We'regoingtobetalkingaboutthereflectionandTherefractionofwaveswhentheyhitaninterfaceotherwiseknownasFresnelequationsifyou'refollowingalonginmybookWhatwe'recoveringisinchapter 8 andsincewe'regoingtocover a wholechapter I'm justgoingtogiveyouthehighlightsandnotgiveyouthederivationAlthoughthederivationsofFresnel's equationsis a reallygoodwaytounderstandifyouyoureallyknowwavesornotSoifyourememberinourpictureWeessentiallyhadanincomingelectricfieldandthatelectricfieldwouldturnonandoffanditwouldessentiallyJigglethecloudsoftheatomtheelectroncloudsoftheatomsupanddownandwesawthatthiscanleadbothtochangeinthephaseconstantwithdistancetheindexofrefractionortherelativepermittivityofthematerialandalsosomeabsorptionorsomelossandIfyouhave a materialwhereyouhavesomeabsorptionand I'm goingtosayit's threehereand a phaseconstantWhichisthreeandthisistheequationthatdescribesourplanewavegoingintothematerialand I'llnotethattheplanewaveisHereatnormalincidencethatessentiallyinthegeneralcaseYou'regoingtoget a changeofthewavelengthofthematerialthatarisesfrom a changeinthevelocityphasevelocityandyou'realsogoingtogetsomekindofexponentialtypeofattenuationasthewavegoesdownintothematerialandYoucancalculatetheabsorptioncoefficientalphaWhichistheattenuationofthewavewithdistanceaswellasthespatialphasechangeofthewaveUsingsomeexpressionsaswesawbeforeandthesearethemostcomplicatedexpressions.
Whatwewanttodotodayistalkaboutessentially a moregeneralcasewhereessentiallyyouhavetheincomingWavewithanincident K vector.
今日お話ししたいのは、より一般的なケースで、入射Kベクトルを持つ入射波がある場合についてです。
We'llcallitcaseof I forinstanceforincidentcominginsomeangleAndasyouwillseewe'regoingtomeasuretheanglesnormaltothesurfacebecausethat's theonlywayyoucanaccuratelymeasurethingsRelatedto a surfaceistothenormalandthatthisincidentwaveisgoingtogiverisetoboththereflectedwaveThat's goingbackoutinthatwayand a transmittedwavethat's goingtogointothemediumthiswaySowe'rewe'rereallyexpandingourpicturehereTothinkabouthowtheseoscillationsoftheatominsidethematerialAren't justcausedbytheincidentwavebutaffectthetransmittedwaveaswellasgiveriseto a wavereflectedfromthesurfaceSoratherthangointo a fullderivation.
LetmejustcoverreallythebasicsbeforewegetintosomecasesSowe'reinterestedinwavesincidentonsurfacesandessentiallywhatwe'regoingtodoiswe'regoingtoassumethatthere's someincomingwaveWith a K vectorcaseof I that's thedirectionIt's cominginandtheincidentangleisdefinedrelativetothesurfacenormalandthatincidentangleisthetasub I righthereAndsothatangleisrightheretheanglebetweenthedashedlineWhichisnormaltothesurfaceandthedirectionoftheincoming K vector?
There's alsosomethingcalledthelawofreflectionwhichsaysthereflectedangleThetasub R isalsoequalTothetasub I soherewehavethetasub R theangleofreflectionandthisanglethetasub R Andthisanglethetasub I areequaltooneanotherinotherwordsformostflatsurfacesTheangleofreflectiondoesn't gooffin a differentdirectionthantheangleofincidenceAnotherthingthat's goingtomakeourlives a loteasieristocharacterizethematerialbytheindexofrefractioninsoessentiallyweHavesomeincidentindexofrefractionThat's theindexofrefractionofthemediumtheincidentwaveisgoingthroughoveronthissideAndwealsowearealsogoingtotalkaboutthetransmittedindexofrefractionthat's theindexofrefractionofthisgraymediumoverherethatthetransmittedwavegoesintoandTheindexofrefractionessentiallyin a generalcaseisgivenbythesquarerootofepsilon R andmusub R ofcoursemostMaterialsarenon-magneticsomusub R isequalto 1 soinmostcasesyoucansimplysaytheindexofrefractionofthesquarerootoftherelativepermittivityandOfcoursesincetherelativepermittivityinfreespaceis 1 theindexofrefractionoffreespaceis 1 aswellHowever, we'vealsoseenthatinlossymaterialsyouhavecomplexpermittivityandinthegeneralcasewecanessentiallyrepresenttheindexofrefractionthiswayis a squarerootofThecomplexepsilondividedbyepsilonnaughtandessentiallythenwecanbasicallytiethatbacktoanalphaand J betatermandTobeperfectlycorrect.
ThisisnotanequalsignthisessentiallysaystheinthecomplexindexofrefractionEssentiallyisrelatedtoalphaandbeta, butisnotequaltoalphaandbetaSoletmejustclarifythatsincethat's a mistakeintheslideNowifthematerialisverylossyorifit's somethinglike a metalthenthenmostoftheincidentradiationisgoingtobeReflected, butifit's a dielectricmaterialthenthetransmittedwavewhichessentiallyhas a k vector K sub T hereiscomingin a differentangleSothetasub T isnotnecessarilyequaltothetasub I orthetasub R thetransmittedwavegoesin a differentdirectionthantheincidentwaveandthereflectedwaveandWecancalculatetheangleofthattransmittedwaveusingsomethingcalledSnell's law, whichisgivenrightherethatthetheIndexofrefractionin I ofthewavecomingintimesthesineoftheincidentangleincidentangleisequaltotheindexofrefractionforthematerialthewavesgettingtransmittedintotimestheangleoftheTransmittedwavesoifweknowinsub I insub T andtheincidentanglethetawecando a simplealgebraicmanipulationtocalculatetheangleoftransmissionAndsothosereallyarethebasicsofwavesandincidentonsurfacesifyouunderstandthesebasicstherestofitAlthoughtherearesomelongequationsfollowsprettystraightforwardlynowthefirstcomplicationwerunacrossisthattheReflectionandtransmissionthatweseedependonthedirectionoftheelectricfieldAndsothefirstdirectionofelectricfieldwe'regoingtotalkaboutiscalled s orperpendicularPolarizationandthewaywewedefinethisiswethinkaswe'vetalkedaboutbeforeof a planeofincidenceandthisplaneofincidenceistheplanethatcontainsboththeincidentandtheIntheplaneofincidencethenormaltothesurfaceliesintheplaneofincidenceandiftheelectricfieldisstickingupoutOftheplaneofincidencesothere's essentially a 90-degreeangleBetweentheelectricfieldandtheplaneofincidencethenwecallit s orperpendicularpolarizationwhereperpendicularmeansIt's perpendiculartotheplaneofincidenceNowiftheelectricfieldisperpendiculartotheplaneofincidenceWecancalculatethemagnitudeofthereflectedfieldandthetransmittedfieldgiventhemagnitudeoftheincidentfieldsoessentiallythisincidentfieldisgoingtocomeinlikethissomeofitsgoingtobounceoffandtherewillalsobeanelectricfieldpointinginthesamedirectionthatwe'regoingtocall e Sub r orthereflectedfieldandtherelationshipbetweentheincidentfield e sub i Righthereandthereflectedfield e sub r righthereisThereflectedfieldandnoticewehavethislittleperpendicularsignthatsaysit's perpendiculartotheplaneofincidenceisequaltothecoefficient r perpendiculartimesthetheincidentfieldandEssentiallywithoutderivingitthereflectioncoefficientforperpendicularlyor s polarizedradiationisgivenbythisitdependsbothontheincidentindexofrefractiontheindexofrefractionofthematerialthatthewavesreflectingfromaregettingtransmittedintoandthecosineofboththeincidentangleandTheanglethewavesbeingtransmittedatandrememberSnell's lawsaysinsub i sineofthetasub i isEqualtotheindexofrefractionofthematerialthatthewavesgettingtransmittedintoTimesthesineofthatangleandsowecalculatethetasub t fromthisequationrighthereSooverallthisgivesthereflectedfieldOnceweknowallofthesethingssimilarly, there's goingtobeanelectricfieldpointinginthesamedirectionThat's transmittedintothematerialWecancallit e sub t hereandtherelationshipbetweenthetransmittedfieldandtheincidentfieldisgivenby a verysimilarexpression, butweusetheterm t perpendicularfortheWefinditdependsontheincidentandtransmittedindicesofrefractiontheincidentandtransmittedanglesAndifweplugitintothisequation, we'llknowwhatfractionoftheelectricfieldgetstransmittedinotherwordstheratioBetweentheincidentfieldthemagnitudeoftheincidentfieldonthetransmittedfieldandbecausethey'revectorsthevectorsaregoingtopointinthesamedirectionNow I'm using a lotofLineshere, butwehavetorememberthattheseareactuallywavescominginasshowninthebottomfiguredownhereSotheyoscillateupanddownandthey'replanewavesSoweassumethateverywhereontheseplanesthatarerepresentedbytheselittlegreensquaresTheelectricfieldis a vectorfieldthevectorsallpointinthesamedirectionButthatdirectioncanvarywithtimeandspaceaswesawinprevioustalksAndagain, theanglesarejustgivenbythe K vectors K incident K transmitted K reflectedwehaveexactlythesamesituationorveryclosetothesamesituationiftheNoticethatinsteadoftheelectricfieldpointingupanddownlikebeforePerpendiculartotheplaneofincidencenowtheelectricfieldliesintheplaneofincidenceandwecallthis P orparallelpolarizationBecausetheelectricfieldisintheplaneofincidence.
これは等号ではなく、複素屈折率を表しています。本来はアルファとベータに関係しますが、アルファとベータに等しくはありません、しかし、誘電体材料であれば、透過波は本質的にkベクトルK sub Tを持ち、異なる角度でやってきます。つまり、θ sub Tは必ずしもθ sub Iやθ sub Rと等しくなく、透過波は入射波や反射波とは異なる方向に進みます、入射する波の屈折率Iに入射角の正弦を乗じたものが、透過する波の屈折率に透過波の角度を乗じたものに等しいので、I下、T下、入射角θがわかれば、簡単な代数操作で透過角
Wehave a reflectedelectricfieldalsolyingintheplaneofincidence e sub R Wehave a transmittedelectricfieldWe'regoingtocall e sub T andtherelationshipbetweentheincidentfieldandthereflectedfieldisgivenbythosesetsofequationsTherelationshipbetweentheincidentfieldandthetransmittedfieldisgivenbythosesetsofequationsagainThefieldspointinprettymuchthesamedirectionTheelectricfieldsinbothcasesareparalleltotheplaneofincidenceExceptthemagnitudeofthefieldtheamountthatgetsthroughvariesdependingonthereflectioncoefficientAndthetransmissioncoefficientagainveryverysimilartheequationsareslightlydifferentbutyoudothecalculationonceyouknowthetheIndexofrefractiononeithersideofthematerialtheincidentangleandfromSnell's lawyoucalculatethetransmittedangleAndagain, let's stressisshowninthebottomthatthisis a wavecominginIt's notjust a linealthoughwecanrepresentitas a lineandthatthewaveessentiallymapsanelectricfieldtoeverypointinspaceGivenbyplanesthatareperpendiculartothepropagationdirectionofthewaveSowhatyou'llseein a lotofbooksorfiguresthatlooklikethisfor s polarizationWehavetheelectricfieldstickingupoutofthescreenrightatyouforperpendicularOrexcusemeforparallelor p polarization.
入射面に横たわる反射電界もある e sub R 透過電界もある e sub T と呼ぶことにする 入射電界と反射電界の関係はこれらの方程式で与えられる 入射電界と透過電界の関係もまたこれらの方程式で与えられる 電界はほとんど同じ方向を向いている 電界の大きさは反射係数と透過係数によって変化する。どちらの場合も電界は入射面に平行です。電界の大きさを除けば、透過する量は反射係数と透過係数によって変化します、また、波の伝搬方向に対して垂直な面によって与えられる電界は、基本的に空間のあらゆる点にマッピングされます
WehavetheelectricfieldintheplaneofthescreenWehaveessentially a distancebetweenpeaksWe'regoingtocallthewavelengthwhenwegointo a materialthewavelengthchangesbecausethephasechangeswhenwegointothematerialandEssentiallythroughSnell's lawwecanrelatetheincidentangleandthetransmittedangleWeknowtheincidentangleisequaltothereflectedangleandwecancalculatethemagnitudesof e sub i wellIfwe'regiven e sub i andwecancalculatethemagnitudesof e sub t and e sub r throughthereflectionandtransmissionCoefficientsforeithertheperpendicularpolarizationcaseortheparallelpolarizationcasequicksummaryAndthesearethetypesoffiguresthatyou'regoingtoseeinmostofyourtextbooksSolet's considerthegrayplanehereinthisfigurebeingtheplaneofincidenceWhathappensifwehaveanelectricfieldthat's notintheplaneofincidenceinotherwordsthisgreenlinethatwe'regoingtoDoesn't stickineitherthe s-planeperpendiculartotheplaneofincidenceorthe p-planeparalleltotheplaneofincidenceThisisprettystraightforward.
このような場合、スネルの法則によって入射角と透過角を関連付けることができます。 入射角と反射角が等しいことが分かっているので、e sub iの大きさを計算することができます。入射角と反射角が等しいことが分かっているので、e sub i の大きさを計算することができます。 e sub t と e sub r の大きさは、垂直偏光の場合と平行偏光の場合の反射係数と透過係数から計算できます。面にも入射面に平行なp面にも刺さらない。
ThisisjustlikeweworkedwiththeparallelizationessentiallyWe'regoingtobreakitintotwocomponentsWe'regoingtosaythereisonecomponentthatis s polarizedorperpendiculartotheplaneofincidenceWe'regoingtoadd a secondcomponent.
WecanapplytheFresnelequationsforthereflectionortransmissionforeachcomponentseparatelyandusesuperpositiontosumupandgettheoverallelectricfieldQuickmessagehereifyouaregiven a problemwhereyouhaveanelectricfieldthat's notpolarizedPerpendicularorparalleltotheplaneofincidenceusesuperpositiontobreakitupEssentiallyifyouknowthisanglethentheparallelcomponentisgoingtobethecosineofthisangle.
Let's callitphithePerpendicularcomponentisgoingtobeproportionaltothesineofphithroughsimplesineofphithroughsimplegeometryandEssentiallywejustdoourcalculationtwiceoneforparalleloneforperpendicularSumitupattheendtogettheoverallfieldsuperpositionworkshereaswellSo I'vetakenyouonkindof a whirlwindtourofFresnel's equationsLet's actuallystopfor a minuteusesomerealnumbersputinsomeRepresentativevaluesandtrytofigureoutwhatwe'redoinghereWhat I'm goingtoassumeisthat I'vegotessentially a blockofmaterialTheindexofrefractiononthissideis 1 soweassumetheincidentwaveisgoingthroughfreespacetheindexofrefraction e subinsub t is 2 inthematerialand I justchosethosenumbers I couldhavechosenanysetofnumbersButthesearetheones I chosesoourincidentwaveisgoingtocomeinhereatsomeanglethetasub I tothenormalandWewanttoknowhowstrongorwhatthemagnitudeofthereflectedandtransmittedfieldsareInessence, what I'vedoneis I wentintothecomputerprogram I liketouseforplottingcalledMATLABand I'veessentiallyplottedtheequationsforthereflectioncoefficientandthetransmissioncoefficientas a functionofincidenceangletheta I andsotheta I isgivenindegreesrighthereandLet's firstlookattheelectricfieldthat's reflectedSoiftheelectricfieldis P polarized, soourelectricfieldvectorpointsinthatdirectionrightthereThenthegreenlineessentiallyrepresentsthestrengthoftheelectricfieldthat's reflectedSoyoucanseeifyoucomeinatnormalincidencein a materialWhoseindicesofrefractionaregivenbyoneontheincidentwaveandtwoonthetransmittedsideabout 0.35 oftheelectricfieldor 35 percentisgoingtobereflectediftheincidenceangleiszeroorthebeamscominginstraightOninthatdirection, orifyou'vecomeouttoabout a 30 degreeanglethathasn't droppedverymuchButbythetimeyougettoabout 50 degreesOnlyaboutmaybe 0.16 oftheincidentfield 16 percentisreflectedAndasyoudropdownto 90 degrees, youcanseethatprettymuchallthelightSo I'vetakenyoukindofon a whirlwindtourofFresnel's equationSolet's stopfor a minuteandessentiallydothecalculationslet's actuallycalculatethereflectionandtransmissioncoefficientsanddoitfor a Setofmaterialswhereessentiallytheincidentwaveiscominginwithanindexofrefraction 1 Sowe'reinfreespaceoveronthissideandtheindexofrefractionofthetransmittedwavefortheindexofrefractionthematerialthewavesgoingintois 2 andEssentiallyifwedothatwecanuseFresnel's equationsgiventheincidentelectricfieldtocalculateRelativelyhowstrongthereflectedandtransmittedelectricfieldsaresolet's firstlookatthereflectedelectricfieldIfwedothiscalculationandessentiallypluginthetasub I and I'verepresentedtheangleofincidencehereindegreesSothisis a 30 degreeangleofincidence a 70 degreeangleofincidence a zerodegreeangleofincidencedownhereisEssentiallygoingtobewhenthewaveiscominginatnormalincidencebecausetheanglebetweentheincident k vectorandthesurfacenormalisZero, there's nodifferencebetweenthemfornormalincidence.
ここではφと呼ぶことにしましょう。垂直成分はφの正弦に比例し、φの正弦は単純な幾何学で計算されます。入射波が自由空間を通過すると仮定して、屈折率e sub in sub tは材料の中で2です、入射角θ Iの関数として反射係数と透過係数の方程式をプロットしました、そこで電場ベクトルはその方向を指しています。緑色の線は基本的に反射される電場の強さを表しています。入射角がゼロの場合、あるいはその方向にまっすぐ入射してくるビームの場合、電界の35パーセント、つまり35パーセントが反射されます。入射野の16パーセン
Youcanseethatfortheelectricfield.
電界を見ればわかるだろう。
That's polarized S orperpendiculartotheplaneoftheelectricfieldinotherwordsWe'retalkingaboutelectricfieldsthatarepointingupinthisdirectioninthiscaseWhatyou'regoingtoseeisthatfornormalincidence?
この場合、電界はこの方向を向いているのです。
You'regoingtoget a reflectioncoefficientofaboutminuspointthreefivenowWhatdoesthatminussignmeanitsimplymeansthere's a hundredandeightydegreephaseshiftintheelectricfieldso s or?
perpendicularpolarizationSees a phaseshiftwhenithitstheinterfaceHoweverastheanglegetslargerandlargerasthetaincidentbecomesfurtherandfurtherawayfromthenormalthereflectioncoefficientGetslargerandlargerorthemagnitudedoesitactuallygetsmoreandmorenegative?
Untildownhereat 90 degreesthereflectioncoefficientiscloseto 1 whichessentiallyissayingThatallthelightisreflectedWhathappensifwetake a lookatthe p polarizedcaseorthewhentheelectricfieldlies?
90度までは反射率は1に近く、すべての光が反射されている。
paralleltotheplaneofincidentInthiscasewegoaheadanderasetheseelectricfieldsandputourelectricfieldsinthatdirectionYoucanseethatwegetabout a thirtyfivepercentreflectionatnormalincidencethatDropsoffuntilatsomeanglewegetzeroreflectionhereAndtheneventuallythereflectioncoefficientbecomesnegativeandwestarttogetthathundredeightydegreephaseshiftagainandsoessentiallywhatthiscurveistellingusisthatit's tellinguswhatthephaseshiftandThestrengthofthereflectedelectricfieldisas a functionoftheincidentangleSimilarlyifwewanttoknowthethestrengthofthetransmittedfieldWecanplotthetransmissioncoefficientas a functionoftheangleofincidenceWhich I'vedonehereforperpendicular s andparallelor p polarizedandessentiallyifwetake a lookatthatMaybe 65% ofthewavegetsthroughThismakes a lotofsensesinceitwasonly 35% overontheothersideButasthewavecomesinatsteeperandsteeperangleseventuallywedroptozerotransmissionandallthewavegetsreflectedItmakes a lotofsense.
There's nophaseshiftifthevalueisnegativeThere's a hundredandeightydegreephaseshiftandthat's prettymuchaboutit.
その値がマイナスであれば位相のずれはない。
We'vegot a fewspecialcases.
いくつか特別なケースがある。
WeneedtotalkaboutOneofthesecasesissomethingcalledtotalinternalreflectionItturnsoutthatifyouhave a wavecomingfrom a materialthathas a highindexofrefractionSointhiscaseinsub I isgreaterthaninsub T AslongasyouranglesofincidencearesmallorthethedirectionofpropagationisIsprettymuchclosetothenormalthenthingscomeoutasyouwouldexpectyouget a reflectedfieldandtransmittedfieldButastheta I increasesandgetsbiggerasyou'regoingfromthematerialwithhigherindextothematerialwithlowerindexessentiallythedirectioncaseof T ofThisvectorisgoingtomovethatwayastheincidentangleincreasesandatsomepointit's goingtoliealongthesurfaceThismeansthatalltheradiationthatcomesinthisdirectionisgoingtogetreflectedYou'regoingtoget a hundredpercentreflectionoftheradiationandthere's goingtobenopropagatingradiationthatgoesoutandthisiscalledtotalinternalreflectionfortheobviousreasonthatthetotalamountoftheradiationgetsReflectedfromthesurfacegoingfrom a materialofhigherindexintolowerindexTheplacethisismostcommonlyusedasanopticalfibersbecausethisiswhatkeepsthelightinsidefiberstogoveryverylongdistancesandEssentiallyifyouwanttocalculatewhattheangleistheincidentangleiswheretotalinternalreflectionStartstooccuryousimplyuseSnell's lawyoubasicallysettheta T isequalto 90 degreesorgreaterandEssentiallyyoucanfindthatcriticalangleLet's callittheta C thereisgiventhebytheequationthesineofthecriticalangleistheInsub T thetransmitteddividedbytheincidentindicesofrefractionthesecondcasewe'vealsomentionedverybriefly, butthat's calledBrewster's angle, andthat's essentiallythepointwheretheReflectioncoefficientoftheparallelor P polarizedelectricfieldcomponentisequaltozerorighthereinthiscaseYougetnoreflectionwhatsoeverAndsofortheta I equaltoBrewsteranglesandBrewsteranglesgivenbythatequationrighttherewheretheta B isBrewsteranglesoforthetasub I equalThetasub B.
ThereisnoreflectionandalloftheradiationistransmittedSoifyouneed a veryveryhightransmissionofradiationThenmakingsureyouareincidentatBrewster's angleisanimportantthingtodonoticethathoweveratBrewster's angleLet's draw a linedownhere, whichinthisparticularcasehappenstobeabout 64 degreesThatthereis a reflectionoftheperpendicular s polarizedlightAndthisiswhypolarizingsunglassesworkbecauseessentiallyRadiationpolarizedinonedirectionoffofshinysurfaceslikepoolsoroceansExperiencesmuchlessreflectionthantheotherotherpolarizedcomponentinotherwordstheparallelcomponenthasverylittlereflectiontheperpendicularcomponenthas a lotbyblockingoutoneofthosedirectionsofopticalradiationyoucanessentiallyreducetheglareandWeuseBrewster's angle a lotwhenwe'rebuildinghigh-poweredthingslikelasersthathavesomuchpowerinsidethemthateven a littlebitofreflectionor a littlebitofabsorptionmightcausetheThedevicetofunctionpoorlyortoheatuporcausesomedamageinternallyandsoweuseBrewster's anglewhenwehavetominimizeReflectionforcertaincases, butthat's beyondthescopeoftheclass