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=2pi/λ.Thisshouldbe180°orpi.Thusweneed2n2(2pi/λ)d=piorForSiO2d=λ4n2=900×10−9m4(1.5)=0.15µµµµmForTiO2d=λ4n2=900×10−9m4(2.3)=0.10µµµµm1.8Thinfilmcoatingandmultiplereflections:Assumethatn1<n2<n3andthatthethicknessofthecoatingisd.Forsimplicity,wewillassumenormalincidence.Thephasechangeintraversingthecoatingthicknessdisφ=(2pi/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-thtermfork>1isAreflectedA0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=mpi,wheremisanodd-integer,orwhenφ=2pin2dλ=m12pileadingtod=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=mpi,wheremisanodd-integer,orwhenSolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7φ=2pin2dλ=m12pileadingtod=mλ4n21.16DiffractionbyalensTheangularpositionθofthefirstdarkringisdeterminedbythediameterDoftheapertureandthewavelengthλ,andisgivenbysinθ=1.22λDSinceθissmall,θ=sinθ=1.22λD=1.22590×10−92×10−2=3.6×10-5rad.FromtheRayleighcriterionthisisalsotheresolvingpower∆θminofthelens.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(pi−2θ)ThephasedifferencebetweentheraysmeetingatCisΦ=kAC−φ−kA′C=k1AC−k1ACcos(pi−2θ)−φ=k1AC[1−cos(pi−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,2pi(2a)n1λcosθm−φm=mpi∴∴∴∴cosθm=λ(mpi+φm)2pin1(2a)=mpi+φmk1(2a)ThenΦm=2k1(a−y)cosθm−φm=2k1(a−y)mpi+φmk1(2a)−φm∴∴∴∴Φm=(1−ya)(mpi+φm)−φm=mpi−ya(mpi+φm)Φm=Φm(y)=mpi−ya(mpi+φm)2.5DielectricslabwaveguideSolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7Givenn1=3.66(AlGaAs),n2=3.4(AlGaAs),2a=2×10-7mora=0.1µm,foronlyasinglemodeweneedV=2piaλn12−n22()1/2<pi2∴∴∴∴λ>2pian12−n22()1/2pi2=2pi(0.1µm)3.662−3.402()1/2pi2=0.542µm.Thecut-offwavelengthis542nm.Whenλ=870nm,V=2pi(1µm)3.662−3.402()1/2(0.870µm)=0.979<pi/2Therefore,λ=870nmisasinglemodeoperation.Forarectangularwaveguide,thefundamentalmodehasamodefielddistance2wo=MFD≈2aV+1V=(0.2µm)0.979+10.979=0.404µm.Thedecayconstantαoftheevanescentwaveisgivenby,α=Va=0.9790.1µm=9.79(µm)-1or9.79×106m-1.Thepenetrationdepthδ=1/α=1/[9.79(µm)-1]=0.102µm.Thepenetrationdepthishalfthecorethickness.2.8Amultimodefiber.Givenn1=1.475,n2=1.455,2a=100×10-6mora=50µmandλ=0.850µm.TheV-numberis,V=2piaλn12−n22()1/2=2pi(50µm)1.4752−1.4552()1/2(0.850µm)=89.47NumberofmodesM,M=V22=89.4722≈4002Thefiberbecomesmonomodewhen,V=2piaλn12−n22()1/2<2.405orλ>2pian12−n22()1/22.405=2pi(50µm)1.4752−1.4552()1/22.405=31.6µµµµmForwavelengthslongerthan31.6µm,thefiberisasinglemodewaveguide.ThenumericalapertureNAisNA=n12−n22()1/2=1.4752−1.4552()1/2=0.242SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7Ifαmaxisthemaximumacceptanceangle,then,αmax=arcsinNAno=arcsin(0.242/1)=14°°°°Modaldispersionisgivenby∆τintermodeL=n1−n2c=1.475−1.4553×108ms−1=66.7psm-1or66.7nsperkmGiventhatσ≈0.29∆τ,maximumbit-rateisBL=0.25Lσtotal≈0.25Lσintermode=0.25(0.29)(66.7nskm−1)=13Mbs-1km(onlyanestimate!)WeneglectedmaterialdispersionatthiswavelengthwhichwouldfurtherdecreaseBL.Materialdispersionandmodaldispersionmustbecombinedbyσtotal2=σintermode2+σmaterial2Forexample,assuminganLEDwithaspectralrmsdeviationσλofabout20nm,andaDm≈−200pskm-1nm-1(atabout850nm)wewouldfindσm=−(200pskm-1nm-1)(20nm)(1km)≈4000pskm-1or4nskm-1,whichissubstantiallysmallerthantheintermodedispersionandcanbeneglected.2.9AsinglemodefiberaGivenn1=1.475,n2=1.455,2a=8×10-6mora=4µmandλ=1.3µm.TheV-numberis,V=2piaλn12−n22()1/2=2pi(4µm)1.4682−1.4642()1/2(1.3µm)=2.094bSinceV<2.405,thisisasinglemodefiber.ThefiberbecomesmultimodewhenV=2piaλn12−n22()1/2>2.405orλ<2pian12−n22()1/22.405=2pi(4µm)1.4682−1.4642()1/22.405=1.13µµµµmForwavelengthsshorterthan1.13µm,thefiberisamulti-modewaveguide.cThenumericalapertureNAisNA=n12−n22()1/2=1.4682−1.4642()1/2=0.108dIfαmaxisthemaximumacceptanceangle,then,αmax=arcsinNAno=arcsin(0.108/1)=6.2°°°°sothatthetotalacceptanceangleis12.4°°°°.eAtλ=1.3µm,fromthefigure,Dm≈−7.5pskm-1nm-1,Dw≈−5pskm-1nm-1.∆τ1/2L=Dm+Dw∆λ1/2SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7=|−7.5−5pskm-1nm-1|(2nm)=15pskm-1+10pskm-1=0.025nskm-1Obviouslymaterialsdispersionis15pskm-1andwaveguidedispersionis10pskm-1Themaximumbit-ratedistanceproductisthenBL≈0.59L∆τ1/2=0.590.025nskm−1=23.6Gbs-1km.MaterialandwaveguidedispersioncoefficientsinanopticalfiberwithacoreSiO2-13.5%GeO2fora=2.5to4µm.0?010201.21.31.41.51.6?0λ(µm)DmDwSiO2-13.5%GeO22.53.03.54.0a(µm)Dispersioncoefficient(pskm-1nm-1)Figure2Q92.18MicrobendinglossRadiusofcurvatureαV1V2<V1Rα1α2R1R2MicrobendinglossαdecreasessharplywiththebendradiusR.(Schematiconly.)Figure2Q18-1FromSection2.9andFigure2.33,wecanrepresentmicrobendinglossαvs.bendradiusRbehaviorschematicallyasinFigure2Q18-1whichincorporatesthediscussionsinthatsection.FromFigure2Q18-1,givenα=α1,RincreasesfromR1toR2whenVdecreasesfromV1toV2.ExpectedR↑withV↓(1)EquivalentlyatoneR=R1α↑withV↓(2)Wecangeneralizebynotingthatthepenetrationdepthintothecladdingδ∝1/V.ExpectedR↑withδ↑(3)EquivalentlyatoneR=R1α↑withδ↑(4)SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7Eqs.(3)and(4)correspondtothegeneralstatementthatmicrobendinglossαgetsworsewhenpenetrationδintocladdingincreases;intuitivelycorrectbasedonFigure2.32.Experimentsshowthatforagivenα=α1,Rincreaseswithdecreasing∆.ObservationR↑with∆↓(5)Considerthepenetrationdepthδintoasecondmedium(Example2.1.3),δ=1α≈λ2pi1n12−n22()1/2=λ2pin11(2∆)1/2∴δ↑with∆↓(6)Thus,δincreaseswithdecreasing∆.Thus.fromEqs.(3)and(6),weexpectExpectedR↑withδ↑with∆↓(7)ThusEq,(7)agreeswiththeobservationinEq.(5).NOTEIfweplot∆vs.Ronalog-logplot,wewouldfindthelineinFigure2Q18-2,thatis,∆∝Rx,x=−0.62.Veryroughly,fromtheoreticalconsiderations,weexpectα∝exp−RRc∝exp−R∆−3/2(8)whereRcisaconstant(“acriticalradiustypeofconstant”)thatisproportionalto∆−3/2.Thus,takinglogs,lnα=−∆3/2R+constant(9)Weareinterestedinthe∆−Rbehaviorataconstantα.Wecanlumptheconstantintolnαandobtain,∆∝R−2/3(10)AsshowninFigure2Q18-2,x=−0.62iscloseto−2/3giventhreepointsandaroughtheory.Log-logplotoftheresultsofexperimentson∆vs.bendradiusRfor1dB/mmicrobendingloss♦♦♦0.0020.01110100Bendradius(mm)∆x=-0.62Figure2Q18-23.2GaAsTheintrinsicconcentrationisni=NcNv()1/2exp−Eg2kBTSolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7sothatni=4.7×1017cm−3()7×1018cm−3()[]1/2exp−1.42eV28.6174×10−5eVK−1()300K()==2.223×1012m−3=2.223×106cm−3Theconductivityisσ=eniµe+µh()=1.608×10−19C()2.223×106cm−3()8500cm2V−1s−1+400cm2V−1s−1()=3.17×10−9Ω−1cm−1Theresistivityisρ=1σ=13.17×10−9Ω−1cm−1()=3.16×108ΩcmTheenergydistancebetweenthebottomoftheconductionbandEcandFermileveloftheintrinsicGaAsEFiisEc−EFi=kBTlnNcnisothatEc−EFi=8.617×10−5eVK−1()300K()ln4.7×1017cm−3()2.223×106cm−3()=0.675eV.IfNcandNvscalewithtemperatureasT3/2thentheintrinsicconcentrationatagiventemperatureTisgivenbytheexpressionni(T)=Nc(300K)Nv(300K)[]1/2T300K3/2exp−Eg2kBT,soat100oC(373K)wewillhaveni(373K)=4.7×1017cm−3()7×1018cm−3()[]1/2×373K300K3/2exp−1.42eV28.6174×10−5eVK−1()373K()=6.61×108cm-3.3.3DirectbandgappnjunctionDe=kTµe/e≈(0.0259V)(7000×10-4m2V-1s-1)=0.01813m2s-1andDh=kTµh/e≈(0.0259V)(310×10-4m2V-1s-1)=0.000803m2s-1Recombinationtimeofelectronsdiffusinginthep-regionisτe=1BNa=1(7.21×10−16m3s−1)(1×1016×106m−3)=138.7nsRecombinationtimeofholesdiffusinginthen-regionisτh=1BNd=1(7.21×10−16m3s−1)(1×1016×106m−3)=138.7nsSolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7Thediffusionlengthsarethen:Le=√[Deτe]=√[0.01813m2s-1)(138.7×10-9s)]=5.02×10-5m,or50.2µmandLh=√[Dhτh]=√[0.000803m2s-1)(138.7×10-9s)]=1.056×10-5m,or10.56µmThediffusioncomponentofthecurrentisI=Idiff=Ιso[exp(eV/(kT))−1]≈Ιsoexp(eV/(kT))forV>>kT/e(≈0.02586V)whereΙso=AJso=Aeni2[(Dh/(LhNd))+(De/(LeNa))]Givenni=1.8×1012m-3,A=0.1×10-6m2,ThusIso=0.1×10−6m2()1.602×10−19C()1.8×1012m−3()20.000803m2s−1()1.056×10−5m()1×1022m−3()+0.1×10−6m2()1.602×10−19C()1.8×1012m−3()20.01813m2s−1()5.02×10−5m()1×1022m−3()∴Iso=2.27×10-21A.TheforwardcurrentduetodiffusionisIdiff=Isoexp[(eV/(kT)]=(2.27×10-21A)exp[(1V)/(0.0259V)]∴Idiff=0.00013Aor0.13mA3.4TheSipnjunctionConsidertemperatureT=300K.kT/e=0.02586V.aThisisap+ndiode:Nd=1015cm-3.Holelifetimeτhinthen-sideisτh=5×10−71+2×10−17Ndopant()=5×10−71+2×10−17×1015cm−3()=490.2nsand,usingthesameequationwithNdopant=1018cm-3,electronlifetimeinthep-sideisτe=23.81ns.I.DiffusioncomponentofdiodecurrentGivenNa=1018cm-3,µe≈250cm2V-1s-1,andwithNd=1015cm-3,µh≈450cm2V-1s-1.Thus:De=kTµe/e≈(0.02585V)(250cm2V-1s-1)=6.463cm2s-1andDh=kTµh/e≈(0.02585V)(450cm2V-1s-1)=11.63cm2s-1Thediffusionlengthsarethen:Le=√[Deτe]=√[6.463cm2s-1)(23.81×10-9s)]=3.92×10-4cm,or3.92µmandLh=√[Dhτh]=√[(11.63cm2s-1)(490.2×10-9s)]=2.39×10-3cm,or23.9µmThediffusioncomponentofthecurrentisI=Idiff=Ιso[exp(eV/(kT))−1]≈Ιsoexp(eV/(kT))forV>>kT/e(≈0.02585V)SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7whereΙso=AJso=Aeni2[(Dh/(LhNd))+(De/(LeNa))]≈Aeni2Dh/(LhNd)asNa>>Nd.Inotherwords,thecurrentismainlyduetothediffusionofholesinthen-region.Thus:Iso=1×10−2cm2()1.60×10−19C()1.45×1010cm−3()211.63cm2s−1()2.39×10−3cm()1×1015cm−3()∴Iso=1.64×10-12Aor1.64pATheforwardcurrentduetodiffusionisIdiff=Isoexp[(eV/(kT)]=(1.64×10-12A)exp[(0.6V)/(0.02585V)]∴Idiff=0.020Aor20mAII.RecombinationcomponentThebuilt-inpotentialis(whereniistheintrinsicconcentration,foundinthetableintheinsidefrontcover):Vo=(kT/e)ln(NdNa/ni2)=(0.02585V)ln[(1018cm-3×1015cm-3)/(1.45×1010cm-3)2]∴Vo=0.755VThedepletionregionwidthWismainlyonthen-side(εrofSiis11.9fromTable5.1).W=2εNa+Nd()Vo−V()eNaNd1/2≈2εVo−V()eNd1/2∴W=211.9()8.854×10−12Fm−1()0.755V−0.6V()1.60×10−19C()1021m−3()1/2i.e.W≈0.451×10-6mor0.451µmRecombinationinthewiderdepletionregioninthen-sideexceedstherecombinationinthenarrowdepletionregioninthep-side.Furtherthewidthofthedepletionregiononthen-sideWn≈W.Therecombinationtimeτrhereisnotnecessarilyτhbutletusassumethatisso(veryroughly).Then,Iro=AeniWn2τh+AeniWp2τe≈AeniW2τh=1×10−2cm2()1.602×10−19C()1.45×1010cm−3()0.451×10−4cm()2490×10−9s()∴Iro=1.070×10-9ATheforwardcurrentduetorecombinationisIrecom=Iroexp(eV/2kT)=(1.070×10-9A)exp[(0.6V)/(2×0.02586V)]∴Irecom=1.17××××10-4Aor0.117mAClearly,thediffusioncomponentdominatestherecombinationcomponent.bThisisasymmetricalpndiode:Nd=Na=1018cm-3.Holelifetimeτhinthen-sideisτh=τe=5×10−71+2×10−17Ndopant()=5×10−71+2×10−17×1018cm−3()=23.81×10-9sSolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7I.DiffusioncomponentWearegiven,µe≈250cm2V-1s-1,andwithNd=1018cm-3,µh≈130cm2V-1s-1.Thus:ThusDe=kTµe/e≈(0.02585V)(250cm2V-1s-1)=6.463cm2s-1andDh=kTµh/e≈(0.02585V)(130cm2V-1s-1)=3.361cm2s-1DiffusionlengthsareLe=√[Deτe]=√[6.463cm2s-1)(23.81×10-9s)]=3.923×10-4cmor3.92µmandLh=√[Dhτh]=√[(3.361cm2s-1)(23.81×10-9s)]=2.829×10-4cmor2.829µmThediffusioncomponentofthecurrentisIdiff=I=Iso[exp(eV/kT)−1]≈Isoexp(eV/kT)forV>>kT/e(≈0.02585V)whereΙso=AJso=Aeni2{[(Dh/(LhNd)]+[De/(LeNa)]}Iso=1×10−2cm2()1.602×10−19C()1.45×1010cm−3()2×3.361cm2s−1()2.829×10−4cm()1018cm−3()+6.463cm2s−1()3.923×10−4cm()1018cm−3()∴Iso=9.554×10-15ATheforwardcurrentduetodiffusionisIdiff=Isoexp(eV/kT)=(9.554×10-15A)exp[(0.6V)/(0.02585V)]∴Idiff=1.15××××10-4Aor0.115mAII.RecombinationcomponentThebuilt-inpotentialisVo=(kT/e)ln(NdNa/ni2)=(0.02585V)ln[(1018cm-3×1018cm-3)/(1.45×1010cm-3)2]∴Vo=0.933VThedepletionregionwidthissymmetricalaboutthejunctionasNa=Nd.W=2εNa+Nd()Vo−V()eNaNd1/2=4εVo−V()eNd1/2∴W=411.9()8.854×10−12Fm−1()0.933V−0.6V()1.602×10−19C()1024cm−3()1/2∴.W≈2.96×10-8mor0.0296µmTherecombinationcurrentpre-exponentialtermisIro=AeWni2τr=1×10−2cm2()1.602×10−19C()0.0296×10−4cm()1.45×1010cm−3()223.81×10−9s()∴Iro=1.44×10-9ATheforwardcurrentduetorecombinationisIrecom=Iroexp(eV/2kT)=(1.44×10-9A)exp[(0.6V)/(2×0.02585V)]∴Irecom=1.58××××10-4Aor0.158mASolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/73.5AlGaAsLEDemitteraWenotethattheemittedwavelengthisrelatedtothephotonenergyEphbyλ=c/υ=hc/Eph.IfwedifferentiateλwithrespecttophotonenergyEphwegetdλdEph=−hcEph2Wecanrepresentsmallchangesorintervals(or∆)bydifferentials,e.g.∆λ/∆Eph≈|dλ/dEph|,then∆λ≈hcEph2∆EphWearegiventheenergywidthoftheoutputspectrum,∆Eph=∆(hυ)≈3kBT.Then,usingthelatterandsubstitutingforEphintermsof&lambd