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Fraunhofer Diffraction: Multiple slits & Circular aperture

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Presentation on theme: "Fraunhofer Diffraction: Multiple slits & Circular aperture"— Presentation transcript:

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

2 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

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

4 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

5 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

6 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)

7 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

8 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

9 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

10 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, _________________

11 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___________

12 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.

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

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

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

16 Fraunhofer diffraction from a circular aperture
Let Then

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

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

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

20 Fraunhofer diffraction from a circular aperture
Now the zeros of J1() occur at, = 0, 3.832, 7.016, , … = 0, 1.22, 2.23, 3.24, … =kR sin = (2/) sin Thus zero at sin  = 1.22/D, 2.23 /D, 3.24 /D, …

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

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

23 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

24 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

25 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


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