光电子学与光子学讲义-作业答案(第1、2章)13版

光电子学与光子学讲义-作业 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 (第1、2章)13版第一章1.10 Refractiveindex(a)Considerlightoffree-spacewavelength1300nmtravelinginpuresilicamedium.Calculatethephasevelocityandgroupvelocityoflightinthismedium.Isthegroupvelocityevergreaterthanthephasevelocity?(b)WhatistheBrewsterangle(thepolarizationangleqp) andthecriticalangle(qc)fortotalinternalreflectionwhenthelightwavetravelinginthissilicamediumisincidentonasilica/airinterface.Whathappensatthepolarizationangle?(c)Whatisthereflectioncoefficientandreflectanceatnormalincidencewhenthelightbeamtravelinginthesilicamediumisincidentonasilica/airinterface?(d)Whatisthereflectioncoefficientandreflectanceatnormalincidencewhenalightbeamtravelinginairisincidentonanair/silicainterface?Howdothesecomparewithpart(c)andwhatisyourconclusion?1.18 Reflectionatglass-glassandair-glassinterfaceArayoflightthatistravelinginaglassmediumofrefractiveindexn1=1.460becomesincidentonalessdenseglassmediumofrefractiveindexn2=1.430.Supposethatthefreespacewavelengthofthelightrayis850nm.(a)WhatshouldtheminimumincidenceangleforTIRbe?(b)Whatisthephasechangeinthereflectedwavewhentheangleofincidenceqi=85°andwhenqi=90°?(c)Whatisthepenetrationdepthoftheevanescentwaveintomedium2whenqi=85°andwhenqi=90°?(d)Whatisthereflectioncoefficientandreflectionatnormalincidence(qi=0°)whenthelightbeamtravelingintheglassmedium(n=1.460)isincidentonaglass-airinterface?(e)Whatisthereflectioncoefficientandreflectanceatnormalincidencewhenalightbeamtravelinginairisincidentonanair/-glassinterface(n=1.460)?Howdothesecomparewithpart(d)andwhatisyourconclusion?1.20TIRandpolarizationatwater-airinterface(1)Giventhattherefractiveindexofwaterisabout1.33,whatisthepolarizationangleforlighttravelinginairandreflectedfromthesurfaceofthewater?(2)consideradiverinseapointingaflashlighttowardsthesurfaceofthewater.Whatisthecriticalangleforthelightbeamtobereflectedfromthewatersurface?1.22phasechangesonTIRConsideralightwaveofwavelength870nmtravelinginasemiconductormedium(GaAs)ofrefractiveindex3.6.Itisincidentonadifferentsemiconductormedium(AlGaAs)ofrefractiveindex3.4,andtheangleofincidenceis80o.Willthisresultintotalinternalreflection?Calculatethephasechangeintheparallelandperpendicularcomponentsofthereflectedelectricfield?1.25Goos-HaenchenphaseshiftArayoflightthatistravelinginaglassmedium(1)ofrefractiveindexn1=1.460becomesincidentonalessdenseglassmedium(2)ofrefractiveindexn2=1.430.Supposethatthefreespacewavelengthofthelightrayis850nm.theangleofincidenceθi=85°.EstimatethelateralGoos-Haenchenshiftinthereflectedwavefortheperpendicularfieldcomponent.RecalculatetheGoos-Haenchenshiftinthesecondmediumhasn2=1(air).Whatisyourconclusion?AssumethatthevirtualreflectionoccursfromavirtualplaneinmediumBatadistancedthatisroughlythesameasthepenetrationdepth.Notethatdactuallydependsonthepolarization,thedirectionofthefield,butwewillignorethisdependence.第二章作业习 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 ：θ2.7Dielectricslabwaveguide ConsideradielectricslabwaveguidethathasathinGaAslayerofthickness0.25μmbetweentwoAlGaAslayers.TherefractiveindexofGaAsis3.6andthatoftheAlGaAslayersis3.40.Whatisthecut-offwavelengthbeyondwhichonlyasinglemodecanpropagateinthewaveguide,assumingthattherefractiveindexdoesnotvarygreatlywiththewavelength?Ifaradiationofwavelength860nm(correspondingtobandgapradiation)ispropagatingintheGaAslayer,whatisthepenetrationoftheevanescentwaveintotheAlGaAslayer?Whatisthemodefieldwidth(MFW)ofthisradiation?Pointouttheeffectofchangeofradiationwavelength(λ)ontheMFW.2.9DielectricslabwaveguideConsideraplanardielectricwaveguidewithacorethickness10μm,n1=1.4446,n2=1.4440.Calculatethe-number,themodeangleθmform=0(useagraphicalsolution,ifnecessary),penetrationdepth,andmodefielddistance(MFW=2α+2δ),forlightwavelengthsof1.0μmand5μm.Whatisyourconclusion?CompareyourMFWcalculationwith.Themodelangleθ0isgivenasθ0=88.85ᵒforλ=1μmandθ0=88.72ᵒforλ=1.5μmforthefundamentalmodem=0.2.10AmultimodefiberConsideramultimodefiberwithacorediameterof60μm,corerefractiveindexof1.47,andacladdingrefractiveindexof1.45bothat870nm.Consideroperatingthisfiberatλ=870nm.(a)Calculatethenumericalaperture.(b)Findoutthenormalizedcore-claddingindexdifference.(c)CalculatetheV-numberforthefiberandestimatethenumberofguidedmodes.(d)Calculatethewavelengthbeyondwhichthefiberbecomessingle-mode.(e)Calculatethemodaldispersionandhencethebitratedistanceproduct.2.12Singlemodefiber Considerafiberwitha-13.5%coreofdiameterof6andrefractiveindexof1.47andacladdingrefractiveindexof1.46bothrefractiveindicesat1300nmwherethefiberistobeoperatedusingalasersourcewithahalfmaximumwidth(FWHM)of2nm..(a)Calculatethe-numberforthefiver.(b)whatisthemaximumalloweddiameterofthecorethatmaintainsoprationsinsingle-mode?(c)Calculatethewavelengthbelowwhichthefiberbecomesmultimode.(d)Calculatethenumericalaperture.(e)Calculatethemaximumacceptanceangle.(f)Obtainthematerialdispersionandwavelengthdispersionandhenceestimatethebitratedistanceproduct()ofthefiber.