Fraunhofer Diffraction: Multiple slits & Circular aperture

Fraunhofer Diffraction: Multiple slits & Circular aperture Mon. Nov. 25, 2002

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

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

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 • 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 • 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 • 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 • 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 • 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 • 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 • 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 • 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  y  P r x  Lens plane

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

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

Fraunhofer diffraction from a circular aperture Let Then

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 • These Bessell functions can be represented as polynomials: • and in particular (for p = 1),

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

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 The central Airy disc contains 85% of the light

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

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 <  f D  

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 • 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

Fraunhofer Diffraction: Multiple slits & Circular aperture