16.2 Two slit interferenceCoherenceTwo-Slit InterferenceThin film InterferenceInterference EffectsInterference 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.CoherenceFor 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 incoherentphaseshiftx2∆φ= πλCoherenceLight fromtwo separatelight bulbs isIncoherentLight from a singlelight bulb passingthrough two slits iscoherentInterferenceδ=0SumConstructive InterferenceλInterferenceδ=2λDestructive Interference2Interferenceδ=λConstructive Interferencepath differencepath difference =δ =r2–r1In phaseCondition for constructive interferencemδλ=Condition for destructive interference1()2mδλ=+m = 0 + 1, + 2,….r1r2Superposition of wavesat A shows interferencedue to path differencesAOrder number mInterference pattern of water wavesConstructiveDestructiveInterference pattern due totwo spherical wavesCoherent wavesInterference patternAmplitude on screenδ=0δ=λδ=-λm=0m=1m=-1Young’s two slit experimentThomas YoungDoes light show wave properties?Path difference from the two slitspath differencedsinδ= θIn the limit L>>d, the rays are nearly parallel3Interference patternBright constructive interferencebrightdsin mθ=λDarkdestructive interferencedarkdsin (m 1/2)θ=+ λm = 0, + 1, + 2, ............dθm=0m=1m= -1maximam= -2m=2Central maximumQuestionLight from a laser is passed through two slits a distance of 0.10 mm apart and is hits a screen 5 m away. The separation between the central maximum and the first bright interference fringe is 2.6 cm. Find the wavelength lf the light.for m= 1237yd (2.6x10 )(0.1x10 )5.2x10 520nmmL (1)(5)−−−λ= = = =dsinθ=mλ sinθ ~ yLfor smallanglesydmL=λsolve for λLydθThin 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 interfacePhase shift due to reflectionn1< n2phase shift=180oReflection with inversionphase shift = 180oPhase shift due to reflectionReflection without inversionPhase shift = zeron1> n2Phase shift = zeroThin film InterferenceFor a thin film in air the phase difference due toreflection is 180oCondition for constructiveinterferenceδ=2t=film11(m ) (m )22nλ+λ= +The wavelength in the film is longer than in air.4Thin film InterferenceFor film in air the phase difference due toreflection is 180oCondition for destructiveinterferenceδ=2t=filmmmnλλ=The wavelength in the film is longer than in air.Soap film interference patternSoap film QuestionA vertical soap film displays a series of colored band due to reflected light. Find the thickness of the film at the position of the 5thgreen band (λ=550 nm, n =1.33)Constructive Interference The 5th band has m=4 (the first is m=0)12t (m )2nλ=+1t(m )22nλ=+971 550x10(4 ) 9.3x10 m 930nm22(1.33)−−=+ = =Anti-reflective CoatingAnti-reflective coatings are usedto reduce reflections at the air-glassinterface.anti-reflectivecoatingno coatingAnti-reflective CoatingAnti-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.n1=1.00 < n2< n3Condition for destructive interference.212( )2=+tmnλtn1n2n3phase shift of 180oatboth interfaces.Total phase difference due to reflection is zero5QuestionAn anti-reflective coating of MgF2 (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 (λ=550 nm).For destructive interference122=tnλ4=tnλ5501004(1.38)nm==Quarter wavelength (in coating) thickness212( )2=+tmnλminimumat m=0Solve for tOptical compact discCd store information in a seriesof pits and bumps in the plastic.The information is read by a reflectedlaser beam. The intensity of the beam is changed byconstructive or destructive interference4=tnλdestructive