Fraunhofer Diffraction: Circular aperture

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Fraunhofer Diffraction Circular aperture

• Wed. Nov. 27, 2002

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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

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Polarization
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Matrix treatment of polarization

• Consider a light ray with an instantaneous
  E-vector as shown

y
Ey
x
Ex
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Matrix treatment of polarization

• Combining the components
• The terms in brackets represents the complex
  amplitude of the plane wave

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Jones Vectors

• The state of polarization of light is determined
  by
• the relative amplitudes (Eox, Eoy) and,
• the relative phases (? ?y - ?x )
• of these components
• The complex amplitude is written as a two-element
  matrix, the Jones vector

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Jones vector Horizontally polarized light
The arrows indicate the sense of movement as the
beam approaches you

• The electric field oscillations are only along
  the x-axis
• The Jones vector is then written,
• where we have set the phase ?x 0, for
  convenience

The normalized form is
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Jones vector Vertically polarized light

• The electric field oscillations are only along
  the y-axis
• The Jones vector is then written,
• Where we have set the phase ?y 0, for
  convenience

The normalized form is
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Jones vector Linearly polarized light at an
arbitrary angle

• If the phases are such that ? m? for m 0,
  ?1, ?2, ?3,
• Then we must have,
• and the Jones vector is simply a line inclined
  at an angle ? tan-1(Eoy/Eox)
• since we can write

?
The normalized form is
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Jones vector and polarization

• In general, the Jones vector for the arbitrary
  case
• is an ellipse

y
Eoy
b
a
?
x
Eox