1.(i) Doppler Broadeningg(ν)0.2e−4 (ln 2)(ν−ν0)2Δ νD2¿0.2 e−2.77(ν−ν0)2ΔνD2Laser oscillation occurs if g(ν)>0.1i.e. e−2.77(ν− ν0)2Δ νD2>0. 5(¿0.10.2)or ν −ν0≤0.5 ∆ νDTotal gain bandwidth ∆ νgain=2×(ν−ν0)=∆ νD=30GHzCavity mode spacing νc=c2 l=3 ×1082× 0.6=0.25GHzSo # of laser modes ¿300.25=120(ii) # of pulses/ sec = cavity frequency =250MHz (c2 l)Pulse duration 1∆ νgain=130 ×10933 ps2.1 The minimum number density is (1.58× 10192+1)cm−3 in order to have population inversion between level 2 and 12.2 Since the cavity loss is negligible, the total loss comes from the mirror reflectivity L≅ 1−R ≅ 0.12.3 Use the formulaNt=(N2−N1)t=8 π n2tspontg(ν)λ2(=1lln(r1r2))And l=10 cmtspont=1× 10−3g(ν)10−12λ 1.06 μm(for Nd : Yag)You can use any value for λ depending on the laser you chooser1=1r2=√0.9To get Nt2.4 Use tc≈nlc(1−R) to calculate cavity lifetime0.13 ×108× 0.1=3 ×10−7Then use Ppeak=nihν2 tc=10 Nt[1 ×10 ]hνtc (volume in cm3) to calculate Pp3(i)(a) Free running laser, multimodeMinimum gain = loss = 0.1+0.5 = 0.6 (0.1 from R)(b) ≈ 100GHz(ii)(a) Range of frequency in which the gain is larger than the loss (0.6) is(∆ ν)gain=(2.43)×100 GHz=80GHzFree spectral range= cavity frequency= c2 nl=3 ×1082=1.5 ×108So # of modes = 80 × 1091.5 ×108= 533(b) Pulse duration = 1(∆ ν)gain180× 109=1.25 ×10−11sPulse energypulse=energy / sec¿of pulses/ sec=11.5 ×108=6.6 × 10−9=6.6 nJ2. Please shift ϕx'−ϕy'=−2 πλ(nx'−ny ')l(a) if Edc=0 ϕx'−ϕy'=2 πλ(0.001)lTo get ϕx'−ϕy'=πSet 2 πλ(0.001)l=πSo l=π2 πλ(0.001)=2000 λ=2mm(b)For this length of crystal, phase shift is π so the output polarization is rotated by 9 0 ° , i.e. parallel to y-axis, i.e. maximum transmissionTo get the first zero transmission, need to rotate the output polarization by another 90 °, i.e. need to impart a phase shift π by the applied electric field Edci.e. 2 r ∙ Edc2 πλl=πEdc=λ4 lr=1.25× 106V