Physics 2102 Jonathan Dowling Lecture 29 Ch 36 Diffraction 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 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 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 p sin to p 1 2 3 L angle of the p th dark fringe can be applied each sub part This procedure locates p p a all of the dark fringes 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 ar The Smaller The Wavelength the Better The Targeting Resolution X band 100m Ka band 1m Kband Laser 1 micron 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 3 4 5 0 0173 the angles where 0 005 0 01 0 015 0 02 0 025 5 0 03 5 0 01741 sin p a radians and p To sin find the maxima one must look near sin p so that a for places where the slope of the diffraction pattern 1 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 Example Diffraction of a laser through a slit 1 2 cm 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 1 y1 0 012 m 0 0060 rad L 2 00 m 6 33 10 7 m 4 a 1 06 10 m 0 106 mm 3 sin 1 1 6 00 10 rad Width of a Single Slit Diffraction Pattern y1 0 y 1 y2 y 3 w p L yp a w 2 L a p 1 2 3 L positions of dark fringes width of diffraction peak from min to min 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 Combined Diffraction and Interferenced So far we have a treated diffraction a and interference independently Interference Only However in a two slit system both phenomena degrees should be present 2 together sin Diffraction Only I 2slit 4 I1slit cos 2 degrees a a y sin L Both d d y sin L degrees Notice that when d a is an integer diffraction minima will fall on top of missing interference maxima Circular Apertures Single slit of aperture a Hole of diameter D 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 The 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 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 Example A 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 Example Solution D The smaller the wavelength or the bigger the telescope opening the better the angular resolution Letters on a newspaper are about x 10mm apart R Orbit altitude R 200km D is telescope diameter Christine s Favorite Formula x R R 1 22 D x D R 1 22 x 200x103m 1 22x500x10 9m 10X10 3m s Angeles from Space rona Declassified Spy Photo rca 1960 s
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