiaann hssllss. gg(νν) = σσ(νν 0 ) = ssssss 1 (539nnμμ)2 8ππ Δνν 2ππ((νν νν 0 ) 2 + (Δνν/2) 2 ) Δν = 10GHz From (7.5.2), we have the laser gain equation. Here we assume The small signal gain coefficient is γγ 0 (νν) = σσ(νν)(nn 2 gg 2 gg 1 NN 1 ) NN 1 0 γγ 0 (νν 0 ) = σσ(νν 0 ) NN 2 = 0.01ssμμ 1 NN 2 = ssμμ 3 2 ππ 10GGGGLL = ssμμ 2 Under steady state condition, the state 2 s rate equation is following. (c) ddnn 2 ddtt = 0 = RR 2 NN 2 ( 1 ττ ττ 20 ) RR 2 = NN 2 ( 1 ττ ττ 20 ) = ss 1 ssμμ 3 From (8.3.7), saturation intensity is defined as II ss = hνν σσττ 2 To find the lifetime at state 2, the system is at equilibrium.

2 II ss = 1 = = 1 ττ 2 ττ 21 ττ nnss nnss = ss 2.3ss ssμμ = WWssμμ 2 ss (d) To achieve the desired carrier concentration at state 2, we nd pumping transition of 5.5eV. (e) PPffμμaa aaffwwssll = 5.5eV RR 2 = 2.99WWssμμ 3 The Q-factor of this system should not change betwn units. So we have the following relation. λλ = λλ λλ = Δνν νν νν = ss λλ = GGLL 10GGGGLL A = A GGLL Wavenumber is defined as the inverse of wavelength. νν = νν = 1 λλ = ssμμ 1 10GGGGLL GGLL ssμμ 1 = 0.333ssμμ 1

3 Problem 4.2 (Verdeyen Problem #8.11) (a) We can apply the procedure in section 8.3 to solve this problem. But in this problem, there is no transition specified from state 2 to state 0. I will assume the time constant is infinitely large, which means there is no transition from state 2 to 0. Assume the pumping rate is a step function and there is no stimulated emission. Hence, (8.3.2a) and (8.3.2b) can be expressed as following. ddnn 2 ddtt = RR 2(tt) NN 2 ττ 2 ddnn 1 ddtt = RR 1(tt) + NN 2 ττ 21 Solve these two differential equations, we get For t ττ 1, we have (8.3.4c). tt NN 2 (tt) = RR 20 ττ 2 (1 ss ττ 2 ) NN 1 (tt) = e tt/ττ 1 ττ 21 NN 1 (tt) = φφ 21 RR 20 ττ 1 {1 + tt tt NN 2 (tt )ssττ 1 ddtt 0 ττ 1 ττ 2 1 ττ tt ss ττ ττ 1 ττ tt ss ττ 2 } 1 2 ττ 2 For equilibrium populations, meaning t, NN 2 (tt) RR 20 ττ 2 = ssμμ 3 and NN 1 (tt) RR 20 ττ 1 = ssμμ 3. (b) It s obvious that this system is not suitable system for a CW laser. For a laser operation, we nd population inversion, which means NN 2 > NN 1. For pulsed laser, we can tolerate that the system only has certain limited time of population inversion. However, CW laser requires the system to be always in population inversion as the laser is turned on. Continuous wave operation means the system is always lasing, which is always in population inversion.

4 (c) Plot above two functions in Mathematica in Log Log scale. Concentration cm N2 t N1 t t s As we can s from the above figure, only when t < ss, there is population inversion NN 2 > NN 1. For a CW operation laser, we want the system to be always in the population inversion state, which is not the case here. To plot the small gain coefficient, we can use (7.5.2). γγ 0 = σσ (NN 2 gg 2 gg 1 NN 1 ) σσ = ssμμ 2 From the figure for this problem, the degeneracy of each energy level are gg 1 = 2JJ = 5 and gg 2 = 2JJ = 3.

5 Problem 4.3 (Verdeyen Problem #8.12) (a) First, I started from (8.3.2a). It has bn specified that we have negligible population in state 1. Hence, the rate equation is the following. ddnn 2 ddtt = RR 2(tt) NN 2 σσ(νν)ii νν NN ττ 2 hνν 2 Assume we re in equilibrium state of the system. ddnn 2 ddtt = 0 NN 2 (tt) = RR 2 (tt) RR 2 (tt)ττ 2 ( 1 ττ + σσ(νν)ii = νν 2 hνν ) (1 + σσ(νν)ii νν hνν ττ 2 ) = RR 2(tt)ττ 2 (1 + II νν IIss ) The spontaneous emission from the side should be always proportional to the population at state 2, meaning PP NN 2. Assume the system is at constant pumping rate. PP = RR 2(tt)ττ 2 (1 + II II ss ) Because the one of the mirror is blocked, there is no cavity to confine the photons to generate stimulated emission, meaning II νν = 0. PP 0 = RR 2(tt)ττ 2 (1 + II = 0 = RR 2 (tt)ττ 2 II ) ss PP = PP 0 (1 + II II ss ) (b) II = 100WW/ssμμ 2 PP PP 0 = II II ss =2 II II ss = 1 II ss = 100WW/ssμμ 2

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10 Problem 4.5 (Verdeyen Problem #8.20) From (8.7.3a), we have output intensity of ASE. II + νν, ll gg = 8ππnn2 hνν 3 NN 2 ddω ss 2 NN 2 gg [GG 0 (νν) 1] 2 gg NN 1 4ππ 1 To determine the FWHM, first we focus on the [GG 0 (νν) 1] term. The reason I ignore 8ππnn2 hνν 3 cc 2 is because it s a monotonically increasing term with frequency. GG 0 (νν) = ss γγ 0(νν)ll gg GG 0 (νν 0 ) = ss γγ 0(νν 0 )ll gg ( νν h /2) 2 γγ 0 (νν)ll gg = γγ 0 (νν 0 ) (νν νν 0 ) 2 + ( νν h /2) 2 ll gg FWHM is defined as the intensity drops to half of the maximum value. Change variable for the above equations. Let [GG 0 (νν) 1] = 1/2 [GG 0 (νν 0 ) 1] νν HH = νν νν 0 νν h /2 1 γγ 0 (νν)ll gg = γγ 0 (νν 0 ) ( νν HH ) ll gg γγ 0 (νν 0 ) ll gg ss( νν HH ) = 1/2 [GG 0 (νν 0 ) 1] γγ 0 (νν 0 ) ll gg ss( νν HH ) = 0.5(ss γγ 0 (νν 0 )ll gg 1) γγ 0 (νν 0 ) ll gg ss( νν HH ) = 0.5(GG 0 (νν 0 ) 1) γγ 0 (νν 0 ) ll gg ss( νν HH ) 2 +1 = 0.5(GG 0 (νν 0 ) + 1) ln (GG 0 (νν 0 )) ( νν HH ) = ln (0.5(GG 0(νν 0 ) + 1)) ( νν HH ) 2 = ln (GG 0 (νν 0 )) ln (0.5(GG 0 (νν 0 ) + 1)) 1 To switch gain to db unit, we nd to change the gain variable. Let s assume a new variable in db unit, t. tt(dddd) = 10 LLffgg(GG 0 (νν 0 ))

11 0.1tt = LLffgg(GG 0 (νν 0 )) tt = GG 0 (νν 0 ) So we substitute this new variable into the equation. ln (10 0.1tt ) νν HH = ln (0.5(10 0.1tt + 1)) 1 Plot this function with Mathematica.

12 Problem 4.6 (Verdeyen Problem #21) (a) From (6.4.2b), we have the photon lifetime in passive cavity. ττ pp = 2nndd/ss 1 RR 1 RR 2 Here we nd to add the additional loss from lens and gain medium. (b) ττ pp = 3dd/ss (0.98) 2 (0.97) 2 = nnssssss From (6.4.4), we can calculate the quality factor. (c) (d) ωω 0 = 2ππff 0 = 2ππss λλ 0 FFFFRR = QQ = ωω 0 ττ pp = LLaadd/ss QQ = ss 3nndd = GGLL Minimum gain has to at least overcome the loss in the cavity. (e) ss γγ 0 (νν 0 ) ll gg RR 1 RR 2 RR 3 TT 1 TT 2 (0.97) 2 > 1 γγ 0 (νν 0 ) > ln (RR 1RR 2 RR 3 TT 1 TT 2 (0.97) 2 ) ll gg γγ 0 (νν 0 ) > ssμμ 1 We have the FSR and optical frequency from above calculations. Now I plot the cavity mode, gain spectrum, and loss level. We can s how many modes are above the loss. There are 9 modes can overcome the loss, which are above threshold.

13 (f) From (7.4.7), simulated emission cross section is defined as following. λλ 2 σσ(νν 0 ) = AA 21 8ππnn 2 gg(νν 0) σσ(νν 0 )= ssμμ 2 (g) From (7.5.4), we have the absorption cross section definition. σσ aaaass (νν) = (gg2/gg1)σσ ssttiiii (νν) σσ aaaass (νν 0 ) = 3 5 σσ ssttiiii(νν 0 ) = ssμμ 2 (h) (1) This problem is very similar to Problem 5.1 in Verdeyen. Let s draw the equivalent lens waveguide. By symmetry, the minimum beam waist, z=0, should appear in the middle of the two lenses, which is exactly the position of the top mirror. (2) Similar to (3.5.1), we nd to find the phase match condition for this cavity. kk 3dd/2 (1 + μμ + aa)ttaann 1 3dd 2LL 0 = qqππ 2ππνν ss kk = 2ππ λλ = 2ππνν ss 3dd 2(1 + μμ + aa)ttaann 1 3dd 2LL 0 = 2qqππ

14 (3) νν ii,pp,qq = ss 6ππdd (2qqππ + 2(1 + μμ + aa)ttaann 1 3dd ) 2LL 0 νν 1,0,qq νν 0,0,qq = LL 0 = 0.58μμ ss 6ππdd 2ttaann 1 3dd 2LL 0 = 58MMGGLL

15 Problem 4.7 (Verdeyen Problem #8.35) (a) NN 2 NN 0 = 1 8 NN 1 NN 0 = 1 4 NN 0 + NN 1 + NN 2 = NN = ssμμ NN 0 = ssμμ 3 NN 0 = ssμμ 3 NN 1 = ssμμ 3 NN 2 = ssμμ 3 (b) Skip because there is no absorption cross section. (c) (1) We can modify (8.3.2ab) to get the rate equations. There is no pumping to state 1, so RR 1 = 0. Neglect the stimulated emission. (2) ddnn 2 ddtt = RR 2 NN 2 NN 2 ττ 2 ddnn 1 ddtt = NN 2 NN 2 NN 1 NN 1 ττ 2 Assume the system is in steady state, meaning all derivatives should be equal to zero. NN 2 NN 2 NN 2 NN 2 ττ 2 = RR 2 ττ 1 ττ 2 = NN 1 NN 1 For optical transparency, it means the populations in both state 1 and 2 should be equal. So there is no absorption. ττ 1 NN 1 NN 1 = RR 2 ττ 1 NN 2 NN 2 = RR 2 ττ 2 NN 2 = NN 1 NN 2 = RR 2 ττ 2 + NN 2 RR 2 ττ 2 + NN 2 NN 1 = RR 2 ττ 1

16 (3) RR 2 = NN 2 NN 1 ττ 1 ττ 2 RR 2 = 1 ττ 1 NN 2 NN 1 1 ττ 2 /ττ 1 From (a), we know the difference betwn equilibrium concentrations of state 2 and 1 are less than zero. NN 2 NN 1 < 0 It s obvious that we nd ττ 2 /ττ 1 > 1 to have a positive pumping rate for optical transparency. Absorption coefficient should be proportional to NN 1 NN 2. Because it nd more populations in the lower state in order to have absorption. αα = σσ(nn 1 NN 2 ) αα = σσ(nn 1 NN 2 + RR 2 (ττ 1 ττ 2 )) For simplicity, I am using following parameters for the plot. NN 1 = 4 NN 2 = 2

17 Problem 4.8 (Verdeyen Problem #9.1) (a) From (7.6.17), we have the Doppler broadening linewidth. (b) νν 0 = νν DD = 8kkTTllnn2 MMss 2 νν 0 TT = = 573KK MM = 20aaμμff ss 6328A = GGLL νν DD = 1.81GGGGLL From (7.2.4b), we have the number of modes per unit volume in blackbody radiation. aa(vv)ddvv = 8ππnn3 vv 2 ddvv ss 3 (c) NN = 8ππnn3 vv 2 νν DD ss 3 = FFFFRR = ss 2nndd = 150MMGGLL There are 13 modes within ± νν DD 2. (d) νν DD 2 = 905MMGGLL Combine the two previous results. We can calculate the probability. (e) and (f) E2 E eV E0 1.95eV 18.71eV aa = 13 = TTLLaannssiittiiffnn ssnnssllgg = = ss UUaaaassLL ssttaattss ssnnssllgg = ss ηη = = 9.44%

18 (g) σσ(νν 0 ) = AA 21 λλ 2 gg(νν 0 ) = 4llnn2 ππ 8ππnn 2 gg(νν 0) = ssμμ 2 1 = νν DD (h) σσ(νν 0 ) = ssμμ 2 γγ 0 = σσ(νν 0 ) NN 2 gg 2 gg 1 NN 1 = 0.05μμ 1 gg 2 = 2JJ = 3 gg 1 = 2JJ = 5 NN 1 = ssμμ 3 NN 2 = ssμμ 3