PHY 2049: Chapter 35 1Paul AveryUniversity of Floridahttp://www.phys.ufl.edu/~avery/[email protected] Interference and DiffractionPart 1: Introduction, Double SltPHY 2049Physics 2 with CalculusPHY 2049: Chapter 35 2Need to Understand Light as Wave!Î(You already have read this material)ÎIndex of refraction Speed of EM wave in medium: Wavelength of light: ÎPropagation of light: Huygens principle (36-2) Explains reflection and refraction Explains interference (from superposition) Explains diffraction (spreading of light around barrier)/nccn=/nnλλ=PHY 2049: Chapter 35 3Interference as a Wave PhenomenonÎInterference of light waves Caused by superposition of waves Intensity can increase or decrease! Contrast with particle model of lightÎEffects and applications Double slit Single slit Diffraction gratings Anti-reflective coatings on lenses Highly reflective coatings for mirrors Iridescent coatings on insects Colors on thin bubbles Interferometry with multiple telescopesPHY 2049: Chapter 35 4Interference from Wave SuperpositionWave 1Wave 2SumBasic rule: Add displacement at every pointPHY 2049: Chapter 35 5Constructive InterferenceÎSame wavelength, phase difference = 0°ÎAmplitude larger: Higher intensitySum()sin( ) 0.5sin( ) 1.5sin( )Exkx kx kx=+ =PHY 2049: Chapter 35 6Destructive InterferenceÎSame wavelength, phase difference = 180° (1/2 λ)ÎAmplitude smaller: Lower intensitySum()sin( ) 0.5sin( ) 0.5sin( )Exkx kx kxπ=+ +=PHY 2049: Chapter 35 7ExamplesÎTwo waves, same λ, with amplitudes 2A and A Initial intensities 4I and I, respectively (I ∝ A2)ÎNo interference Combined intensity: Inew= 4I + I = 5IÎMaximum constructive interference (φ = 0) New amplitude: Anew= 2A + A = 3A New intensity: Inew= 9IÎMaximum destructive interference (φ = π) New amplitude: Anew= 2A – A = A New intensity: Inew= IPHY 2049: Chapter 35 8General Treatment of InterferenceÎMost interference is partial Amplitudes for 2 waves are generally different Phase difference : 0 < φ < 180°ÎAdditional considerations Wavelengths can be different Multiple waves may interfere (e.g., diffraction grating) But easy to accommodate: just sum over all waves()12, cos( ) cos( )E x t E kx t E kx tωωφ=−+−+(),cos( )iiiiiExt E kx tωφ=−+∑PHY 2049: Chapter 35 9Interference and Path Length Two sources, spaced 3 wavelengths apart, emit waves with the same wavelength and phase. In how many places on the circle will the net intensity be a relative maximum?Answer = 12Can you see why?Key idea: Path difference leadsto phase differenceHint: Start at far right and movecounterclockwise towards top,noting path length changes.PHY 2049: Chapter 35 10Interference and Path Length ÎTwo sources, separated by 4λ, emit waves at same wavelength and phase. Find relative minima on +x axis. Solution: path differencemust be a half-multiple of λx()()22124Lx xnλλΔ= + −= +4λ()2121621nxnλ−+=+ΔL= 7λ/2x= 0.54λn= 3ΔL= 5λ/2x= 1.95λn= 2ΔL= 3λ/2x= 4.58λn= 1ΔL= λ/2x= 15.8λn= 04 valuesPHY 2049: Chapter 35 11Double Slit InterferenceÎIncident light Light waves strike 2 narrow slits close together Light goes through both slits, diffracts in all directionsÎInterference At certain angles, waves constructively interfere ⇒ brighter At other angles, waves destructively interfere ⇒ darker Light wavesScreenDiffracted lightInterferenceSlitsPHY 2049: Chapter 35 12Basic Requirements for Two Slit SetupÎLight beam strikes normal to slitsÎLight beam illuminates both slits equallyÎLight beam is in phase at both slits: coherent Young used small slit in front of 2 slits to get coherence Modern versions use laser for coherence (much brighter)PHY 2049: Chapter 35 13Two Slit Analysis()12sin Maximumsin Minimumdmdmθλθλ==+θdPath difference = d sinθPHY 2049: Chapter 35 14Example 1: d = 5λ--±1.1±90±1.0±5±64.2±0.9±53.1±0.8±4±44.4±0.7±36.9±0.6±3±30±0.5±23.6±0.4±2±17.5±0.3±11.5±0.2±1±5.7±0.1000θminsinθminθmaxsinθmaxm()()()()1122Max sin / 0.2Min sin / 0.2md mmdmθλθλ===+=+PHY 2049: Chapter 35 15Intensity vs Angle for d = 5λPHY 2049: Chapter 35 16Example 2: d= 2.0 μm, λ = 550 nm ÎHow many bright fringes? Where are they?Îm can equal 0, ±1, ±2, ±3 ⇒ 7 maxima()sin / 0.275md mθλ==θ= 55.6°sinθ= 0.825m = ±3θ= 33.4°sinθ= 0.55m = ±2θ= 16.0°sinθ= 0.275m = ±1θ= 0sinθ= 0m = 0PHY 2049: Chapter 35 17Intensity vsθfor d= 2.0 μm, λ = 550 nmPHY 2049: Chapter 35 18Calculating Double Slit IntensityÎAssumptions Each slit acts as a source of waves Waves radiate equally in all directionsθdPath difference = d sinθPHY 2049: Chapter 35 19Double Slit Intensity (2)ÎAdd amplitudes, include phase difference Assume equal size slit widths Phase difference from path difference: 2π×# wavelengths We ignore x dependence here (analysis does not depend on it)()cos( ) cos( )EtEtEtωωφ=++2sindπθφλ=PHY 2049: Chapter 35 20Double Slit Intensity (3)ÎIntensity is time average of amplitude squared Consider single wave of amplitude E = E0cosωt Intensity from time average of E2: <…> is time average over a period ()()()22tot22 2 22 222cos( ) cos( )cos cos2coscosIKEtEtKE t KE tKE t tωωφωωφωωφ=++=++++We work this out on next page22 2 2212cosIKE t KEω==PHY 2049: Chapter 35 21Double Slit Intensity (4)2sindπθφλ=()()221cos21cos21cos cos cos2ttttωωφωωφφ=+=+=()()22tot 02102201cos 2 1cos4cossin4cosIKE IIdIφφφπθλ=+=+=⎛⎞=⎜⎟⎝⎠2sindπθφλ=PHY 2049: Chapter 35 22Double Slit Intensity (5)ÎSo the intensity isÎMaxima occur when argument inside cos() is nπÎMinima occur when argument inside cos() is (n+1/2)π()204cos sin/II
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