fraunhofer diffraction: circular aperture
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1
Fraunhofer Diffraction: Circular aperture
Wed. Nov. 27, 2002
2
Fraunhofer diffraction from a circular aperture
x
y
P
Lens plane
r
dxdyeCE ikrP
3
Fraunhofer diffraction from a circular aperture
22 yR
22 yR
Do x first – looking downPath length is the same for all rays = ro
dyyReCE ikrP
222
Why?
4
Fraunhofer diffraction from a circular aperture
Do integration along y – looking from the side
-R
+R
y=0
ro
r = ro - ysin
P
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Fraunhofer diffraction from a circular aperture
R
R
ikyikrP dyyReCeE o 22sin2
sinkRR
y
)1(sin
kRRkky
)2(1 222222 RRRyR
)3(dyRd
Let
Then
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Fraunhofer diffraction from a circular aperture
1
1
22 12 deRCeE iikrP
o
The integral 1
1
1
21J
de i
where J1() is the first order Bessell function of the first kind.
7
Fraunhofer diffraction from a circular aperture
These Bessell functions can be represented as polynomials:
and in particular (for p = 1),
0
2
!!
21
k
pkk
P pkkJ
!4!3
2
!3!2
2
!2
21
2
642
1
J
8
Fraunhofer diffraction from a circular aperture
Thus,
where = kRsin and Io is the intensity when =0
2
12
J
II o
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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 atsin = 1.22/D, 2.23 /D, 3.24 /D, …
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-10 -8 -6 -4 -2 0 2 4 6 8 10
0.5
1.0
-10 -8 -6 -4 -2 0 2 4 6 8 10
0.5
1.0
Fraunhofer diffraction from a circular aperture
12J
2
12
J
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#) > f# > 1
Cannot focus any wave to spot with dimensions <
D
f
-10
-8-6
-4-2
02
46
810
0.5
1.0
13
-10
-8-6
-4-2
02
46
810
0.5
1.0
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
-10
-8-6
-4-2
02
46
810
0.5
1.0
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
D
22.1min
radXX 6
5
min 10510
105
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Polarization
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Matrix treatment of polarization
Consider a light ray with an instantaneous E-vector as shown
tkEjtkEitkE yx ,ˆ,ˆ,
x
y
Ex
Ey
y
x
tkzioyy
tkzioxx
eEE
eEE
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Matrix treatment of polarization Combining the components
The terms in brackets represents the complex amplitude of the plane wave
tkzio
tkziioy
iox
tkzioy
tkziox
eEE
eeEjeEiE
eEjeEiE
yx
yx
~
ˆˆ
ˆˆ
18
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
ioy
oxiyi
oy
iox
oy
oxo eE
Ee
eE
eE
E
EE x
x
~
~~
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Jones vector: Horizontally polarized light
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
0
1
00~
~~
AAeE
E
EE
xiox
oy
oxo
x
y
The arrows indicate the sense of movement as the beam approaches you
The normalized formis
0
1
20
x
y
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
1
000~
~~
AAeEE
EE
yioyoy
oxo
The normalized formis
1
0
21
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
oy
oxm
y
x
E
E
E
E1
sin
cos1~
~~ m
oy
oxo A
E
EE
x
y
The normalized form is
22
Jones vector and polarization
In general, the Jones vector for the arbitrary case
is an ellipse
i
oy
oxo eE
EE~
a
b
Eox
Eoy
x
y
22
cos22tan
oyoxEE
EE oyox
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