PHY 2049: Chapter 37 1Paul AveryUniversity of Floridahttp://www.phys.ufl.edu/~avery/[email protected] Interference and DiffractionPart 2: Single SlitPHY 2049Physics 2 with CalculusPHY 2049: Chapter 37 2Single Slit SetupLight wavesScreenDiffracted lightInterferenceSlitPHY 2049: Chapter 37 3Single Slit Analysissin Minimumamθλ=θaWaves from slit interferem = ±1, ±2, ±3, …Unlike, 2 slit, this is condition for a minimum!PHY 2049: Chapter 37 4Example 1: a = 5λ±90±1.0±5±53.1±0.8±4±36.9±0.6±3±23.6±0.4±2±11.5±0.2±1θminsinθminm()Min sin / 0.2ma mθλ==Total of 5 mimimaPHY 2049: Chapter 37 5Intensity vs Angle for a = 5λPHY 2049: Chapter 37 6Calculating Intensity For Single SlitÎFollow similar method to double slit Each sub-element of slit acts as a source of waves Waves radiate equally in all directionsθaWaves from slit interferePHY 2049: Chapter 37 7Single Slit Intensity (2)ÎAdd amplitudes by integration over slit (0 < y < a) Include phase difference vs y: (Phase difference from path difference: 2π×# wavelengths)()0cos( )2sin/ayyEt t dyyωφφπθλ∝+=∫2sin/yyφπθλ=()()0cos( 2 sin / )sin 2 / sinsinaEt t y dyta tωπ θλωπλ ωθ∝++−∝∫PHY 2049: Chapter 37 8Single Slit Intensity (3)ÎIntensity is time average of amplitude squared()()22tot22222222sin( ) sinsinsin sinsin sin2sin sinsinttIKKKttKttωφ ωθωφωθθωφ ωθ+−⎛⎞=⎜⎟⎝⎠=++−+We work this out on next page2sin/aφπθλ=PHY 2049: Chapter 37 9Single Slit Intensity (4)()()221sin21sin21sin sin cos2ttttωωφωωφ φ=+=+=()22122tot222sin1cossin sinKKIφφθθ=−=2tot max2sinIIαα=sin /aαπθλ=3 termsSumPHY 2049: Chapter 37 10Single Slit Intensity (5)ÎSo the intensity isÎMinima occur when argument inside sin() is mπ()()22tot maxsin sin / / sin /II a aπθλ π θλ=sinamθλ=PHY 2049: Chapter 37 11Height of Maxima for Single SlitÎMaxima occur approximately when α = (m+1/2)π In addition to central max, intensity I0, which occurs at α = 0ÎHeight of maxima ≈ I0/ (m+1/2)2π2m = 1 I1= 0.045I0 m = 2 I2= 0.016I0 m = 3 I3= 0.0083I0 m = 4 I4= 0.0050I0 m = 5 I5= 0.0033I0ÎSo height of maxima shrink
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