Problem 3.1 (Verdeyen 5.13) First, I calculate the ABCD matrix for beam traveling through the lens and space.

Transcription

1 Problem 3. (Verdeyen 5.3) First, I calculate the ABCD matrix for beam traveling through the lens and space. T = dd 0 0 dd 2 ff 0 = dd 2 dd ff 2 + dd ( dd 2 ff ) dd ff ff Aording to ABCD law, we can have the following relation for beam waist. qq 2 = AAqq + BB CCqq + DD qq 2 (zz) = zz + jj qq (zz) = zz + jj Since it requires mode-matching, the two minimum beam waists should satisfy the ABCD law. So we have qq 2 (0) = AAqq (0) + BB CCqq (0) + DD = AA jj + BB CC jj + DD = jj AA jj + BB = DD jj CC B = CC A = DD BB CC = AA DD = Substitute the above ABCD into the equations. = ffdd 2 + dd (ff dd 2 ) zz dd 2 02 ff ff dd 2 = dd = ff dd ff (ff dd ) + ff = dd 2 = ff ff (ff dd ) + dd ff ff (ff dd ) = ff ff ff + ff dd + dd (ff dd ) ff ff ff + 2ff dd zz zz 2 02 dd 0 zz = 0 dd 2 2ff dd + ff ff ff = 0

2 2ff zz ± 4(ff 0 zz ) zz ( ff ff ff 0 zz ) 0 dd = 2 = ff ± ff 2 ( ff ff ff ) = ff ± ff 2 (( ) 2 ff 2 ( )) = ff ± ff 2 ( ) 2 + ff 2 ( ) = ff ± ff 2 ( ) 2 dd 2 = ff (ff dd ) = ff ff ff ± ff 2 ( ) 2 = ff ff 2 ( ) 2 = ff ± ff 2 ( ) 2

3 Problem 3.2 (Verdeyen 6.) (a) dd = 3 4 RR 2 Γ 2 = 0.99 Γ 2 2 = 0.97 From (6.5.3), we have the resonant frequency. νν mm,pp,qq = 2nndd Now we re considering TEM0,0. (b) () FSR FFFFRR = νν mm,pp,qq = + mm + pp [qq + ( dd ) /2 ] ππ RR 2 2nndd νν 0,0,qq = + mm + pp [qq + ] 3 2nndd [qq + 3 ] RR 2 = 2mm dd = 3 2 mm λλ = 5000A = 500nnmm 2nndd = 3 08 mm/ 2 3 = 0 8 = 00MMMMzz 2 mm (2) Cavity Q From (6.3.5), we have (3) Photon lifetime From (6.4.2b), we have (4) Finess ττ pp = LLLLLLhtt ffffffqqffffnnff = λλ = 6 04 MMzz FFFFRR LLLLLLhtt ffffffqqffffnnff = λλ λλ = λλ = 5000A = A QQ = 4ππnndd λλ = RR RR 2 2nndd/ RR RR 2 = = 25.89nnff

4 From (6.3.8), we have FF = 2ππ RR RR 2 =

5 Problem 3.3 (Verdeyen ) 6.5 νν 0 = MMzz λλ 0 = = 600nnmm νν FFFFRR = 25MMMMzz = dd =.2mm 2nndd FF = FFFFRR νν /2 = 25MMMMzz 2.5MMMMzz = From (6.4.4), we have photon lifetime. 6.0 ττ pp = QQ = νν 0 = 5 04 MMzz = νν /2 2.5MMMMzz QQ = 2ππνν 0 2ππ MMzz = nnff From p.54, the requirement of a laser to oscillate is its lifetime is negative, meaning photons are growing in the cavity. From (6.3.8), we can calculate the RR RR 2. 2ππ FF = RR RR 2 RR RR 2 = 2ππ FF RR RR 2 = 2ππ FF = ττ pp = 2nndd/ GGRR GGRR 2 < 0 GGRR GGRR 2 < 0 GG > RR RR 2 =.06947

6 Problem 3.4 (Verdeyen 6.9) (a) It s a stable cavity. 0.5 TT = = AA + DD = 3 4 (b) νν mm,pp,qq = 2nndd νν mm,pp,qq = For TEM0,0,q, its frequency is For TEM,0,q, its frequency is (c) + mm + pp [qq + ( dd ) /2 ] ππ RR 2 dd = 0.75 RR mm + pp [qq ] 2nndd ππ νν 0,0,qq = νν,0,qq = νν,0,qq νν 0,0,qq = From (6.6.2a), the TEM0,0,q transmission is 2nndd [qq + ππ ] 2nndd [qq + 2 ππ ] 2nndd = 33.3MMMMzz ππ TT = ff 2(aa ωω )2 Less than 0.% loss, which means TT = 99.9% = TT = ff 2 aa ωω 2 = aa ωω = The beam waist is largest at the spherical mirror. From (5.3.9), we have the beam waist. ππωω 2 λλ = ddrr dd RR

7 (d) ωω = 0.498mmmm aa = mmmm DDLLaammffttffff = 2aa =.85mmmm From (8..2), we have the condition to overcome the loss and induce oscillation in the cavity. γγ 0 (νν) 2ll gg ln ( RR RR 2 ) 2ll gg ln RR RR 2 = 0.5 ln 0.95 = mm = mm

8 Problem 3.5 (Verdeyen 6.25) (a) From (3.5.), we know the phase shift of a mode inside the cavity. kkdd ( + mm + pp)ttaann dd zz 0 = qqππ But this equation is for one flat mirror and a curved mirror cavity. Here we have two flat mirrors and a lens in between. We need to modify the equation for phase matching. (b) kkdd ( + mm + pp)ttaann dd + kkdd 2 ( + mm + pp)ttaann dd 2 = qqππ νν = kk = 2ππ λλ/nn = 2ππnn /νν = 2ππννnn 2(dd + dd 2 ) qq + ( + mm + pp)(ttaann dd + ttaann dd 2 ) From (6.4.2b), we have photon lifetime. (c) ττ pp = 2nn(dd + dd 2 )/ TTRR TTRR 2 = 28.8nsec λλ 0 = 545A From (6.3.5), we have the quality factor. (d) QQ = 4ππnn(dd + dd 2 ) λλ 0 From (6.4.5), linewidth can be obtained. = TTRR TTRR 2 ττ pp = 2nn(dd + dd 2 )/ GGTTRR GGTTRR 2 = 99.4nsec νν /2 = 2ππττ pp = 5.648MMMMzz

9 Problem 3.6 (Verdeyen 7.4) (a) (7.3.4) LL = LL 2 n = At equilibrium, the time rate of change should be zero. NN 2 BB 2 ρρ(νν) = NN AA 2 + BB 2 ρρ(νν) = ff hνν/kkkk Neglect the stimulated emission process, which means BB 2 = 0. BB 2 ρρ(νν) AA 2 = ff hνν/kkkk ρρ(νν) = AA 2 BB 2 ff hνν/kkkk Aording to Einstein s theory, we can get the coefficient. AA 2 = AA 2 = 8ππnn2 nn gg hνν 3 BB 2 BB 2 3 ρρ(νν) = 8ππnn2 nn gg hνν 3 3 ff hνν/kkkk This is the low temperature approximation for the original blackbody radiation. when ff hνν kkkk, it means T is small. ff hνν kkkk ~ff hνν/kkkk (b) Similar to the argument in the previous problem. If hνν, then it automatically satisfies that ff hνν kkkk. And it leads to ff hνν kkkk We have the energy density in the previous problem. In order to match the results, we require with n = and nn gg ~nn. ~ff hνν/kkkk ρρ(νν) = AA 2 BB 2 ff hνν/kkkk kkkk AA 2 = 8ππnn2 nn gg hνν 3 BB 2 3 = 8ππhνν3 3

10 (c) From the above derivation, we have the Wein s distribution. ρρ WWWWWWWW (νν) = 8ππhνν3 3 From (7.2.5), we have Rayleigh-Jeans distribution. ff hνν/kkkk ρρ Rayleigh Jeans (νν) = 8ππνν2 3 kktt And Planck s distribution is the following. ρρ Planck (νν) = 8ππhνν3 3 ff hνν kkkk By observations, the Wein and Rayleigh-Jeans are both approximations under extreme conditions. For Wein s distribution, we have shown that it is an extreme case when hνν kkkk. For Rayleigh-Jeans distribution, it is the other condition hνν kkkk. Note that lim xx 0 WW xx lim ρρ hνν Planck (νν) = 8ππνν2 kktt. kkkk 3 xx ρρ Planck (νν) = 8ππνν2 kktt 3 hνν/kktt ff hνν kkkk =. Then we have the Rayleigh-Jeans distribution

11 Problem 3.7 (Verdeyen 7.4) (a) First we need to write down the energy (here we re using frequency because frequency is proportional to energy) distribution function of each group of atoms. As described in the problem, we should use Lorentzian function. The lineshape function can be written down in the following equation. νν h LL(νν) = pp(ff)ddff 2ππ[(ff νν) 2 + ( νν h /2) 2 ] I ignore the Planck constants here because we only concern about the shape, not the exact number. From the figure of p(f), we can modify the above lineshape function. Let xx = ff νν LL(νν) = νν 0 + νν ss /2 νν 0 νν ss /2 νν h ddff νν ss 2ππ[(ff νν) 2 + ( νν h /2) 2 ] = νν νν 0 + νν ss /2 h ddff 2ππ νν ss [(ff νν) 2 + ( νν h /2) 2 ] νν 0 νν ss /2 Note: LL(νν) = νν h 2ππ νν ss νν 0 + νν ss 2 +νν νν 0 νν ss 2 +νν ddxx [xx 2 + ( νν h /2) 2 ] xx 2 + aa 2 ddxx = aa ttaann xx + CC aa LL(νν) = (b) νν h 2 [ttaann 2 νν 2ππ νν ss νν h νν 0 + νν + νν ss h 2 ttaann ( 2 (νν νν 0 + νν νν ss h 2 ))] Without loss of generality, I am going to assume the following parameters for the plot. g.0 νν ss = aanndd νν 0 = h s h s h s

12 Problem 3.8 (Verdeyen 7.0) (a) (b) ddnn 2 ddtt = PP 2 NN 2 ττ 2 ddnn ddtt = NN 2 ττ 2 NN ττ Solve the differential equation by Mathematica. (c) NN 2 (tt) = PP 2 ττ 2 ( ff tt ττ 2 ) ddnn ddtt = PP 2 ff tt ττ 2 NN NN (tt) = PP 2 ττ ( ττ ff tt ττ + ττ 2 ff ττ ττ 2 ττ ττ 2 We have the lifetime of each state and the pumping strength. Carrier concentration cm ττ = 2μμ ττ 2 = μμ PP 2 = 0 20 mm 3 ττ tt ττ 2 ) t s N t N2 t Inversion happens when N2 state has more carriers, which means NN 2 > NN. I used Mathematica to get a numerical solution. (d) Steady state means t. δt = 2.972µs lim NN (tt) = PP 2 ττ = mm 3 tt lim NN 2(tt) = PP 2 ττ 2 = 0 4 mm 3 tt