# Chapter 10 Diffraction December 3 Fraunhofer diffraction: the

Chapter 10 Diffraction December 3 Fraunhofer diffraction: the single slit 10.1 Preliminary considerations Diffraction: The deviation of light from propagation in a straight line. There is no essential physical distinction between interference and diffraction. Huygens-Fresnel Principle: Every unobstructed point of a wave front serves as a source of spherical wavelets. The amplitude of the optical field at any point beyond is the superposition of all these wavelets, taking into account their amplitudes and phases. Fraunhofer (far field) diffraction: Both the incoming and outgoing waves approach being planar. a2/R where R is the smaller of the two distances from the source to the aperture and from the aperture to the observation point. a is the size of the aperture. The diffraction pattern does not change when moving the observation plane further away. Fresnel (near field) diffraction: The light source or the plane of observation is close to the aperture. General case of diffraction. The diffraction pattern changes when the observation plane moves. P S a R1 R2 1 Mathematical criteria for Fraunhofer diffraction: The phase for the rays meeting at the observation point is a linear function of the aperture variables. S P y' Waves from a point source: Harmonic spherical wave: y' sin A A E ( r, t ) cos(kr t ), or E ( r, t ) ei ( kr t ) r r y

A is the source strength. P (x,y) D/2 Coherent line source: r dy' x exp(ikr ) E ( x, y ) L dy ' D/2 r D/2 -D/2 L is the source strength per unit length. This equation changes a diffraction problem into an integration (interference) problem. 2 y 10.2 Fraunhofer diffraction 10.2.1 The single slit The slit is along the z-axis and has a width of D. P (x,y) D/2 y' r ( y ' ) R 2 y '2 2 Ry ' sin cos2 2 R y ' sin y ' 2R r

R x -D/2 exp(ikr ) In the phase, r is approximated by R-y' sinif D2/R<<1. E ( x, y ) L dy ' D/2 Fraunhofer diffraction condition r D / 2 exp[ik ( R y ' sin )] L dy ' D/2 R In the amplitude, r is approximated by R. D sin[(kD / 2) sin ] L exp(ikR ) R ( kD / 2) sin D sin L exp(ikR ), (kD / 2) sin R The overall phase is the same as a point source 2 sin at the center of the slit. I ( ) I (0) Integrate over z gives the same function. D/2 3 sin I ( ) I (0)

2 y kD sin 2 P (x,y) D/2 y' dI 2 sin ( cos sin ) I (0) 0 d 3 r R x -D/2 minima sin 0, , 2 , 3 ... tan , 0, 1.43 , 2.46 , 3.47 ,... maxima sin I ( ) I (0)

Width of the central peak 2 2 2 kD / 2 D Widths of the side peaks 2 I/I(0)= 0.047 0.016 kD / 2 D Example 10.1 4 Phasor model of single slit Fraunhofer diffraction: rolling paper 5 Read: Ch10: 1-2 Homework: Ch10: 2,7,8,9 Due: December 10 6 December 5 Double slit and many slits 10.2.2 The double slit L b/2 E ( x, z ) R

b/2 z exp[ik ( R z ' sin )]dz ' a b / 2 L exp[ik ( R z ' sin )]dz ' R a b / 2 b sin L exp(ikR )1 exp( ika sin ) R R-a sin P (x,z) b a R x 2 b Let I 0 L , intensity at the axis when there is only one slit. (We now define I EE *.) R (ka / 2) sin (kb / 2) sin sin I ( ) 4 I 0 2

cos 2 The result is a rapidly varying double-slit interference pattern (cos 2) modulated by a slowly varying single-slit diffraction pattern (sin2/2). 7 sin I ( ) 4 I 0 2 cos 2 2 sin( b sin / ) 2 a sin I ( ) I (0) cos b sin / Single-slit diffraction Two-slit interference Fringes Envelope Question: Which interference maximum coincides with the first diffraction minimum? b sin a m a sin m b Half-fringe (split fringe) may occur there. Our author counts a half-fringe as 0.5 fringe.

half-fringe 8 10.2.3 Diffraction by many slits P (x,z) z R-2a sin in R-a s R b a x C L / R, F ( z ' ) exp[ik ( R z ' sin )] b/2 a b / 2 b/2 a b / 2 E ( x, z ) C F ( z ' )dz 'C 2 a b / 2 F ( z ' )dz ' C 2a b / 2 ( N 1) a b / 2 F ( z ' )dz '... C

( N 1) a b / 2 F ( z ' )dz ' sin exp(ikR )1 exp( i 2 ) exp( i 4 ) ... exp[ i 2( N 1) ] (ka / 2) sin sin 1 exp( i 2 N ) bC exp(ikR ) 1 exp( i 2 ) (kb / 2) sin bC 2 sin sin N I ( ) I 0 sin 2 9 sin I ( ) I 0 2 sin N sin

(ka / 2) sin (kb / 2) sin 2 Principle maxima: 0, , 2 , ... Minima (totally N-1): 2 3 ( N 1) , , , ... N N N N Subsidiary maxima (totally N-2): 3 5 (2 N 3) , , ..., 2N 2N 2N sin 2 sin N

sin a 4b N 6 2 Example 10.3 10 Phasor model of three-slit interference: rotating sticks 11 Read: Ch10: 2 Homework: Ch10: 14,15,17 Due: December 10 12 December 7 Rectangular aperture and circular aperture 10.2.4 The rectangular aperture Coherent aperture: E A Y y exp(ikr ) dS r Aperture dS=dydz P(Y,Z) r R

R X 2 Y 2 Z 2 x r X 2 (Y y ) 2 ( Z z ) 2 R 1 ( y 2 z 2 ) / R 2 2(Yy Zz ) / R 2 z X Z R 1 2(Yy Zz ) / R 2 R 1 (Yy Zz ) / R 2 E (Y , Z ) Fraunhofer diffraction condition A exp(ikR) exp ik (Yy Zz ) / R dS R Aperture 13 Y y Rectangular aperture: dS=dydz P(Y,Z) r R b

a z E (Y , Z ) x Z A exp(ikR ) exp ik (Yy Zz ) / R dS R Aperture A exp(ikR ) a /2 b/2 exp ik (Yy Zz ) / R dydz a /2 b /2 R ka Z ' ab A exp(ikR ) sin ' sin ' 2 R , R ' ' ' kb Y

2 R sin ' I (Y , Z ) I (0) ' 2 sin ' ' ka sin Z ), 2 kb ( sin Y ). 2 ( 2 14 sin ' I (Y , Z ) I (0) ' Y minimum: Z minimum:

2 sin ' ' 2 ' kaZ / 2 R, ' kbY / 2 R. 2R kb 2R ' kaZ / 2 R , 2 , ... Z m ka ' kbY / 2 R , 2 , ... Y m 15 10.2.5 The circular aperture Importance in optical instrumentation: The image of a distant point source is not a point, but a diffraction pattern because of the limited size of the lenses. Yy Zz q sin sin q cos cos q cos( ) exp(ikR ) E (Y , Z ) A exp ik (Yy Zz ) / R dS R Aperture A exp(ikR ) 2 a 0 0 exp ik q / R cos( )dd R 0 exp(ikR ) 2 a A

0 0 exp ik q / R cos dd R a exp(ikR ) A 2 J 0 ( kq / R )d 0 R exp(ikR ) J (kaq / R ) A 2a 2 1 R kaq / R 4 A2 A2 J 1 (kaq / R ) E 2 R kaq / R 2 2 Y y P(Y,Z) a x z q R Z Bessel functions:.

1 2 exp(iu cos v)dv 0 2 i m 2 J m (u ) exp[i (mv u cos v)]dv 0 2 d m [u J m (u )] u m J m 1 (u ) du J (u ) 1 lim 1 u 0 u 2 J 0 (u ) 2 2 J (kaq / R ) 2 J1 (ka sin ) I ( ) I (0) 1 I ( 0 ) ka sin kaq / R 16 2 2 J (ka sin ) I ( ) I (0) 1 ka sin 2 J0(u)

J1(u) u I ( ) / I (0) q1 0.018 Radius of Airy disk: ka sin 3.83 J1 (kaq / R ) 0 kaq1 / R 3.83 q1 1.22 R f , qlens 1.22 for a lens 2a D P D f Example 10.6 17 Read: Ch10: 2 Homework: Ch10: 25(Optional),28,40 Due: December 10 18

## Recently Viewed Presentations

• What is competitive advantage? A competitive advantage is an advantage over your competition achieved by offering consumers greater value, either by means of lower prices or by providing higher quality, more unique products/services that warrants higher prices.
• School teachers were strict and were allowed to hit their students or make them wear a dunce hat if they were bad or said the wrong answer. Facts About School. In the New England colonies, children were taught to read...
• PROJECT GOALS. Temple's broad . data science question: Investigate how the specification of concepts change over time across 4 historical Encyclopedia Britannicas (1797-1911)
• The duty to assist agencies of law enforcement. How does common law differ from positive law? On which early legal system is the U.S. legal system based? What are the four sources of law? Which source of law in the...
• What is dementia? Dementia is a progressive condition, caused by structural & chemical changes in the brain (Alzheimer Society UK, 2007). Dementia is characterised by memoryimpairment, cognitive impairment, progressive decline in functioning & behavioural changes.
• History of Design Drawing A means for communication Design is used to relay a visual concept Change Technology has made advances in fabricating and design techniques that are clearly seen looking back through history Before the Industrial Revolution Cave men...
• Tree service. The Tree Service Solution. Epsilon's superior engineering delivers the most effective combination of power, performance, speed, and quality. Designed for constant use in the harshest environments - The main lift cylinder is protected by a heavy-duty steel guard,...
• MIPS instructions Author: zhenghao Last modified by: zhenghao Created Date: 1/23/2009 3:21:14 AM Document presentation format: On-screen Show (4:3) Other titles: Arial Calibri Office Theme MIPS instructions Writing a bubble sort code The code we wrote in the class