Lecture 39: FRI 24 APRPowerPoint PresentationSlide 3ExampleSlide 5Slide 6Radio WavesInterference: ExampleSlide 9Slide 10Slide 11Slide 12Reflective CoatingsSlide 14Slide 15Slide 16Michelson-Morley ExperimentSlide 18Gravitational Waves Interferometry: an International DreamLecture 39: FRI 24 APRLecture 39: FRI 24 APRPhysics 2102Jonathan DowlingCh. 35: InterferenceCh. 35: InterferenceChristian Huygens 1629-1695The Lunatic Fringe:The waves arriving at the screen fromthe two slits will interfere constructivelyor destructively depending on the differentlength they have travelled.θsindL =Δinteger 0,1,2...m :veConstructi =,=Δ λmLinteger 0,1,2...m )21( :eDestructiv =,+=Δ λmLλθ md =sin :veConstructiMaximum fringe at =0, (central maxima)other maxima at⎟⎠⎞⎜⎝⎛=dmλθ arcsinSimilarly for dark fringes: d sin = (m+1/2) 1 2 0 02 cos 2 cosE E E E Eφβ⎛ ⎞= + = =⎜ ⎟2⎝ ⎠r r⎟⎠⎞⎜⎝⎛2==φ22020cos4EEIIθλπλπφ sindL⎟⎠⎞⎜⎝⎛2=Δ⎟⎠⎞⎜⎝⎛2=The ReturnOf the Phasor!Phase Difference2=360°=180°ExampleExampleRed laser light (=633nm) goes through two slits 1cm apart, and produces a diffraction pattern on a screen 55cm away. How far apart are the fringes near the center?For the spacing to be 1mm, we need d~ L /mm=0.35mmIf the fringes are near the center, we can usesin ~ , and then m=dsin~d => =m/d is the angle for eachmaximum (in radians)q= l/d =is the “angular separation”.The distance between the fringes is thenx=L=L/d=55cmx633nm/1cm=35 mExampleExampleIn a double slit experiment, we can measure the wavelength of the light if we know the distances between the slits and the angular separation of the fringes. If the separation between the slits is 0.5mm and the first order maximum of the interference pattern is at an angle of 0.059o from the center of the pattern, what is the wavelength and color of the light used?d sin=m => =0.5mm sin(0.059o)= 5.15 10-7m=515nm ~ green ExampleExampleA double slit experiment has a screen 120cm away from the slits, which are 0.25cm apart. The slits are illuminated with coherent 600nm light. At what distance above the central maximum is the average intensity on the screen 75% of the maximum? I/I0=4cos22 ; I/Imax=cos2 =0.75 => =2cos–1 (0.75)1/2=60o=/3 rad=(2d/ )sin => = sin-1(/2d) 22 rad (small!)y=L48mRadio WavesRadio WavesTwo radio towers, separated by 300 m as shown the figure, broadcast identical signals at the same wavelength.A radio in the car, which traveling due north, receives the signals. If the car is positioned at the second maximum, what is the wavelength of the signals? Where will the driver find a minimum in reception?dsin=m => = dsin/2sin = 400/(4002+10002)1/2=0.37=55.7m“Dark” fringes: dsin=(m+1/2) sin=(m+1/2)/d=2.5x55.7m/300m =0.464 =>= 28otan=y/1000m => y=1000m x tan(28o)=525mInterference: ExampleInterference: ExampleA red light beam with wavelength =0.625m travels through glass (n=1.46) a distance of 1mm. A second beam, parallel to the first one and originally in phase with it, travels the same distance through sapphire (n=1.77). •How many wavelengths are there of each beam inside the material? In glass, g=0.625m/1.46= 0.428 m and Ng=D/ g=2336.45In sapphire, s=0.625m/1.77= 0.353 m (UV!) and Ns=D/ s=2832.86•What is the phase difference in the beams when they come out?The difference in wavelengths is Ns-Ng=496.41. Each wavelength is 360o, so N=496.41 means =Nx360o=0.41x360o=148o•How thick should the glass be so that the beams are exactly out of phase at the exit (destructive interference!)N=D/ s D/ g= (D/ )(n2-n1)=0.31 (D/ )=m+1/2A thickness D=(m+0.5) 2.02 m would make the waves OUT of phase.For example, 1.008 mm makes them in phase, and 1.010 mm makes them OUT of phase.Thin Film Interference:The patterns of colors that one seesin oil slicks on water or in bubblesis produced by interference of thelight bouncing off the two sides ofthe film.To understand this we need to discuss the phase changes that occur when a wave moves from one mediumto the another where thespeed is different. Thiscan be understood witha mechanical analogy.Reflection, Refraction and Changes of Phase:Consider an UP pulse moving in a rope, that reaches a juncturewith another rope of different density. A reflected pulse is generated.The reflected pulse is also UP if the speed of propagation in the rope of the right is faster than on the left. (Low impedance.)The reflected pulse is DOWN if the speed of propagation in the right is slower than on the left. (High impedance.)The extreme case of ZERO speed on the right corresponds to a ropeanchored to a wall. (Highest impedance.)If we have a wave instead of a pulse “DOWN” means 180 degrees OUT of phase, and UP means 360° or IN PHASE.Thin FilmsFirst reflected lightray comes from firstinterface, second from second. These rays interfere with each other.How they interfere will depend on the relative indices of refraction.In the example above the first ray suffers a 180 degree phase change(1/2 a wavelength) upon reflection. The second ray does not change phase in reflection, but has to travel a longer distance to come back up. The distance is twice the thickness of the layer of oil. For constructive interference the distance 2L must therefore be ahalf-integer multiple of the wavelength, i.e. 0.5 , 1.5 ,…,(0.5+2n).22number odd2 :phaseIn nLλ×=2integer2 :phase-AntinLλ×=n1n2n3If the film is very thin, then the interference is totally dominated by the180° phase shift in the reflection. At the top the film is thinnest (due to gravity it lumps at the bottom), so one sees thefilm dark at the top.This film is illuminated with white light, therefore we see fringes ofdifferent colors corresponding to the various constructive interferencesof the individual components of the white light, which change as wego down. The thickness increases steeply as we go down, which makesthe width of the fringes become narrower and narrower.Thin Films: Soap BubblesAir: n=1AirSoap: n>1180°0°Reflective CoatingsReflective CoatingsTo make mirrors that reflects light of only a given wavelength, a coating of a specific thickness is used so that there is constructive interference of the given wavelength.
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