1 6.2 Two slit interference Coherence TwoSlit Interference Thin film Interference Interference Effects Interference is a general property of waves. A condition for interference is that the wave source is coherent. Interference between two waves gives characteristic interference patterns due to constructive and destructive interference. Coherence For two waves to show interference they must have coherence. Two waves are coherent if one wave has a constant phase relation to the other coherent incoherent phase shift x 2 D f = p l Dx Coherence Light from two separate light bulbs is Incoherent Light from a single light bulb passing through a small slit is coherent Laser light is coherent Interference d=0 Sum Constructive Interference l Interference d= 2 l Destructive Interference2 Interference d=l Constructive Interference path difference path difference =d =r 2 – r 1 In phase Condition for constructive interference m d l = Condition for destructive interference 1 ( )2 m d l = + m = 0 + 1, + 2,…. r 1 r 2 Superposition of waves at A shows interference due to path differences A Order number m Coherent waves barrier Interference pattern of water waves Constructive Destructive Interference pattern due to two spherical waves Coherent waves Interference pattern Amplitude on screen d=0 d=l d=l m=0 m=1 m=1 Young’s two slit experiment Thomas Young Does light show wave properties? Light shows wave properties Path difference from the two slits path difference dsin d = q In the limit L>>d, the rays are nearly parallel perpendicular to r 1 and r 2 Not to scale3 Interference pattern Bright constructive interference bright dsin m q = l Dark destructive interference dark dsin (m 1/ 2) q = + l m = 0, + 1, + 2, ............ maxima m=2 Central maximum d q m=0 m=1 m= 1 m= 2 Wavelength of light Light from a laser is passed through two slits a distance of 0.10 mm apart and is hits a screen 5.0 m away. The separation between the central maximum and the first bright interference fringe is 2.6 cm. Find the wavelength of the light. for m= 1 l = yd mL dsinq=ml sinq ≈ θ ≈ y L for small angles y d m L = l solve for l L y d q First maxima m=1 - - - = = = 2 3 7 (2.6x10 m)(0.1x10 m) 5.2x10 m 520nm (1)(5.0m) 2.5cm 5.0 m 0.10mm Thin film interference Thin film interference • In thin film interference the phase difference is due to reflection at either side of a thin film of transparent material. • The phase difference is due to two factors: – Path difference through the film (corrected for the change in speed of light in the material) – Phase shift due to reflection at the interface Phase shift due to reflection n 1 < n 2 phase shift=180 o Reflection with inversion phase shift = 180 o Phase shift due to reflection Reflection without inversion Phase shift = zero n 1 > n 2 Phase shift = zero4 Thin film Interference For a film in air the phase difference due to reflection is 180 o. Condition for destructive interference d=2t= film m m n l l = The wavelength in the film is shorter than in air. If the path difference (2t) is negligible then there is destructive interference. Destructive interference occurs when the path length difference equals integral multiples of the wavelength. Interference between light reflected from Top and bottom surfaces. m=0, 1, 2 , 3 …………. Thin film Interference for a soap film in air Condition for constructive interference d=2t= film 1 1 (m ) (m ) 2 2 n l + l = + For constructive interference the path difference must be half integral multiples of the wavelength to make up for the phase shift on reflection. Soap film Question A vertical soap film displays a series of colored band due to reflected light. Find the thickness of the film at the position of the 5 th green band (l=550 nm, n =1.33) Constructive Interference The 5th band has m=4 (the first is m=0) 1 2t (m )2 n l = + 1 t (m )2 2n l = + 9 7 1 550x10 (4 ) 9.3x10 m 930nm 2 2(1.33) - - = + = = Antireflective Coating Antireflective coatings are used to reduce reflections at the airglass interface. antireflective coating no coating Antireflective Coating Antireflective coatings consists of a thinlayer of material with a refractive index in between that of air and glass. Destructive interference between light reflected at the two surfaces reduces the intensity of reflected light. n 1 =1.00 < n 2 < n 3 What is the condition for destructive interference? 2 1 2 ( )2 = + t m n l t n 1 n 2 n 3 • There is a phase shift of 180 o at both interfaces. • The phase difference due to reflection is zero • The path difference must be a halfintegral number of wavelengths.5 Question An antireflective coating of MgF 2 (n=1.38) is used on a glass surface to reduce reflections. Find the minimum thickness of the coating that can be used for green light (l=550 nm). For destructive interference 1 2 2 = t n l 4 = t n l = = 550 100 4(1.38) nm nm Quarter wavelength (in coating) thickness 2 1 2 ( )2 = + t m n l minimum at m=0 Solve for t Optical compact disc A CD stores information in a series of pits and bumps in the plastic. The information is read by a reflected laser beam. The intensity of the beam is changed by destructive interference of the reflected light 4 = t n l destructive