1. 1 Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture Fri. Nov. 22, 2002

2. 2 Fraunhofer diffraction limit If aperture is a square -  X  The same relation holds in azimuthal plane and  2 ~ measure of the area of the aperture Then we have the Fraunhofer diffraction if, Fraunhofer or far field limit

3. 3 Fraunhofer, Fresnel limits The near field, or Fresnel, limit is See 10.1.2 of text

4. 4 Fraunhofer diffraction Typical arrangement (or use laser as a source of plane waves) Plane waves in, plane waves out S f1f1 f2f2  screen

5. 5 Fraunhofer diffraction 1. Obliquity factor Assume S on axis, so Assume  small ( < 30 o ), so 2. Assume uniform illumination over aperture r’ >>  so is constant over the aperture 3. Dimensions of aperture << r r will not vary much in denominator for calculation of amplitude at any point P consider r = constant in denominator

6. 6 Fraunhofer diffraction Then the magnitude of the electric field at P is,

7. 7 Single slit Fraunhofer diffraction y = b y dy P  roro r r = r o - ysin  dA = L dy where L   ( very long slit)

8. 8 Single slit Fraunhofer diffraction Fraunhofer single slit diffraction pattern

9. 9 Single Slit Fraunhofer diffraction: Effect of slit width Minima for sin  = 0  =  p  = k(b/2)sin  or, sin  =  p( /b) First minima at sin  =  /b  b

10. 10 Single Slit Fraunhofer diffraction: Effect of slit width Width of central max  2 ( /dimension of aperture) This relation is characteristic of all Fraunhofer diffraction If b is very large  0 and a point source is imaged as a point If b is very small (~ )  /2 and light spreads out across screen (diminishes at large angles for to F(  )

11. 11 Diffraction from an array of N slits, separated by a distance a and of width b y=0 y=a y=a+b y=2a y=2a+b y=3a y=3a+b y=(N-1)a y=(N-1)a + b y=b P     

12. 12 Diffraction from an array of N slits It can be shown that, where,

13. 13 Diffraction and interference for N slits The diffraction term Minima for sin  = 0  =  p  = k(b/2)sin  or, sin  =  p( /b) The interference term Amplitude due to N coherent sources Can see this by adding N phasors that are 2  out of phase. See Hecht Problem 10.2

14. 14 Interference term Maxima Maxima occur at  =  m  (m = 0,1, 2, 3,..) To see this use L’Hopital’s rule _______ Thus maxima occur at sin  =  m /a This is the same result we have derived for Young’s double slit Intensity of principal maxima, I = N 2 I o i.e. N times that due to one slit

15. 15 Interference term Minima Minima occur for  =  /N, 2  /N, … (N-1)  /N and when we add m  For example, _______________________ Thus principal maxima have a width determined by zeros on each side Since  = (  / )a sin  =  /N The angular width is determined by sin  = /(Na) Thus peaks are N times narrower than in a single slit pattern (also a > b)

16. 16 Interference term Subsidiary or Secondary Maximum Subsidiary or Secondary Maximum Now between zeros must have secondary maxima Assume these are approximately midway Then first at [ m+3/(2N) ]  Then it can be shown that

17. 17 Single slit envelope Now interference term or pattern is modulated by the diffraction term which has zeros at  =(  b/ )sin  =  p  or, sin  =  p /b But, sin  =  m /a locate the principal maxima of the interference pattern

18. 18 Single slit envelope Thus at a given angle a/b=m/p Then suppose a/b = integer For example, a = 3b Then m = 3, 6, 9, interference maxima are missing

19. 19 Diffraction gratings Composed of systems with many slits per unit length – usually about 1000/mm Also usually used in reflection Thus principal maxima vary sharp Width of peaks Δ  = (2/N)  As N gets large the peak gets very narrow For example, _________________

20. 20 Diffraction gratings Resolution Resolution Imagine trying to resolve two wavelengths 1  2 Assume resolved if principal maxima of one falls on first minima of the other See diagram___________

21. 21 Diffraction gratings m 1 = a sin  m 2 = a sin  ’ But must have Thus m( 2 - 1 )= a (sin  ’ - sin  ) = ( 1 /N) Or mΔ = /N Resolution, R = /Δ = mN E.g.

22. 22 Fraunhofer diffraction from a circular aperture x y   P  Lens plane r

23. 23 Fraunhofer diffraction from a circular aperture Do x first – looking down Path length is the same for all rays = r o  Why?

24. 24 Fraunhofer diffraction from a circular aperture Do integration along y – looking from the side -R +R y=0 roro  r = r o - ysin  P   

25. 25 Fraunhofer diffraction from a circular aperture Let Then

26. 26 Fraunhofer diffraction from a circular aperture The integral where J 1 (  ) is the first order Bessell function of the first kind.

27. 27 Fraunhofer diffraction from a circular aperture These Bessell functions can be represented as polynomials: and in particular (for p = 1),

28. 28 Fraunhofer diffraction from a circular aperture Thus, where  = kRsin  and I o is the intensity when  =0

29. 29 Fraunhofer diffraction from a circular aperture Now the zeros of J 1 (  ) occur at,  = 0, 3.832, 7.016, 10.173, …  = 0, 1.22 , 2.23 , 3.24 , …  =kR sin  = (2  / ) sin  Thus zero at sin  = 1.22 /D, 2.23 /D, 3.24 /D, …

30. 30 Fraunhofer diffraction from a circular aperture The central Airy disc contains 85% of the light

31. 31 Fraunhofer diffraction from a circular aperture D  sin  = 1.22 /D

32. 32 Diffraction limited focussing sin  = 1.22 /D The width of the Airy disc W = 2fsin   2f  = 2f(1.22 /D) = 2.4 f /D W = 2.4(f#) > f# > 1 Cannot focus any wave to spot with dimensions < D f  

33. 33 Fraunhofer diffraction and spatial resolution Suppose two point sources or objects are far away (e.g. two stars) Imaged with some optical system Two Airy patterns  If S 1, S 2 are too close together the Airy patterns will overlap and become indistinguishable S1S1 S2S2 

34. 34 Fraunhofer diffraction and spatial resolution Assume S 1, S 2 can just be resolved when maximum of one pattern just falls on minimum (first) of the other Then the angular separation at lens, e.g. telescope D = 10 cm = 500 X 10 -7 cm e.g. eye D ~ 1mm  min = 5 X 10 -4 rad