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κL = 61.21Misalkan:n o= 3,n1 = 0,1, Λ =0,2μmRBand gap0.80.60.40.2λ = 2 Λ = 1,2μm; f =Bno B25 0 THzn1 −1κ ≈ G = 0,5μm ; L 12μm2n=o0-15 -10 -5 0 5 10 15δLLebar pitanoδ cδ ≈ ( ω − ωB) → ω − ωB=cnf−fB=20THzo=1,25 μmδL=-15 sd 15 δ=-1,25 sd 1,25 μm -1-1x3x1038ms−1= 1,25 x1014s-1Δf=2x20THz= 40THz ; fB=250THz2cλf = → Δλ= Δf;λc12 −140 x10s x(1,2μm)Δλ=8 −13x10ms2=0,19 μm;λB= 1,2μm92

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