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6.2 Two slit interferenceInterference EffectsCoherenceSlide 4InterferenceSlide 6Slide 7path differenceSlide 9Interference pattern due to two spherical wavesYoung’s two slit experimentPath difference from the two slitsInterference patternQuestionClicker Question 1Clicker Question 2Thin film interferencePhase shift due to reflectionSlide 19Thin film InterferenceSlide 21Soap film interference patternSoap filmSlide 24Anti-reflective CoatingSlide 26Slide 27Optical compact disc6.2 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 incoherentphaseshiftx2Df = plCoherenceLight fromtwo separatelight bulbs isIncoherentLight from a singlelight bulb passingthrough two slits iscoherentInterference=0SumConstructive InterferenceInterference=2lDestructive InterferenceInterference=Constructive Interferencepath differencepath difference = =r2 – r1In phaseCondition for constructive interferencemd l=Condition for destructive interference1( )2md l= +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 screen=0==-m=0m=1m=-1Young’s two slit experimentThomas YoungDoes light show wave properties?Path difference from the two slitspath differencedsind= qIn the limit L>>d, the rays are nearly parallelInterference patternBright constructive interferencebrightdsin mq = lDarkdestructive interferencedarkdsin (m 1/ 2)q = + lm = 0, + 1, + 2, ............dm=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= 12 37yd (2.6x10 )(0.1x10 )5.2x10 520nmmL (1)(5)- --l = = = =dsin=m sin ~ yLfor smallanglesyd mL= lsolve for LydClicker Question 1In a two slit interference experiment, how does the separation between peaks in the interference pattern change if the distance between slits is increased?A. IncreaseB. DecreaseC. Stays the sameD. IndeterminateClicker Question 2In a two slit interference experiment how does the distance between the peaks on the screen change if the wavelength of the light is increased?A. increasesB. decreases C. stays the sameD. indeterminateThin 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 constructiveinterference=2t=film1 1(m ) (m )2 2 nl+ l = +The wavelength in the film is longer than in air.Thin film InterferenceFor film in air the phase difference due toreflection is 180oCondition for destructiveinterference=2t=filmm mnll =The wavelength in the film is longer than in air.Soap film interference patternSoap filmQuestionA vertical soap film displays a series of colored band due to reflected light. Find the thickness of the film at the position of the 5th green band (=550 nm, n =1.33)Constructive Interference The 5th band has m=4 (the first is m=0)12t (m )2 nl= +1t (m )2 2nl= +971 550x10(4 ) 9.3x10 m 930nm2 2(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= +t mnltn1n2n3phase shift of 180o atboth interfaces.Total phase difference due to reflection is zeroQuestionAn 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 (=550 nm).For destructive interference 122=tnl4=tnl5501004(1.38)nm= =Quarter wavelength (in coating) thickness 212 ( )2= +t mnlminimumat 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=tnldestructive


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UCSD PHYS 1C - Two Slit Interference

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