Fraunhofer Diffraction Multiple slits Circular aperture Mon Nov

Fraunhofer Diffraction: Multiple slits & Circular aperture Mon. Nov. 25, 2002 1

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=3 a+b y=3 a y=2 a+b y=2 a y=a+b y=a P a y=b y=0 2

Diffraction from an array of N slits n It can be shown that, n where, 3

Diffraction and interference for N slits The diffraction term n Minima for sin = 0 n = p = k(b/2)sin n or, sin = p( /b) The interference term n Amplitude due to N coherent sources n Can see this by adding N phasors that are 2 out of phase. See Hecht Problem 10. 2 4

Interference term Maxima occur at = m (m = 0, 1, 2, 3, . . ) n To see this use L’Hopital’s rule _______ n Thus maxima occur at sin = m /a n This is the same result we have derived for Young’s double slit n Intensity of principal maxima, I = N 2 Io n i. e. N times that due to one slit n 5

Interference term n Minima occur for = /N, 2 /N, … (N-1) /N 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 n n sin = /(Na) n Thus peaks are N times narrower than in a single slit pattern (also a > b) 6

Interference term n n n Subsidiary or Secondary Maximum Now between zeros must have secondary maxima Assume these are approximately midway Then first at [ m+3/(2 N) ] Then it can be shown that 7

Single slit envelope n Now interference term or pattern is modulated by the diffraction term n which has zeros at =( b/ )sin = p or, sin = p /b But, sin = m /a locate the principal maxima of the interference pattern n n 8

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

Diffraction gratings Composed of systems with many slits per unit length – usually about 1000/mm n Also usually used in reflection n Thus principal maxima vary sharp n Width of peaks Δ = (2/N) n As N gets large the peak gets very narrow n For example, _________ n 10

Diffraction gratings Resolution n Imagine trying to resolve two wavelengths 1 2 n Assume resolved if principal maxima of one falls on first minima of the other n See diagram______ n 11

Diffraction gratings m = a sin n n n n 1 m 2 = a sin ’ But must have Thus m( 2 - 1 )= a (sin ’ - sin ) = ( 1/N) Or mΔ = /N Resolution, R = /Δ = m. N E. g. 12

Fraunhofer diffraction from a circular aperture y x P r Lens plane 13

Fraunhofer diffraction from a circular aperture Do x first – looking down Path length is the same for all rays = ro Why? 14

Fraunhofer diffraction from a circular aperture Do integration along y – looking from the side P +R y=0 -R ro r = ro - ysin 15

Fraunhofer diffraction from a circular aperture Let Then 16

Fraunhofer diffraction from a circular aperture The integral where J 1( ) is the first order Bessell function of the first kind. 17

Fraunhofer diffraction from a circular aperture n These Bessell functions can be represented as polynomials: n and in particular (for p = 1), 18

Fraunhofer diffraction from a circular aperture n Thus, n where = k. Rsin and Io is the intensity when =0 19

Fraunhofer diffraction from a circular aperture n Now the zeros of J 1( ) occur at, = 0, 3. 832, 7. 016, 10. 173, … = 0, 1. 22 , 2. 23 , 3. 24 , … =k. R sin = (2 / ) sin • Thus zero at sin = 1. 22 /D, 2. 23 /D, 3. 24 /D, … 20

Fraunhofer diffraction from a circular aperture The central Airy disc contains 85% of the light 21

Fraunhofer diffraction from a circular aperture D sin = 1. 22 /D 22

Diffraction limited focussing n n sin = 1. 22 /D The width of the Airy disc W = 2 fsin 2 f = 2 f(1. 22 /D) = 2. 4 f /D W = 2. 4(f#) > f# > 1 n Cannot focus any wave to spot with dimensions < f D 23

Fraunhofer diffraction and spatial resolution n Suppose two point sources or objects are far away (e. g. two stars) Imaged with some optical system Two Airy patterns ¨ S 1 If S 1, S 2 are too close together the Airy patterns will overlap and become indistinguishable S 2 24

Fraunhofer diffraction and spatial resolution n 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, n e. g. telescope D = 10 cm = 500 X 10 -7 cm n e. g. eye D ~ 1 mm min = 5 X 10 -4 rad n 25