Fraunhofer Diffraction: Multiple slits

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Fraunhofer Diffraction Multiple slits Circular
aperture

• Mon. Nov. 25, 2002

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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
y3ab
P
?
y3a
y2ab
?
y2a
yab
?
ya
a
yb
?
y0
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Diffraction from an array of N slits

• It can be shown that,
• where,

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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

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Interference term

• Maxima occur at ? ?m? (m 0,1, 2, 3, ..)
• To see this use LHopitals rule _______
• Thus maxima occur at sin ? ?m?/a
• This is the same result we have derived for
  Youngs double slit
• Intensity of principal maxima, I N2Io
• i.e. N times that due to one slit

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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 gt b)

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Interference term

• Subsidiary or Secondary Maximum
• Now between zeros must have secondary maxima
• Assume these are approximately midway
• Then first at m3/(2N) ?
• Then it can be shown that

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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

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Single slit envelope

• Thus at a given angle a/bm/p
• Then suppose a/b integer
• For example, a 3b
• Then m 3, 6, 9, interference maxima are missing

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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 ?? (2/N)?
• As N gets large the peak gets very narrow
• For example, _________________

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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___________

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Diffraction gratings

• m?1 a sin ?
• m?2 a sin ?
• But must have
• Thus m(?2 - ?1 ) a (sin? - sin?) (?1/N)
• Or m?? ?/N
• Resolution, R ? /?? mN
• E.g.

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Fraunhofer diffraction from a circular aperture
?
y
?
P
r
x
?
Lens plane
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Fraunhofer diffraction from a circular aperture
Path length is the same for all rays ro
Do x first looking down
Why?
?
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Fraunhofer diffraction from a circular aperture
Do integration along y looking from the side
?
P
R
?
y0
ro
?
?
-R
r ro - ysin?
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Fraunhofer diffraction from a circular aperture
Let
Then
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Fraunhofer diffraction from a circular aperture
The integral
where J1(?) is the first order Bessell function
of the first kind.
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Fraunhofer diffraction from a circular aperture

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

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Fraunhofer diffraction from a circular aperture

• Thus,
• where ? kRsin? and Io is the intensity when ?0

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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,

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Fraunhofer diffraction from a circular aperture
The central Airy disc contains 85 of the light
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Fraunhofer diffraction from a circular aperture
D
?
sin? 1.22?/D
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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)? gt ? f gt 1
• Cannot focus any wave to spot with dimensions lt ?

f
D
?
?
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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
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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