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湖北大学:《大学物理》第十七章(17-3) 薄膜干涉

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2.薄膜干涉(Thin- Film Interference) Color of oil slick Color of soap bubble a broad Source
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2.薄膜干涉(Thin- Film Interference) Color of oil slick Color of soap bubble a broad Source transparent thin-film

2. 薄膜干涉(Thin-Film Interference) Color of oil slick Color of soap bubble a broad source transparent thin-film S1 S2

a DB= ABsin i n> n D 2etgrsin i n18=2n2 AC-n, DB+ B n e =2n 2n,etgrsin i cos r AD可看作同相面 n,sn l-n2 sin 6=n2(AC+CB)=2n2 SIn i 2n,e sin i+ coSr cosr (n, DB 2n.e (1¥si2r)+ cosr AC= CB cosr one cosr+

a b n1 n2 n1 B C A D n2> n1 AD可看作同相面 ( )  = n2 AC +CB r e AC CB cos  = = i r e DB = ABsin i 2 2 2 1   = n AC − n DB + n etgr i r e n 2 sin cos = 2 2 − 1 2  + n sin i n sin r  1 = 2 ) 2 ( 1  − n DB − = 2etgrsin i i r r n e r e n sin cos sin 2 cos 2 = 2 − 1 2 (1 sin ) cos 2 2 2  = − r + r n e 2 2 cos 2  = n e r + 2  + i

a n> n 2n (1-si2r)+ D 2 1 2n.+ B n e n sin l=n sin r C n The conditions for a or:8=2en,-ni sini+ milnor a maximuin intensity: o=2evn2-ni sin t+ k k=1.23 (max) (2k+ k=0.1.2.3.…(min)

a b n1 n2 n1 B C A D n2> n1 i r e n sin i n sin r  1 = 2 2 (1 sin ) cos 2 2 2   = − r + r n e 2 2 cos 2  = n e r + 2 2 sin 2 2 1 2 2   = e n − n i + or: 2 2 sin 2 2 1 2 2   = e n − n i + 2 (2 1)  k + k ={ k =1.2.3.  k = 0.1.2.3.  (max) (min) The conditions for a minimum or a maximum intensity:

Equal inclination interference (等倾干涉) Shown as the fis sereen a broad lens SoUrce 2 3 thi wo rays“1”、“2” are not coherent light!

Equal inclination interference i i i i (等倾干涉) Shown as the fig. r thin film lens e a broad source “1” “2” “3” “4” two rays“1”、“2” are not coherent light! i screen

k+1 onclusion 1) the same i-angle corresponds to same grade of interference (equal inclination interference等倾干涉) 2) the smaller i-angle is, the greater optical path difference is, and the higher k is Discuss 1)

i i' i i' r Conclusion: 1)the same i-angle corresponds to same grade of interference (equal inclination interference等倾干涉) 。 2)the smaller i-angle is, the greater optical path difference is, and the higher k is. k k +1 Discuss: 1)

D)when thin film thickness increases frings extend outwardly δ=2eVn2-n2si2i+=kk=1.2.3.…max) when thin film thickness e increases frings extend outwardly

1)when thin film thickness increases: frings extend outwardly 2 2 sin 2 2 1 2 2   = e n − n i + = k k =1.2.3.  (max) i i' i i' when thin film thickness frings extend outwardly e increases i , :

2)when thin film thickness decreased frings contract] inwardly e 6=2en2-n2sin2i+=kk=1.2.3.…明纹 frings extend outwardly e decreases frings contract inward

2)when thin film thickness decreases: i i' i i' 2 2 sin 2 2 1 2 2   = e n − n i + = k k =1.2.3.  明纹 frings extend outwardly frings contract inwardly e decreases i , frings contract inwardly e

3)if white light falls on, colour fringes appear from red to violet 8=2evn2-nf i+o=kn 2 k=1.2.3.…(max) ek一定,λ大,δ大,Vm-nsm2大 i小,条纹靠中心 3)等倾干涉定域在无限远,只能通过透镜或将 眼调到聚焦无限远才能看到

3) if white light falls on, colour fringes appear from red to violet. 2 2 sin 2 2 1 2 2   = e n − n i + = k k =1.2.3.  (max) e.k 一定,  大,  大, n n i 2 2 1 2 2 − sin 大 i 小,条纹靠中心 3)等倾干涉定域在无限远,只能通过透镜或将 眼调到聚焦无限远才能看到

2 屏 3 扩展 透镜 “4”光线之间虽 光源 非相干光,但 在同一倾角下 2 加强则同时加 3” 强。减弱则都 减弱 薄膜 4)扩展光源成为观察等倾干涉条纹的有利条件

i i i i 薄膜 透镜 屏 e 扩展 光源 “1” “2” “3” “4” 4)扩展光源成为观察等倾干涉条纹的有利条件。 “1”、“2”、 “3” “4”光线之间虽 非相干光,但 在同一倾角下, 加强则同时加 强。减弱则都 减弱

Equal thickness interference(等厚干涉) 8=2evn2-nf i+ sini=o (2k+ 0Vm=1 Suppose: the thickness corresponding each fringes are respectively: k k k=.1.2.3.…(max) 6=2en+ 2((2k+1)k=0.1.2.3.…(min) (1)

Equal thickness interference (等厚干涉)  Suppose:the thickness corresponding each fringes are respectively: e1 .e2 .e3 ek  2 2 2   = en + 2 (2 1)  k + k ={ k = .1.2.3.  k = 0.1.2.3.  (max) (min) ……(1) n =1 6.wedge(劈尖干涉) 2 2 sin 2 2 1 2 2   = e n − n i + 2 (2 1)  k + k ={ sin i = 0

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