chapter2 solutionWord下载.docx
- 文档编号:15892716
- 上传时间:2022-11-16
- 格式:DOCX
- 页数:15
- 大小:153.91KB
chapter2 solutionWord下载.docx
《chapter2 solutionWord下载.docx》由会员分享,可在线阅读,更多相关《chapter2 solutionWord下载.docx(15页珍藏版)》请在冰豆网上搜索。
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
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- chapter2 solution
![提示](https://static.bdocx.com/images/bang_tan.gif)