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