Lecture 41: WED 29 APRPHYS 2102-2 FINALPowerPoint PresentationAngles of the Secondary MaximaExample: Diffraction of a laser through a slitWidth of a Single-Slit Diffraction PatternSlide 7ExerciseCombined Diffraction and InterferenceDiffraction GratingsCircular AperturesThe Rayleigh CriterionSlide 13Example SolutionSlide 15Slide 16Review of Interference from Thin FilmsReflection Phase ShiftsEquations for Thin-Film InterferenceFilm Thickness Much Less Than lColor Shifting by Morpho Butterflies and Paper CurrenciesProblem Solving Tactic 1: Thin-Film EquationsLecture 41: WED 29 APR Lecture 41: WED 29 APR Physics 2102Jonathan DowlingCh. 36: DiffractionCh. 36: DiffractionQuickTime™ and a decompressorare needed to see this picture.PHYS 2102-2 FINAL•5:30-7:30PM FRI 08 MAY•COATES 143•1/2 ON NEW MATERIAL•1/2 ON OLD MATERIAL•Old Formula Sheet:http://www.phys.lsu.edu/classes/spring2007/phys2102/formulasheet.pdfX-band: =10cmK-band: =2cmKa-band:=1cmLaser:=17m7Radar: 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⎛ ⎞=⎜ ⎟⎝ ⎠Example: Diffraction of a laser through a slit Light from a helium-neon laser ( = 633 nm) passes through a narrow slit and is seen on a screen 2.0 m behind the slit. The first minimum of the diffraction pattern is observed to be located 1.2 cm from the central maximum. How wide is the slit?11(0.012 m)0.0060 rad(2.00 m)yLθ = = =7431 1(6.33 10 m)1.06 10 m 0.106 mmsin (6.00 10 rad)aλ λθ θ−−−×= ≅ = = × =×1.2 cmWidth of a Single-SlitDiffraction Pattern; 1, 2,3, (positions of dark fringes)pp Ly paλ= = L2(width of diffraction peak from min to min)Lwaλ=w-y1y1y2y30X-band: =10cmYou are doing 137 mph on I-10 and you pass a little old lady doing 55mph when a cop, Located 1km away fires his radar gun, which has a 10 cm opening. Can he tell you from the L.O.L. if the gun Is X-band? What about Laser?QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.1m1m10 m1000mw =2λLa=2×0.1m×1000m0.1m=2000mNo! Beam is much wider than car separation — too wide to tell who is who.No! Beam is much wider than car separation — too wide to tell who is who.w =2λLa=2×0.000001m×1000m0.1m=0.02mLaser-band: =1mYes! Beam width is much less than car separation.Yes! Beam width is much less than car separation.Exercise Two single slit diffraction patterns are shown. The distance from the slit to the screen is the same in both cases. Which of the following could be true?(a) The slit width a is the same for both; .(b) The slit width a is the same for both; .(c) The wavelength is the same for both; width a1<a2.(d) The slit width and wavelength is the same for both; p1<p2.(e) The slit width and wavelength is the same for both; p1>p2. (degrees) (degrees) (degrees)Combined Diffraction and Interference So far, we have treated diffraction and interference independently. However, in a two-slit system both phenomena should be present together.( )( )222slit 1slitsin4 cos ;sin ;sin .I Ia ayLd dyLαδαπ πα θλ λπ πδ θλ λ⎛ ⎞=⎜ ⎟⎝ ⎠= == =daa Notice that when d/a is an integer, diffraction minima will fall on top of “missing” interference maxima.Interference OnlyDiffraction OnlyBothA device with N slits (rulings) can be used to manipulate light, such as separate different wavelengths of light that are contained in a single beam. How does a diffraction grating affect monochromatic light?Diffraction GratingsFig. 36-17Fig. 36-18sin for 0,1, 2 (maxima-lines)d m mθ λ= = K(36-11 )Circular Apertures When light passes through a circular aperture instead of a vertical slit, the diffraction pattern is modified by the 2D geometry. The minima occur at about 1.22/D instead of /a. Otherwise the behavior is the same, including the spread of the diffraction pattern with decreasing aperture.Single slit of aperture aHole of diameter DThe Rayleigh Criterion The Rayleigh Resolution Criterion says that the minimum separation to separate two objects is to have the diffraction peak of one at the diffraction minimum of the other, i.e., 1 22 D.Example: The Hubble Space Telescope has a mirror diameter of 4 m, leading to excellent resolution of close-lying objects. For light with wavelength of 500 nm, the angular resolution of the Hubble is = 1.53 x 10-7 radians.ExampleA spy satellite in a 200km low-Earth orbit is imaging the Earth in the visible wavelength of 500nm. How big a diameter telescope does it need to read a newspaper over your shoulder from Outer Space?1 22D (The smaller the wavelength or the bigger the telescope opening — the better the angular resolution.)Letters on a newspaper are about x= 10mm apart.Orbit altitude R7=7200km & D is telescope diameter.Formula:x= R7=R(1.22/D)D7=7R(1.22/x)=7(200x103m)(1.22x500x10–9m)/(10X10–3m)=112.2mExample SolutionRxLos Angeles from Space!Corona Declassified Spy Photo:Circa 1960’sSome Professor’sHouse: Google Maps2009Review of Interference from Thin FilmsFig. 35-1512?φ =0θ ≈QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.Soap BubbleOil SlickReflection Phase Shiftsn1n2n1 > n2n1n2n1 < n2Reflection Reflection Phase ShiftOff lower index ---> 0Off higher index ---> 2 10(35-15)High-To-Low: Phase Shift — NO!Low-To-High: Phase Shift — Pi!High-To-Low: Phase Shift —
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