chapter2 solutionWord下载.docx

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Solution

a）Ray1experiencesaphaseshiftk1AB–2φ（atB）

Ray2experiencesaphaseshiftk1A΄B΄（atB΄）

Thephasedifferencebetweenthetworaysisk1（AB-A΄B）–2φ

Forawaveguidemodethisphasedifferencemustbem2π

Fromthegeometry:

ABcosθ=2a

A΄B΄=A΄C+CB΄

A΄C=ACcos（π-2θ）

CB΄=CBcos（π-2θ）

ThereforeA΄B΄=A΄C+CB΄=（AC+CB）cos（π-2θ）=ABcos（π-2θ）

Thismeansthatthephaseshiftbetweenthetwobeams

k1（AB-A΄B΄）–2φ=k1[AB-ABcos（π-2θ）]–2φ=k1AB[1+cos（2θ）]

=k1AB（2cos2θ）–2φ

orthephaseshiftisequaltok1[2a/cosθ]2cos2θ–2φ=k1（4acosθ）–2φ

Foramodethisphaseshiftmustbem2π

Ork14acosθ–2φ=m2π

k12acosθ–φ=mπ

2πn12a/λcosθm–φm=mπwhichisthetext’s（page52equation3）waveguidecondition.

b）c）Fromthegeometrywehavethefollowing:

（a-y）/AC=cosθ

andA'

C/AC=cos（π-2θ）

ThephasedifferencebetweentheraysmeetingatCis

Φ=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-φ

Given,

Then

Additionalproblem:

Inclasswetracedthepropagationofa1V/mwavebetweentwomirrorsthateachhavereflectivityr.Wekepttrackofthephaseshiftbetweenthetwomirrors（separatedbyadistanceL）andfoundthatthetotalelectricalfieldE（phasor）inthecavitywasgivenby1/[1-r2exp（-2jkL）].IfrisrealandR=r2,showthatthetotalintensityincavitycanbewrittenas（constant）/[（1-R）2+4Rsin2kL]（seeequation3onzpage30ofthetext）.Itwillhelpyoutorememberthatintensityisproportionalto|E|2and|E|2=E

E*whereE*isthecomplexconjugateofE.

2.2TEfieldpatterninslabwaveguideConsidertwoparallelrays1and2interferingintheguideasinFigure2.4.Giventhephasedifference（asinQuestion2.1）

betweenthewavesatC,distanceyabovetheguidecenter,findtheelectricfieldpatternE（y）intheguide.Plotthefieldpatternforthefirstthreemodestakingaplanardielectricguidewithacorethickness20μm,n1=1.455n2=1.440,lightwavelengthof1.3μm.

ThetwowavesinterferingatCareoutphasebyΦ,

whereAisanarbitraryamplitude.Thus,

or

=Eocos（ωt+Φ'

）

inwhichcos（ωt+Φ'

）isthetimedependentpartthatrepresentsthewavephenomenon,andthecurlybracketscontaintheeffectiveamplitude.Thus,theamplitudeEois

ToplotEoasafunctionofy,weneedtofindφmform=0,1and2,thefirstthreemodes.FromExample2.1.1inthetextbook,thewaveguideconditionis

wecannowsubstituteforφmwhichhasdifferentformsforTEandTMwavestofind,

TEwaves

TMwaves

TheabovetwoequationscanbesolvedgraphicallyasinExample2.1.1tofindθmforeachchoiceofm.Alternativelyonecanuseacomputerprogramforfindingtherootsofafunction.Theaboveequationsarefunctionsofθmonlyforeachm.Usinga=10μm,λ=1.3μm,n1=1.455n2=1.440,theresultsare:

TEModesm=0m=1m=2

θm（degrees）88.8487.6786.51

φm（degrees）163.75147.02129.69

TMModesm=0m=1m=2

φm（degrees）164.08147.66130.60

ThereisnosignificantdifferencebetweentheTEandTMmodes（thereasonisthatn1andn2areveryclose）.

WecansetA=1andplotEovs.yusing

withtheφmandmvaluesinthetableabove.

Figure2Q2

2.3TEandTMModesindielectricslabwaveguideConsideraplanardielectricguidewithacorethickness20μm,n1=1.455n2=1.440,lightwavelengthof1.3μm.Giventhewaveguidecondition,Eq.（3）in§

2.1,andtheexpressionsforphasechangesφandφ'

inTIRfortheTEandTMmodesrespectively,

and

usingagraphicalsolutionfindtheangleθforthefundamentalTEandTMmodesandcomparetheirpropagationconstantsalongtheguide.

FromExample2.1.1inthetextbook,thewaveguideconditionis

TEModesm=0

θm（degrees）88.8361

βm=k1sinθm7,030,883