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UVA PHYS 632 - Lecture 15 Interference Chp. 35

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Lecture 15 Interference Chp. 35Huygen’s Principle, Wavefronts and CoherencePowerPoint PresentationSlide 4Once light is in phase there are three ways to get light out of phaseSlide 6Concept of path length difference, phase and index of refractionWave reflects 180 degrees out of phase when n1 < n2Thin film Interference Phenomenon: ReflectionThin film Interference Phenomenon TransmissionSlide 12Slide 13Slide 14Slide 15Solve for xSlide 17Slide 18Slide 19Lecture 15 Interference Chp. 35Opening Demo •Topics–Interference is due to the wave nature of light–Huygen’s principle, Coherence–Change in wavelength and phase change in a medium –Interference from thin films–Examples –Young’s Interference Experiment and demo–Intensity in double slit experiment–Warm-up problem•DemosHuygen’s Principle, Wavefronts and Coherence€ E = Emsin(2πλ− 2πf t)k€ E = Emsin(kx −ωt)Examples of coherence are: Laser light Small spot on tungsten filament WavefrontMost light is incoherent: Two separate light bulbs Two headlight beams on a car Sun is basically incoherentInterference is the combination of two or more waves to form a composite wave, based on the principle of superpositionIn order to form an interference pattern, the incident light must satisfy two conditions: (i) The light sources must be coherent. This means that the plane waves from the sources must maintain a constant phase relation. For example, if two waves are completely out of phase with φ = π , this phase difference must not change with time. (ii) The light must be monochromatic. This means that the light consists of just one wavelength λ = 2π/k . Light emitted from an incandescent lightbulb is incoherent because the light consists of waves of different wavelengths and they do not maintain a constant phase relationship. Thus, no interference pattern is observed.Once light is in phase there are three ways to get light out of phase1. Rays go through different material with different index of refraction2. Reflection from a medium with greater index of refraction3. The selected rays travel different distances.Now lets look at examplesIn Phase Out of Phase by 180 degrees or  radians or/2In betweenConcept of path length difference, phase and index of refractionPath length difference = Phase difference = c = f Vacuum Vacuumvn= f nf /n n=c/vRays are in phase if where m     Rays are out of phase if where m=1,2,31 is the same as radian (rad),  is the same as  rad, etc.λλλλ222111LnLNLnLNnn====( )λ1212nnLNN−=−( )λ12nnL −( )λλmnnL=−12( )λλ⎟⎠⎞⎜⎝⎛+=−2112mnnLWave reflects 180 degrees out of phase when n1 < n2n2n1airwaterEair 1.0air 1.00soap 1.30Leye21Reflection180 deg phase changeThin film Interference Phenomenon: Reflection€ E = Emsin(kx −ωt)€ E1= Emsin(kx −ωt + π)E1= −Emsin(kx −ωt)€ E2= Emsin(k(x + 2L) −ωt)€ Suppose the soap thickness L is such that2L =12 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟λn € E2= Emsin(kx + k2L −ωt) =Emsin(kx +2πλλ2−ωt) =E2= Emsin(kx −ωt + π)€ 2L = m +12 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟λn2 where m = 0, 1, 2, ...Constructive InterferenceNow consider the path length differences€ λn=λnFirst consider phase change upon reflectionShow wave demosoap 1.30Thin film Interference PhenomenonTransmissionair 1.0air 1.00LeyeTransmission21€ 2L = m( )λn λ =2nLmm = 1,2,3,4….Constructive interferenceNo phase changes upon reflectionn =1.3039. A disabled tanker leaks kerosene (n = 1.20) into the Persian Gulf, creating a large slick on top of the water (n = 1.30).(a) If you are looking straight down from an airplane while the Sun is overhead at a region of the slick where its thickness is L=460 nm, for which wavelength(s) of visible light is the reflection brightest because of constructive interference? Path difference between ray 1and ray 2 = 2L. Phase changes cancel outFor constructive interference path differencemust = integral number of wavelengthsair 1.0Water 1.30Kerosene 1.20L21180 deg phase change€ λ =2n2Lm=2 1.20( )460nm( )m=1104, 552, 368 nmfor m =1, m = 2, and m = 3 respectivelyWe note that only the 552 nm wavelength falls within the visible light range.€ 2L = m( )λn2 λ =2n2Lm(b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest? (Hint: use figure (a) with appropriate indices of refraction.) Scuba diverFor transmission, ray 2 undergoes 180 deg phase shift upon reflection at theKerosene-water interface. Therefore, for constructive interference 2L= integral number of wavelengths in n2 plus half a wavelength.air 1.0Water 1.30Kerosene 1.20L21€ 2L = m +12 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟λn2 where m = 0, 1, 2, ...Solve for λ€ 2L = m +12 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟λn2 where m = 0, 1, 2, ...λn=λn2We note that only the 441.6 nm wavelength (blue) is in the visible range, € 2L = m +12 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟λn2 Solving for λλ =4n2L2m +1€ λ =4n2L2m +1=4(1.2)(460)1= 2208 nm m = 0€ λ =4n2L2m +1=4(1.2)(460)3= 736 nm m =1€ λ =4n2L2m +1=4(1.2)(460)5= 441.6 nm m = 2Visible spectrum is 430 nm - 690 nmThe wave from S1 travels a distance x and the wave from S2 travels a distance .22xd +27. S1 and S2 in Fig. 36-29 are point sources of electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d = 4.00 m, and they emit at the same power. (a) If a detector is moved to the right along the x-axis from source S1, at what distances from S1 are the first three interference maxima detected?xdetector.22xd +The path difference is € d2+ x2− x€ d2+ x2x€ path difference = d2+ x2− x = mλ m =1,2,3..The solution for x of this equation isFor constructive interference we haveSolve for x€ x =d2− m2λ22mλ for m =1, 2, 3,.. € d2+ x2− x = mλd2+ x2= mλ + x Now square both sides€ d2+ x2= ( mλ + x)2 d2+ x2= m2λ2+ 2mλx + x2 Now cancel x2€ d2= m2λ2+ 2mλx solve for x.2222λλmmdx−=€ For m = 3 x =16 − 3( )22( )3( )=1.17m.m=3What about m = 4 ? This corresponds to x=0. Path difference =4 meters.€ x =16 − m22mm=2€ For m = 2 x =16 − 2( )22( )2( )= 3.0m.m=1€ For m =1 x =16 − 1( )22( )1( )= 7.5m.Although the amplitudes are the same at the sources, the waves travel different distances to get to the points of minimum intensity and each


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