Physics 2102 Jonathan Dowling Lecture 26 TUE 27 APR 2010 Ch 36 Diffraction QuickTime and a decompressor are needed to see this picture EXAM III AVG 60 STD 15 A 100 85 B 84 75 C 74 45 D 44 40 F 39 0 Michelson Interferometers As we saw in the previous example interference is a spectacular way of measuring small distances like the thickness of a soap bubble since we are able to resolve distances of the order of the wavelength of the light for instance for yellow light we are talking about 0 5 of a millionth of a meter 500nm This has therefore technological applications In the Michelson interferometer light from a source at the left in the picture hits a semiplated mirror Half of it goes through to the right and half goes upwards The two halves are bounced back towards the half plated mirror interfere and the interference can be seen by the observer at the bottom The observer will see light if the two distances travelled d1 and d2 are equal and will see darkness if they differ by half a wavelength Michelson Morley Experiment Michelson won the Nobel prize in 1907 for his optical precision instruments and the spectroscopic and metrological investigations carried out with their aid The interpretation of these results is that there is no displacement of the interference bands The result of the hypothesis of a stationary ether is thus shown to be incorrect A A Michelson Am J Sci 122 120 1881 The largest Michelson interferometer in the world is in Livingston LA in LSU owned land it is operated by a project funded by the National Science Foundation run by Caltech and MIT and LSU collaborates in the project http www ligo la caltech edu Mirrors are suspended with wires and will move detecting ripples in the gravitational field due to astronomical events Gravitational Waves Interferometry an International Dream GEO600 British German Hannover Germany TAMA Japan Mitaka LIGO USA Hanford WA and Livingston LA AIGO Australia Wallingup Plain 85km north of Perth VIRGO French Italian Cascina Italy Things You Should Learn from This Lecture 1 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 Single Slit Diffraction When 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 2 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 locates all be applied to each sub part This procedure th p fringes sin p p p 1 2 3 L angle of the p dark fringe of the dark a Conditions for Diffraction Minima p sin p p p 1 2 3 L a angle of the p th dark fringe Pairing 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 the Diffraction 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 the Diffraction Pattern 2 E 2r sin 2 Emax r r Emax E Emax sin sin 2 Emax 2 a sin 2 2 I CE 2 sin I I max Minima m or a sin m Diffraction Patterns 1 633 nm 0 8 a 0 25 mm 0 6 0 5 mm 0 4 1 mm 0 2 0 03 a sin 2 mm 0 02 0 01 0 radians 2 sin I I max Blowup 0 01 0 02 0 03 The wider the slit opening a or the smaller the wavelength the narrower the diffraction pattern Radar The Smaller The Wavelength the Better The Targeting Resolution X band 10cm Ka band 1cm K band 2cm Laser 1 m Angles of the Secondary Maxima The diffraction minima are precisely at the angles where sin p a and p so that sin 0 0 05 0 04 1 0 03 633 nm p a 0 2 mm 2 sin I I max 1 2 0 02 p a Max 0 00475 0 0045 3 0 00791 0 0077 8 0 0109 9 3 0 01108 However the 0 01 0 0141 diffraction maxima 2 4 0 01424 7 are not quite at the 3 4 5 0 0173 angles where sin 0 005 0 01 0 015 0 02 0 025 5 0 03 5 0 01741 p a radians and p To find maxima one must look near sin p a for so that sinthe 1 places where the slope of the diffraction pattern goes to zero i e where d sin 2 d 0 This is a transcendental equation that must be solved numerically The table gives the Max solutions Note that Max p a
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