Chapter 11. Fraunhofer DiffractionChapter 11. Fraunhofer ...

1 Chapter 11. Fraunhofer DiffractionChapter 11. Fraunhofer DiffractionLast lecture Numerical aperture of optical fiber Allowed modes in fibers Attenuation Modal distortion, Material dispersion, Waveguide dispersionThis lecture Diffraction from a single slit Diffraction from apertures : rectangular, circular Resolution : diffraction limit Diffraction from multiple-slitsDiffraction regimesDiffraction regimesFraunhofer DiffractionFraunhofer DiffractionFraunhofer diffraction Specific sort of diffraction far-field diffraction plane wavefront Simpler mathsFresnel Diffraction This is most general form of diffraction No restrictions on optical layout near-field diffraction curved wavefront Analysis difficultFresnel DiffractionFresnel DiffractionScreenObstruction11-1. Fraunhofer Diffraction from a Single Slit11-1. Fraunhofer Diffraction from a Single Slit Consider the geometry shown below. Assume that the slit is very long in the direction perpendicular to the page so that we can neglect diffraction effects in the perpendicular direction.

2 R0 ()() contribution to the electric field amplitudeat point P due to the wavelet emanating fromthe element ds in the slit is given bydEdEi krtrLet rr for the source element ds at sThen for any elementdEdEr = == = + (){}0expikrt + why?r0 0,.,, ..sinLLWe can neglect the path differencein the amplitude term but not in the phase termWe letwhere E is the electric field amplitude assumed uniform over the width of the sThe path differenceslitSubstituting we obtaidndEEs = =(){}() ( )()()/200/200/200/2expsinexpexpsinexpsin expsinbLLPPbbLPbEdsEdEi k rstEi krti k sdsrriksEIntegrating we obtainEi krtrik =+ = = ()() ( )()() ( )()000000expexpexpsin1sin2expexpexp2exp2 LPLPLE valuating with the integral limits we obtainiiEEikrtrikwherekbRearranging we obtainEbEikrtiiriEbikrtri = = = ()()00222*000220sin2sinexp11 sinsin22 LLPPE biikrtrThe irradiance at point P is given byEbI=cE EcIr = == 2102sin( )

3 , =sinII ckb =00sinsinc10,lim sinclim1sin0,,1,2,1sin2 Thefunction isforThe zeroes of irradiance occur whenor wkhenmbm === == ==K2102sin( ), =sinII ckb =The angular width of the central maximum :()112sinsinmmb =+= =,,2/,122sinIn terms of the length y on the observation screenand in terms of wavelengthkwe can writeybybffZeroes in the irradiance pattern will occur whenbymfThe maximum in the irradiance pattern is af =ytmfyb === = = maxima are found fromdd = == = = 0yfFraunhofer Diffraction pattern from a Single Slit2102sin(sin )II c kb =2 LWLb = = 11-2. Beam spreading due to diffraction11-2. Beam spreading due to diffraction16-3. Rectangular Apertures16-3. Rectangular Apertures()20sincsinWhen tof the rectangular aperture are comparable,a diffraction pattern is observed in both the x - and y - dimensions, governed in each dimension by the formula we have already developed : he length a and width bII =()21sin2 cwherekaZeroes in the irradiance pattern are observed whenmfmfyorxba ===xySquare AperturesSquare AperturesCircular AperturesCircular AperturesxdsdA= =AreaiskApdAerEE sin02222 +=xsR222sRx =dssRerEERRiskAp22sin02 = sin ,/kRRsv=={} = = )(21210221102 JrREdvverREEAviAp(the first order Bessel function of the first kind) Fraunhofer Diffraction from Circular Apertures: Bessel FunctionsFraunhofer Diffraction from Circular Apertures.

4 Bessel kDkR(first zero)()()212()0 JII = = )(2102 JrREEAp0)at (or, 0 when 21)(1= minimum in the Airy pattern is atDkDkD == Fraunhofer Diffraction from Circular Apertures: The Airy PatternFraunhofer Diffraction from Circular Apertures: The Airy Pattern()I D =: Airy pattern: Far-field angular radiusAiry pattern and Airy discAiry pattern and Airy disc()()212()0 JII = Airy discComparison : Slit and Circular AperturesComparison : Slit and Circular AperturesCircular aperture(Airy function)Single slit(sinc function)Sin 0 /D2 /D3 /D /D 2 /D 3 /DIntensity16-4. Resolution16-4. Resolution Ability to discern fine details of object Lord Rayleigh in 1896 resolution is a function of the Airy disc. Two light sources must be separated by at least the diameter of first dark band. Called Rayleigh CriterionImage blurringdue todiffractionRayleigh Criterion : Two light sources must be separated by at least the diameter of first dark limitRayleigh LimitRayleigh LimitResolution limit of a lens:(f = focal length) = = = minx The resolution of a microscope isroughly equal to the Fraunhofer Diffraction from double Slits11-5.

5 Fraunhofer Diffraction from double SlitsNow for the double slit we can imagine that we placean obstruction in the middle of the single slit. Then all that we have to do to calculate the field from the double slit is to change the limits of() ( )()()()( )()()()()()()/20/20/20/20/200/2expexpsin expexpsinexpsinexpexpsinabLPababLababLPa b integration. EEikrtiksdsrEikrtiksdsrIntegrating we obtainiksiEEikrtrik + ++ = + = + ()()()()()()()()()()/2/20000sinsinexpsin sinexpexpsin22sinsinexpexp22expexpex2aba bLLPksikikrtik a bik a bErikik a bik a bbikrtEEiri + + = + + = () ( ) ( ) () ( ){}pexpexpexpexpsinsinii iiiwherek aandk b + ==() ( )() ( )()()()()0022*20020expexp2 cosexpexp2 sinexp2cos2 sin2114sin4cos224 LPLPPBut we know thatiiiiiSubstituting we obtainbikrtEEiriThe irradiance at point P is given byEbI=cE Ecr + = = = = 22200020sin14cos, 2 LEbI= IwhereIcr = Fraunhofer Diffraction from a double SlitFraunhofer Diffraction from a double Slit2202sin4cosThe irradiance at point P from a doubleslit is given by the product of thediffraction pattern from single slit and the interference pattern from a double slitII = Fraunhofer Diffraction from a double SlitFraunhofer Diffraction from a double SlitSingle SlitDouble Slit11-6.

6 Fraunhofer Diffraction from Many Slits (Grating)11-6. Fraunhofer Diffraction from Many Slits (Grating)Now for the multiple slits we just need to againchange the limits of integration. For N even slits with width b evenly spaced a distance a apart, wecan place the origin of the coordinate system at t()( )()(){/221/2021/210expexpsinjNjabLPjabjh e center obstruction and label the slits with the index j (Note that the diagram does not exactly correspond with this). EEikrtiksdsr = + = = ()()()}()()()()()()()21/221/221/2/201021 /221/21/2expsinexpsinexpsinexpsinsinjabj abjabjNLPjjabjabjabiksdsIntegrating we obtainiksEEikrtrikiksik + + == + + = + ()()()()()2/2010exp21sin21sinexpexpsin22 21sin21sinexpexp22jNLjikrtikja bikja bErikikja bikja b == + = + ()() ()()()()() ( ){}()/201000sinsinexpexp21expexpexp21exp exp2expjNLPjLPSubstuting using k aandk band rearranging we obtainbikrtEEijiiijiiriWe can rewrite this asbikrtEEr ==== = + = () () (){}()(){}()() ( ) ( )( )

7 {}/21/2010/20102sinexp 2 1exp2 12sinexpRe exp21sinexpRe expexp 3exp 5exp1jNjjNLjjNLjiijijiEbikrtijrEbikrtiii iNrThe last te ====== + = = ++++ L() ( ) ( )( ){}/212*000sinRe expexp 3exp sin22jNjLPPPrm is a geometric series that converges toNii iiNThe details of the last step are outlined in the book The irradiance at poP is given byEbIcEEcr ==++++ = == L22sinN 220sinsinsinPNII = Fraunhofer Diffraction from Multiple SlitsFraunhofer Diffraction from Multiple Slits22222*000011sinsinsinsin22sinsinsin , .,''sinlimLPPPmThe irradiance at point P is given byEbNNIcEEcIrNWhenmthe termis a maximum For this condition from L Hospital s rule === =()()sinsincoslimlimsinsincos1sinsin0,1, 2,2.

8 MmdNNNNdNddThe principal maxima the irradiance pattern occur forppkaammNNFor large N the principal maxima are bright and well separated This ana === ==== == K,sinlysis givesus the grating equationam =2sin 2sinsin NN2 ma=sinm=1m=2m=0 Diffraction grating equationDiffraction grating equation ma=sinm=1m=2m=0m=1 Fraunhofer Diffraction from Multiple SlitsFraunhofer Diffraction from Multiple SlitsN = 2N = 3N = 4N = 5