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LSU PHYS 2102 - Diffraction

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Lecture 27: THU 29 APR 2010Lecture 27: THU 29 APR 2010Physics 2102Jonathan DowlingCh. 36: DiffractionCh. 36: DiffractionX-band: λ=10cmK-band:λ=2cmKa-band:λ=1cmLaser:λ=1!µm!Radar: 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 diffractionminima are precisely atthe angles wheresin θ = p λ/a and α = pπ(so that sin α=0). However, thediffraction maxima arenot quite at the angleswhere sin θ = (p+½) λ/aand α = (p+½)π(so that |sin α|=1).0.017350.0174150.014170.0142440.010990.0110830.007780.0079120.004530.004751θMax(p+½) λ/ap12345λ = 633 nma = 0.2 mm To find the maxima, one must look near sin θ = (p+½) λ/a, for placeswhere the slope of the diffraction pattern goes to zero, i.e., whered[(sin α/α)2]/dθ = 0. This is a transcendental equation that must be solvednumerically. The table gives the θMax solutions. Note that θMax < (p+½) λ/a.θ (radians)2maxsinI I!""# $=% &' (Example: Diffraction of a laserthrough a slit Light from a helium-neon laser (λ = 633 nm) passes through a narrowslit and is seen on a screen 2.0 m behind the slit. The first minimum ofthe diffraction pattern is observed to be located 1.2 cm from the centralmaximum. How wide is the slit?11(0.012 m)0.0060 rad(2.00 m)yL!= = =7431 1(6.33 10 m)1.06 10 m 0.106 mmsin (6.00 10 rad)a! !" "###$= % = = $ =$1.2 cmWidth of a Single-SlitDiffraction Pattern; 1, 2,3, (positions of dark fringes)pp Ly pa!= = !2(width of diffraction peak from min to min)Lwa!=w-y1y1y2y30X-band: λ=10cmYou are doing 137 mph on I-10 and you pass a little old lady doing 55mphwhen 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?1m1m10 m1000mw =2!La=2 " 0.1m " 1000m0.1m= 2000mNo! Beam is much wider than carseparation —too wide to tell who is who.w =2!La=2 " 0.000001m " 1000m0.1m= 0.02mLaser-band: λ=1µmYes! Beam width is much less thancar separation.Exercise Two single slit diffraction patterns are shown. The distancefrom the slit to the screen is the same in both cases. Which of the following could be true?λ1λ2(a) The slit width a is the same for both; λ1>λ2.(b) The slit width a is the same for both; λ1<λ2.(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.θ (degrees)θ (degrees)θ (degrees)Combined Diffraction andInterference So far, we have treateddiffraction and interferenceindependently. However, ina two-slit system bothphenomena should bepresent together.( )( )222slit 1slitsin4 cos ;sin ;sin .I Ia ayLd dyL!"!# #! $% %# #" $% %& '=( )* += == =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 separatedifferent wavelengths of light that are contained in a single beam. How does adiffraction grating affect monochromatic light?Diffraction GratingsFig. 36-17Fig. 36-18sin for 0,1, 2 (maxima-lines)d m m! "= = …(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 occurat about 1.22λ/D instead of λ/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 ResolutionCriterion says that the minimumseparation to separate two objectsis to have the diffraction peak ofone at the diffraction minimum ofthe other, i.e., Δθ = 1.22 λ/D.Example: The Hubble Space Telescopehas a mirror diameter of 4 m, leading toexcellent resolution of close-lyingobjects. For light with wavelength of500 nm, the angular resolution of theHubble is Δθ = 1.53 x 10-7 radians.ExampleA spy satellite in a 200km low-Earth orbit is imaging theEarth in the visible wavelength of 500nm.How big a diameter telescope does it need to read anewspaper over your shoulder from Outer Space?Δθ = 1.22 λ/D (The smaller the wavelength or the bigger thetelescope opening — the better the angular resolution.)Letters on a newspaper are about Δx!= 10mm apart.Orbit altitude R!=!200km & D is telescope diameter.Formula:Δx!= RΔθ!=!R(1.22λ/D)D!=!R(1.22λ/Δx)=!(200x103m)(1.22x500x10–9m)/(10X10–3m)=!12.2mExample SolutionRΔxΔθLos Angeles from Space!Corona Declassified Spy Photo:Circa 1960’sSomeProfessor’sHouse:


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