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Download (等倾干涉) — equal thickness interference.
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30-3. Interference in Thin Films (P691) 1. Equations for Thin-Film Interference Both rays 1 & 2 are derived from the same incident beam n2 (ab bc ) n1ad P 1 i d. a L /2 (n2>n1;n1>n2) 0 (n1<n2<n3;n1>n2>n3) 2 i .b . c n1 n2 n1 (Pay more attention on the half-wavelength loss ! ) we get 2 L n22 n12 sin2 i ( i , L) which depends on the thickness L of film and incident angle i. Bright fringe 2 m ( m 1 , 2 , 3 , ) ( i , L) 2m22 (m 1,2,3, ) Dark fringe ((22mm''11)) (m ' 0 , 1 , 2 , ) ( m ' 0 , 1 , 2 , ) 2. Discussion: 22 (1) If L= constant, optical path difference depends on the incident angle i — equal inclination interference. (等倾干涉) The rays with the same incident angle i form a fringe with the same order. (2) If i = constant, Optical path difference depends on the thickness — equal thickness interference. (等厚干涉) The same order of fringe corresponds to the same thickness L. (3) It’s a method dividing amplitude to get coherent light. 3. Wedge-shaped film (劈尖) (P692): Incoming ray is perpendicular to a thin wedge which is put in the air. Bright fringes 2 m ( m 1 , 2 , 3 , ) 2 2n 2 L Dark fringes 2 (2m'1) (m' 0,1,2, ) 2 : 10 ~ 10 rad 4 Supplementary explanation: ray1 ray2 5 air a (1) The zeroth dark fringe on the edge shows the half-wavelength loss. n2 · air L (2) The same fringe corresponds to the equal thickness—— equal thickness interference. (3) Separation of fringes: L 2n2 l B D l 2n2 sin Lm Lm+1 ,l L l sin (4) For multi-chromatic incident light, a series of colorful fringes is seen next to one another. L L 2n2 ( n2=1 for air ) (5) Applications: l • measure small length: Scale: wavelength of light Magnification coefficient: 2n sin 1 2n sin • test flatness of surface (检查表面平整度): The pattern observed shows when plates are optically flat. 等厚条纹 待测工件 The pattern observed shows when plates are not so flat. 例 工件上放平板玻璃成空气劈尖,单色光垂直照射, 干涉条纹如图。工件表面上纹路凹还是凸?纹深? 解:等厚线:纹路是凹的 a 2 H b2 a sin H b sin ab b a 平板玻璃 工件 H H lk 纹路处 4. Newton’s Ring (P692): The interference only occurs for rays reflected by the top and button surfaces of the air film. m 1, 2, 3 Bright 2m 2 2e 2 (2m '1) m ' 0, 1, 2 Dark 2 ( 2m 1) R Newton’s ring rB 2 , m 1,2,3, C R e rD m' R , m' 0,1,2, (1) It is a typical case of equal thickness interference (2) The dark central point is the result of half-wavelength loss. Unequal separation r B O A 30-4. Michelson Interferometer (P695) M1&M2are 2 mirrors M1 is the virtual image of M1 formed by the splitter mirror. M2 M1 interference in the thin air film between M1 and M2, d2 S G2 2 compensator 1 E 2 d m d m 2 M1 1 beam splitter M1//M2 equal inclination interference M1//M2 equal thickness interference Applications: d1 2 G1 等厚条纹 EXAMPLE: A thin film with index of refraction n = 1.40 is placed in one arm of a Michelson interferometer, perpendicular to the optical path. If this causes a shift of 7.0 fringes of the pattern produced by light of = 589 nm, what is the film thickness? Solution: The change of optical path difference due to the insertion of the film of thickness L is: 2(n 1) L Δ m Δm (589nm )(7.0) L 5.2 m 2(n 1) 2(1.40 1) •Questions (思考题) •P697 13; P69715 •Problems (练习题) • P699 23; P69925; P69932