光电子学与光子学的原理及应用s.o.kasap-课后答案

SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/71.4AntireflectioncoatinggggForlighttravelinginmedium1incidentonthe1-2interfaceatnormalincidence,r12=n1-n2n1+n2=n1-n1n3n1+n1n3=1-n3n11+n3n1Forlighttravelinginmedium2incidentonthe2-3interfaceatnormalincidence,r23=n2-n3n2+n3=n1n3-n3n1n3+n3=n1n3-1n1n3+1=1-n3n11+n3n1thus,r23=r12Significance?Foranefficientantireflectioneffect,wavesA(reflectedat1-2)andB(reflectedat2-3)inFigure1Q4belowshouldinterferewithnear“totaldestruction”.Thatmeanstheyshouldhavethesamemagnitudeandthatrequiresthatthereflectioncoefficientbetween1and2shouldbethesameasthatbetween2and3;r12=r23.Thus,thelayer2canactasanantireflectioncoatingifitsindexn2=(n1n3)1/2.Thiscanbeachievedbyr12=r23.Thebestantireflectioncoatinghastohavearefractiveindexn2suchthatn2=(n1n3)1/2=[(1)(3.5)]1/2=1.87.Givenachoiceoftwopossibleantireflectioncoatings,SiO2witharefractiveindexof1.5andTiO2witharefractiveindexof2.3,bothareclose.ThephasechangeforwaveBgoingthroughthecoatingofthicknessdis2k2dwherek2=n2koandko=wavevectorinfreespace=2π/λ.Thisshouldbe180°orπ.Thusweneed2n2(2π/λ)d=πorForSiO2d=λ4n2=900×10-9m4(1.5)=0.15μμμμmForTiO2d=λ4n2=900×10-9m4(2.3)=0.10μμμμm1.8Thinfilmcoatingandmultiplereflections:Assumethatn1n2n3andthatthethicknessofthecoatingisd.Forsimplicity,wewillassumenormalincidence.Thephasechangeintraversingthecoatingthicknessdisϕ=(2π/λ)n2dwhereλisthefreespacewavelength.Thewavehastobemultipliedbyexp(-jϕ)toaccountforthisphasedifference.Thecoefficientsaregivenby,r1=r12=n1-n2n1+n2=-r21,r2=r23=n2-n3n2+n3andt1=t12=2n1n1+n2,t2=t21=2n2n1+n2,t23=2n3n2+n3,Consider1-t1t2,SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/71-t1t2=1-4n1n2(n1+n2)2=n12+n12+2n1n2-4n1n2(n1+n2)2=n12+n12-2n1n2(n1+n2)2=n1-n2n1+n22=r12(1)TheamplitudeofthereflectedbeamisAreflected=A1+A2+A3+...i.e.Areflected/A0=r1+t1t2r2e-j2ϕ-t1t2r1r22e-j4ϕ+t1t2r12r23e-j6ϕ+...(2)sothatthek-thtermfork1isAreflectedA0k=-t1t2r1-r1r2e-j2ϕ()k(3)sothatthereflectioncoefficientisr=AreflectedA0=r1-t1t2r1-r1r2e-j2ϕ()kk=1∞∑SincetheEq.(2)isageometricserieswithtermsgivenbyEq.(3),thesummationissimple,r=r1-t1t2r1-r1r2e-j2ϕ()1--r1r2e-j2ϕ()=r1+t1t2r2e-j2ϕ1+r1r2e-j2ϕ(4)UsingEq.(1),Eq.(4),r=r11+r1r2e-j2ϕ()+1-r12()r2e-j2ϕ1+r1r2e-j2ϕi.e.r=r1+r2e-j2ϕ1+r1r2e-j2ϕ(5)TheamplitudeofthetransmittedbeamisCtransmitted=C1+C2+C3+...i.e.Ctransmitted/A0=t1t23e-jϕ-t1t23r1r2e-j3ϕ+t1t23r12r22e-j5ϕ+...(6)sothatthek-thtermisCtransmittedA0k=-t1t23ejϕr1r2-r1r2e-j2ϕ()k(7)sothatthetransmissioncoefficientist=CtransmittedA0=-t1t23ejϕr1r2-r1r2e-j2ϕ()kk=1∞∑=t1t23ejϕr1r2r1r2e-j2ϕ1+r1r2e-j2ϕ(8)SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7i.e.t=t1t23e-jϕ1+r1r2e-j2ϕ1.9Antireflectioncoating:ConsiderthetransmissioncoefficientobtainedinQuestion1.8,t=t1t2e-jϕ1+r1r2e-j2ϕr1andr2arepositivenumbers.Tomaximizetweneedexp(-j2ϕ)=-1.whichmeansthatexp(-j2ϕ)=cos(-2ϕ)+jsin(-2ϕ)=-1.Thiswillbesowhen2ϕ=mπ,wheremisanodd-integer,orwhenϕ=2πn2dλ=m12πleadingtod=mλ4n2Inadditionweneedr1r2=1.Considerchoosingn2=(n1n3)1/2.Forlighttravelinginmedium1incidentonthe1-2interfaceatnormalincidence,r1=r12=n1-n2n1+n2=n1-n1n3n1+n1n3=1-n3n11+n3n1Forlighttravelinginmedium2incidentonthe2-3interfaceatnormalincidence,r2=r23=n2-n3n2+n3=n1n3-n3n1n3+n3=n1n3-1n1n3+1=1-n3n11+n3n1thus,r2=r1whichconfirmsthatweneedn2=(n1n3)1/2.ThereflectioncoefficientfromQuestion1.8isr=r1+r2e-j2ϕ1+r1r2e-j2ϕThisiszero(noreflection)whenthenumeratoriszero,thatisr1=-r2exp(-j2ϕ)Themagnitudeofexp(-j2ϕ)isunityandsincer1andr2arepositivequantities,wemusthavetwoconditionstoobtainzerointhenumerator:Condition1:r2=r1.Thisrequiresn2=(n1n3)1/2asderivedabove.Condition2:exp(-j2ϕ)=cos(-2ϕ)+jsin(-2ϕ)=-1whichwillbesowhen2ϕ=mπ,wheremisanodd-integer,orwhenSolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7ϕ=2πn2dλ=m12πleadingtod=mλ4n21.16DiffractionbyalensTheangularpositionθofthefirstdarkringisdeterminedbythediameterDoftheapertureandthewavelengthλ,andisgivenbysinθ=1.22λDSinceθissmall,θ=sinθ=1.22λD=1.22590×10-92×10-2=3.6×10-5rad.FromtheRayleighcriterionthisisalsotheresolvingpowerΔθminofthelens.Iff=focallengthofthelens,theradiusrofthecentralAirydiskisdeterminedbyθ=r/f∴r=fθ=(40×10-2m)(3.6×10-5rad)=1.44×10-5mor14.4μm.Fornearlyallpracticalpurposes,this29μmdiameterspotatthefocalplaneisapoint.2.1DielectricslabwaveguideFromthegeometrywehavethefollowing:(a-y)/AC=cosθandΑ′C/AC=cos(π-2θ)ThephasedifferencebetweentheraysmeetingatCisΦ=kAC-ϕ-kA′C=k1AC-k1ACcos(π-2θ)-ϕ=k1AC[1-cos(π-2θ)]-ϕ=k1AC[1+cos(2θ)]-ϕ=k1[(a-y)/cosθ][1+2cos2θ-1]-ϕ=k1[(a-y)/cosθ][2cos2θ]-ϕ=2k1(a-y)cosθ-ϕGiven,2π(2a)n1λcosθm-ϕm=mπ∴∴∴∴cosθm=λ(mπ+ϕm)2πn1(2a)=mπ+ϕmk1(2a)ThenΦm=2k1(a-y)cosθm-ϕm=2k1(a-y)mπ+ϕmk1(2a)-ϕm∴∴∴∴Φm=(1-ya)(mπ+ϕm)-ϕm=mπ-ya(mπ+ϕm)Φm=Φm(y)=mπ-ya(mπ+ϕm)2.5DielectricslabwaveguideSolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7Givenn1=3.66(AlGaAs),n2=3.4(AlGaAs),2a=2×10-7mora=0.1μm,foronlyasinglemodeweneedV=2πaλn12-n22()1/2π2∴∴∴∴λ2πan12-n22()1/2π2=2π(0.1μm)3.662-3.402()1/2π2=0.542μm.Thec