fraunhofer diffraction

Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture : 

1 Fraunhofer Diffraction: Single, multiple slit(s) & Circular aperture Rahul Katiyar (B.Tech.)

Fraunhofer diffraction limit : 

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

Fraunhofer, Fresnel limits : 

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

Fraunhofer diffraction : 

4 Fraunhofer diffraction Typical arrangement (or use laser as a source of plane waves) Plane waves in, plane waves out S f1 f2  screen

Fraunhofer diffraction : 

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

Fraunhofer diffraction : 

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

Single slit Fraunhofer diffraction : 

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

Single slit Fraunhofer diffraction : 

8 Single slit Fraunhofer diffraction Fraunhofer single slit diffraction pattern

Single Slit Fraunhofer diffraction: Effect of slit width : 

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

Single Slit Fraunhofer diffraction: Effect of slit width : 

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()

Diffraction from an array of N slits, separated by a distance a and of width b : 

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     

Diffraction from an array of N slits : 

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

Diffraction and interference for N slits : 

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

Interference term : 

14 Interference term 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 = N2Io i.e. N times that due to one slit

Interference term : 

15 Interference term 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)

Interference term : 

16 Interference term 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

Single slit envelope : 

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

Single slit envelope : 

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

Diffraction gratings : 

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, _________________

Diffraction gratings : 

20 Diffraction gratings 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___________

Diffraction gratings : 

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.

Fraunhofer diffraction from a circular aperture : 

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

Fraunhofer diffraction from a circular aperture : 

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

Fraunhofer diffraction from a circular aperture : 

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

Fraunhofer diffraction from a circular aperture : 

25 Fraunhofer diffraction from a circular aperture Let Then

Fraunhofer diffraction from a circular aperture : 

26 Fraunhofer diffraction from a circular aperture The integral where J1() is the first order Bessell function of the first kind.

Fraunhofer diffraction from a circular aperture : 

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

Fraunhofer diffraction from a circular aperture : 

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

Fraunhofer diffraction from a circular aperture : 

29 Fraunhofer diffraction from a circular aperture Now the zeros of J1() 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, …

Fraunhofer diffraction from a circular aperture : 

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

Fraunhofer diffraction from a circular aperture : 

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

Diffraction limited focussing : 

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  

Fraunhofer diffraction and spatial resolution : 

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 S1, S2 are too close together the Airy patterns will overlap and become indistinguishable S1 S2 

Fraunhofer diffraction and spatial resolution : 

34 Fraunhofer diffraction and spatial resolution Assume S1, S2 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