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Chapter 17 Wave Optics Part 1 Interference of light Colorful interfering phenomena Soap film under sunlight Oil film under sunlight Interference fringes produced by wedgeshaped air film between two flat glass plates Newton’s rings (Equal thickness fringes) Equal inclination fringes §17-1 Monochromaticity and Coherence of Light 1.The emitting process of light The life time of electrons on the excited state (激发态) is 11 8 ~ 10 10 s Stimulating pumping Stimulation Radiation The persistent of light is 8 ~ 10 s Lighting is intermittent(间歇) Optical wave train Different wave trains are not coherent light. · · independent(from different atoms) independent(from different time of same atom) 2. Laser (from stimulated radiation 受激辐射) E2 f= (E2-E1)/h f f Identical wave train f E1 (frequency, vibrating direction, phase) Coherent light 3. Monochromaticity(单色性) •Monochromatic light:has single frequency •Frequency width(频宽) f •wavelength width(线宽) Good monochromaticity: f, are smaller. •Spectrum curve (光谱曲线) :The intensity distribution of light with wavelength ( or frequency ) Good monochromaticity I Bad monochromaticity I 4. Coherency Coherent lights: Identical frequency, vibrating direction, phase Can produce coherent superposition The intensity distribution of resultant light in crossing space: I I1 I 2 2 I1 I 2 cos If I1=I2, I 4 I1 cos 2 2 The intensity of light varies between bright and dark alternately. If I1 and I2 are not coherent light, then I I1 I 2 No varying 5. The methods to obtaining coherent light: •principle:By using some arrangements, divide the light wave emitted by an identical point from monochromatic source into two beams. And superposing these two beams in the space. •methods: The way of division of wavefront: The way of division of amplitude.: The way of division of wavefront: S1 S d S2 The way of division of amplitude.: Study by yourself §17-2 Two beams interference I. Young’s double-silt experiment 1. The experiment arrangement and principle. 2.The positions of bright fringes and dark fringes. The path difference: r ? The distribution characteristics of the fringes and the distance between them. 3. Analyze the distribution characteristics of the fringes of the white light. II. Fresnel’s double-mirror experiment III. Lloyd’s mirror experiment A SS C d M M’ B S S’ E’ E Half-wave loss §17-3 Optical path and optical path difference, Property of thin lens S nr nvt v--speed of light in medium I. Optical path c n t ct n =Distance of light traveling through vacuum at the same time For a monochromatic light, f is same in different medium, but and v are different. Let in vacuum: , c In the medium with refractive index n:v , c Then ' , v n n II. Optical path difference n2r2 n1r1 The phase difference: 2 :the wavelength in vacuum ' III. The half-wave loss of reflecting light n1<n2 Phase shift Has half-wave loss n1>n2 No half-wave loss The transmitting ( refracting ) light never have half-wave loss. IV. A lens does not cause any additional optical path difference or phase shift. AF, CF travel a larger distance in the air and a shorter distance in the lens. BF A B C is inverse. AF= CF= BF ,they are converging on point F and forming a bright image point. F §17-4 Interference by division of amplitude I. Equal-inclination interference 1. The interference of reflecting light The optical path difference is 2n2 AB n1 AD 2 e AB cos e D i A C B n1 n2 n1 n1 AD AC sin i 2e tg sin i And n1 sin i n2 sin e D i A C B e 2n2 2n1etg sin i cos 2 2n2e 2 (1 sin ) cos 2 n1 n2 n1 n1 2n2e cos 2 n2 cos n2 1 sin n2 n1 sin i 2 2 2 2e n n sin i 2 2 2 2 1 2 The conditions that occur bright and dark fringes are k 1,2, --Bright fringes k 2k 1 2 k 0,1,2, --Dark fringes Discussions: (i ) ,As long as the lights coming from different area of the source have same incident angle i, they have same , and belong to same interference level k --Equal inclination interference S ii i i The interference fringes are series of concentric circles. They are alienation(稀 疏)near the center, and dense near the edge. R larger, i larger, R smaller, k smaller i smaller, k larger screen lens S Reflecting plate i film i screen lens S Reflecting plate i film i When e ,we have k , the circle fringes are produced from the center. When e ,we have k ,the circle fringes are swallowed at the center. At the center, ----Bright k 2n2e ( 2k 1) ----Dark 2 2 2. The interference of transmitting light The optical path difference is ' 2e n22 n12 sin2 i e C B k (2k 1) 2 n1 n2 n1 n1 ----bright ----dark k 0,1,2 The reflecting light and transmitting light are compensative each other. [Example] A thin oil film (n= 1.30) is illuminated by the white light. Someone observes the reflected light by the film. When the observing direction has the angle 300 with respect to the normal direction of the film, the film appears blue ( 4800Å). Find the minimum thickness of the oil film. If the film is observed at the normal direction of the film, what color does the film appear? Solution : According to 2e n n sin i k 2 k=1 e=emin 2 2 emin 2 1 2 ( 2 1) 4 n sin i 2 2 ( 2 1) 4.8 10 4 1.3 0.5 2 7 1.0 10 m 2 7 The film is observed at the normal direction of the film, i=0: 2ne 2 k 4ne 5.20 10 2k 1 2k 1 7 For k =1, 5.20 10 m --green For k =2, 1.733 10 m 7 7 --Ultraviolet II. Application 1.Transmission enhanced film (anti-reflecting film) (增透膜) n0 < n < n e n0 1 n 1.38 MgF2 n' 1.5 glass The two beams reflected by the upper and bottom interface of the MgF2 film all have halfwave loss. The optical path difference between and is 2ne ( 2k 1) 2 --reflecting beams are destructive. --transmitting beams are constructive. The or e n0 1 n 1.38 MgF2 n' 1.5 glass minimum thickness of MgF2:e nemin 4 min 4n ne --optical thickness 2. Reflection enhanced film(增反膜) Considering the reflected beams by each interface. For the first film, nH 2.40 nL 1.38 nL nH nL 2nH e1 2 k k 1,2, k=1 nH e1 4 For the second film, ZnS MgF2 2nLe2 2 k nH nL k 1,2, k=1 nLe2 4 nH n' 1.5 III. Equal-thickness interference 1. Wedged interference Assume the incident beams are perpendicular to the interfaces of the film. 2ne 2 --bright k ( 2k 1) --dark 2 n medium wedge At the contact edge(e=0) , k =0, appears dark fringe. For air wedge, 2e 2 Discussion air wedge The points with identical thickness e have the same interference level k. --Equal thickness interference The fringes are the straight lines parallel to the edge. They have same distance. And there is a dark fringe at the contact edge(e =0) because of half-wave loss. The thickness difference between two adjacent bright(or dark) fringes: l 2nek 2 k ek 1 2nek 1 2 ( k 1) e ek 1 ek For air wedge, ek 2n e 2 e The distance between two bright (or dark) fringes: l ek e e l 2n sin ek 1 e --identical distance When the upper interface of the wedge film is moved upwards, the distance of the fringes is constant, but all the fringes move toward the contact edge. A e B A e [Example]To determine the thickness d of the SiO2 over the Si precisely, it is usually corroded to a wedge shape. The light with =5893Å is incident normally from above. There are 7 bright fringes over the length of the film. Calculate d=? (Si: n1=3.42,SiO2: n2=1.50 are known.) n2 d n1 Si SiO2 Solution: The optical difference between two rays reflected by the upper and lower surface of SiO2 is n2 d n1 Si SiO2 k 2n2e k k 0,1, ek 2n2 At the contact edge, k=0,the bright fringe at d should correspond to k=6 6 6 d e6 1.1786 10 m 2n2 2. Newton’s ring R At points with thickness e , 2ne 2 --bright k ( 2k 1) --dark Plane-convex n r 0 e Plane glass 2 At the center(e =0): corresponding to a dark point with k =0. Discussion: is determined by e --equal-thickness interference The fringes are series of concentric circles. -- Newton’s ring the radius r of the circle: r R R e 2eR e 2 as 2 e R 2 2 R r 2 r e 2R n 0 Bright circle, rk 2k 1R 2n k 1,2 kR k 0,1,2 rk n the distance between two adjacent circles : Dark circle, ( k 1) R kR r n n 1 k 1 k R -- alienation(稀疏)near the n center, dense near the edge. When the Plane-convex lens moves upwards, the circle fringes are swallowed at the center. A B [Example] A drop of oil is on a plane glass. When a monochromatic light with =5760Å is incident on it normally, the interference fringes produced by the reflected lights are shown in figure. The center point of the oil is dark. Find Is the bright or dark fringe at the edge of the oil? The maximum thickness of the oil film =? If the oil spreads gradually, How do the fringes change? (oil: n2=1.60, glass: n3=1.50) n2 n3 Solution Because n1<n2,n2>n3, there is half-wave loss in . i.e., 2n e 2 2 At the edge of oil, e=0, we have 2 i.e., satisfies dark fringes condition, ( 2k 1) 2 So There is a dark circle fringe at the oil edge, corresponding to k=0. The center dark point corresponds to k=4. k emax 2n2 10 4 5700 10 7 7.2 10 m 2 1.6 When the oil spreads gradually, the dark circle fringe located in the edge spreads gradually, the center point changes from dark to bright, and to dark alternately, the level k of the fringe becomes less and less. §17-5 Michelson’s interferometer I. Instrument M ' 2 M1 is fixed, M2 can be moved M 1 d G1 Beam spliter G2 M 2 Compensating plate Discussion M1and M2are perpendicular to each other-equal inclination interference produced by air -- concentric circles film. Assume the thickness of the air film between M1and M2 is d Then, at the center of the fringes 2d k M2 moves d : ' 2( d d ) k' d ( k' k ) n 2 2 n--The number of the fringes out of or in the center. M1and M2 are not perpendicular to each other -- equal thickness interference produced by air wedge. II. Application d n 2 Measure length d—known , read out n, Measure ----read out n and d, Measure out d Look refractive index n—known ,read d n 1e for “ether(以太)” -- “ zero”result Michelson’s interferometer be used to measure the refractive index of gas. Michelson’s interferometer be used to measure flow field.