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ID:37291529
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页数:18页
时间:2019-05-20
《光电子学与光子学的原理及应用s.o.kasap 课后答案》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/71.4AntireflectioncoatingggForlighttravelinginmedium1incidentonthe1-2interfaceatnormalincidence,n31-n1-n2n1-n1n3n1r12===n1+n2n1+n1n31+n3n1Forlighttravelinginmedium2incidentonthe2-3interfaceatnormalincidence,n1n3-11-n2-n3n
2、1n3-n3n3n1r23====n2+n3n1n3+n3n1+11+n3n3n1thus,r23=r12Significance?Foranefficientantireflectioneffect,wavesA(reflectedat1-2)andB(reflectedat2-3)inFigure1Q4belowshouldinterferewithnear“totaldestruction”.Thatmeanstheyshouldhavethesamemagnitudeandthatrequiresthatthereflectioncoefficientbetween1and2s
3、houldbethesameasthatbetween2and3;r12=r23.Thus,thelayer2canactasanantireflection1/2coatingifitsindexn2=(n1n3).Thiscanbeachievedbyr12=r23.1/2Thebestantireflectioncoatinghastohavearefractiveindexn2suchthatn2=(n1n3)=1/2[(1)(3.5)]=1.87.Givenachoiceoftwopossibleantireflectioncoatings,SiO2witharefracti
4、veindexof1.5andTiO2witharefractiveindexof2.3,bothareclose.ThephasechangeforwaveBgoingthroughthecoatingofthicknessdis2k2dwherek2=n2koandko=wavevectorinfreespace=2p/l.Thisshouldbe180°orp.Thusweneed2n2(2p/l)d=por-9l900´10mForSiO2d===0.15mmmm4n4(1.5)2-9l900´10mForTiO2d===0.10mmmm4n4(2.3)21.8Thinfilm
5、coatingandmultiplereflections:Assumethatn16、aregivenby,n-nn-n1223r=r==-r,r=r=11221223n+nn+n12232n2n2n123andt=t=,t=t=,t=,11222123n+nn+nn+n121223Consider1-t1t2,SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7224nnn+n+2nn-4nn1-tt=1-12=1112121222(n+n)(n+n)1212(1)222n+n-2nnn-n=1112=12=r221(n+n)n+n1212Theamplit7、udeofthereflectedbeamisAreflected=A1+A2+A3+...-j2f2-j4f23-j6fi.e.Areflected/A0=r1+t1t2r2e-t1t2r1r2e+t1t2r1r2e+...(2)sothatthek-thtermfork>1isAreflectedt1t2-j2fk=-(-r1r2e)(3)Ar0k1sothatthereflectioncoefficientis¥Areflec
6、aregivenby,n-nn-n1223r=r==-r,r=r=11221223n+nn+n12232n2n2n123andt=t=,t=t=,t=,11222123n+nn+nn+n121223Consider1-t1t2,SolutionsforOptoelectronicsandPhotonics:PrinciplesandPracticesChapter1/2/3/7224nnn+n+2nn-4nn1-tt=1-12=1112121222(n+n)(n+n)1212(1)222n+n-2nnn-n=1112=12=r221(n+n)n+n1212Theamplit
7、udeofthereflectedbeamisAreflected=A1+A2+A3+...-j2f2-j4f23-j6fi.e.Areflected/A0=r1+t1t2r2e-t1t2r1r2e+t1t2r1r2e+...(2)sothatthek-thtermfork>1isAreflectedt1t2-j2fk=-(-r1r2e)(3)Ar0k1sothatthereflectioncoefficientis¥Areflec
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