Lecture 40: MON 27 APRThings You Should Learn from This LectureSingle Slit DiffractionAnalyzing Single Slit DiffractionConditions for Diffraction MinimaPairing and InterferenceCalculating the Diffraction PatternCalculating the Diffraction Pattern (2)Diffraction PatternsPowerPoint PresentationAngles of the Secondary MaximaLecture 40: MON 27 APR Lecture 40: MON 27 APR Physics 2102Jonathan DowlingCh. 36: DiffractionCh. 36: DiffractionQuickTime™ and a decompressorare needed to see this picture.Things You Should Learn from This Lecture1. When light passes through a small slit, is spreads out and produces a diffraction pattern, showing a principal peak with subsidiary maxima and minima of decreasing intensity. The primary diffraction maximum is twice as wide as the secondary maxima.2. We can use Huygens’ Principle to find the positions of the diffraction minima by subdividing the aperture, giving min= ±p /a, p = 1, 2, 3, ... .3. Calculating the complete diffraction pattern takes more algebra, and gives I=I0[sin()/]2, where = a sin()/.4. To predict the interference pattern of a multi-slit system, we must combine interference and diffraction effects.Single Slit DiffractionWhen light goes through a narrow slit, it spreads out to form a diffraction pattern.Analyzing Single Slit Diffraction For an open slit of width a, subdivide the opening into segments and imagine a Hyugen wavelet originating from the center of each segment. The wavelets going forward (=0) all travel the same distance to the screen and interfere constructively to produce the central maximum. Now consider the wavelets going at an angle such that a sin a . The wavelet pair (1, 2) has a path length difference r12 = , and therefore will cancel. The same is true of wavelet pairs (3,4), (5,6), etc. Moreover, if the aperture is divided into p sub-parts, this procedure can be applied to each sub-part. This procedure locates all of the dark fringes.thsin ; 1, 2, 3, (angle of the p dark fringe)p pp paλθ θ= ≅ = LConditions for Diffraction Minimathsin ; 1, 2, 3,(angle of the p dark fringe)p pp paλθ θ= ≅ = LPairing and Interference Can the same technique be used to find the maxima, by choosing pairs of wavelets with path lengths that differ by ? No. Pair-wise destructive interference works, but pair-wise constructive interference does not necessarily lead to maximum constructive interference. Below is an example demonstrating this.Calculating theDiffraction Pattern We can represent the light through the aperture as a chain of phasors that “bends” and “curls” as the phase between adjacent phasors increases. is the angle between the first and the last phasor.Calculating theDiffraction Pattern (2)( )2 sin / 2E rθβ=max max/ ; /E r r Eβ β= =( )maxmaxsinsin / 2/ 2EE Eθαββ α= =sin2aβ πα θλ≡ =2maxsinI Iθαα⎛ ⎞=⎜ ⎟⎝ ⎠: or sinMinima m a mα π θ λ=± =±2I CE=Diffraction Patternssinaπα θλ=2maxsinI Iθαα⎛ ⎞=⎜ ⎟⎝ ⎠-0.03-0.02-0.01 0 0.01 0.02 0.030.20.40.60.81 = 633 nma = 0.25 mm0.5 mm1 mm2 mm (radians) The wider the slit opening a, or the smaller the wavelength , the narrower the diffraction pattern.BlowupX-band: =10cmK-band: =2cmKa-band:=1cmLaser:=1BmBRadar: The Smaller The Wavelength the Better The Targeting Resolution0.005 0.01 0.015 0.02 0.025 0.030.010.020.030.040.05Angles of the Secondary Maxima The diffraction minima are precisely at the angles wheresin = p /a and = p(so that sin =0). However, the diffraction maxima are not quite at the angles where sin = (p+½) /aand = (p+½)(so that |sin |=1).p(p+½) /aMax1 0.004750.004532 0.007910.007783 0.011080.010994 0.014240.014175 0.017410.0173512345= 633 nma = 0.2 mm To find the maxima, one must look near sin = (p+½) /a, for places where the slope of the diffraction pattern goes to zero, i.e., whered[(sin )2]/d= 0. This is a transcendental equation that must be solved numerically. The table gives the Max solutions. Note that Max (p+½)/a. (radians)2maxsinI Iθαα⎛ ⎞=⎜ ⎟⎝
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