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1 P1X: Optics, Waves and Lasers Lectures, 2005-06. Lecture 5: Interference and diffraction of light (II) Interference in Thin Films (Y&F 35.4) o Thin films on a substrate undergo the phenomenon of multiple wave interference. o Examples are the bright bands of colour from a thin layer of oil floating on water, the patterns in soap bubbles, antireflective coatings in glasses. Thin film of oil on water n2n2 n1n1 n3n3 t d n 1 < n 2 < n 3

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2 P1X: Optics, Waves and Lasers Lectures, 2005-06. Geometric description of thin film interference o Interference between bc and ef: At b there is a reflection: change of phase = ( r = /2) if n 1 <n 2 At d there is a reflection: change of phase = ( r = /2) if n 2 <n 3 So constructive interference when: bc + /2 - (bd + de +ef + /2) = m f e b n2n2 n1n1 n3n3 t a c d n 1 < n 2 < n 3 o If light perpendicular: bd + de = 2t bc=ef Therefore: 2t=m for constructive interference m=0,1,2,… 2t=(m+1/2) for destructive interference m=0,1,2,... 2 = wavelength in medium n 2

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3 P1X: Optics, Waves and Lasers Lectures, 2005-06. Note: When light is reflected off a medium that has higher refractive index than the initial medium, there is a phase shift of rads (equivalent to /2). If the second medium has a smaller refractive index, there is no phase shift. If the two refractive indices are equal there is no reflection.

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4 P1X: Optics, Waves and Lasers Lectures, 2005-06. Examples of thin film interference o Non-reflective coatings: We want to create a thin film on a substrate that is non-reflective, so we want destructive interference. If we have a quarter-wave plate: t= 4 then: Path difference = 2t = /2 (destructive interference) v 1 =c/n 1 so: f 1 = c/n 1 v 2 =c/n 2 so: f 2 = c/n 2 If n 1 = 1 (air) then: 2 = 1 /n 2 therefore: 2t= 2 /2 = 1 /(2n 2 ) o The optical path-length between two interfering waves is: 2t n 2 = 1 /2 2 = c/ (f n 2 ) = 1 (n 1 /n 2 ) physical path-length

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5 P1X: Optics, Waves and Lasers Lectures, 2005-06. o Example 35-8 (Y&F): Non reflective coatings for lenses are normally designed for 550 nm (central yellow-green part of the visible spectrum). Magnesium fluoride (MgF 2 ) with n=1.38 is used as a lens coating material for a glass of n=1.52. What thickness should the coating be? The wavelength of light in the coating: 2 = 1 /n 2 = 550 nm/1.38 = 400 nm Therefore: t = 2 /4 = 400/4 = 100 nm. MgF 2 n 2 =1.38 Air n 1 =1 Glass n 3 =1.52 t d n 1 < n 2 < n 3

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6 P1X: Optics, Waves and Lasers Lectures, 2005-06. o Interference in an air wedge between two glass plates: Ray 1: reflected at A (no phase change) Ray 2: travels path ABC (phase change of at B) o Constructive interference: 2t - /2 = m 2t = (m+1/2) o Destructive interference: 2t + /2 = (m+1/2) 2t =m Difference is ABC = 2t - /2 (it is small). C t B A n2n2 n1n1 n3n3 h L xmxm 21 o Distance x m from line of contact to mth dark fringe: Since 2t m =m and x m = t m (L/h) then: x m = (m /2)(L/h)

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7 P1X: Optics, Waves and Lasers Lectures, 2005-06. Newtons rings: interference between a curved lens (radius R) and a flat glass plate inside a medium of refractive index n (e.g. n=1 for air and n= 1.33 for water). Observation of concentric dark and light fringes with centre in the centre of the lens.

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8 P1X: Optics, Waves and Lasers Lectures, 2005-06. Phase change of (equivalent to /2) at the bottom surface reflection: r 1 -r 2 = 2t therefore: Optical path length = n(r 1 -r 2 ) = 2tn + /2 Dark rings (destructive interference): dark centre ring 2tn + /2 = (m+1/2) Therefore: Bright rings (constructive interference): 2tn + /2 = m Therefore: x m 2 = R 2 - (R-t) 2 = R 2 - (R 2 -2Rt+t 2 ) = 2Rt -t 2 When R>>t then x m 2 ~ 2Rt xmxm t R n

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