Physics 2102 Jonathan Dowling Lecture 40 MON 27 APR Ch 36 Diffraction QuickTime and a decompressor are needed to see this picture 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 10cm Ka band 1cm Kband 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 a 0 2 mm 2 sin I I max p p 1 2 0 02 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
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