Download E9160: Born Approximation for Scattering and Fraunhofer Diffraction

E9160: Born Approximation for Scattering and Fraunhofer Diffraction
Submitted by: Irina Tkachenko and Irina Segal
The problem:
(1) Write the Born approximation for the scattered amplitude in a way similar to Fraunhofer
diffraction approximation.
(2) Find Fraunhofer approximation for diffraction where half of the rays pass through a glass board
(where the phase shift due to the passage in the glass is given)
(3) Find the exact calculation for the optical system in literature
(4) Use the results of the previous questions to find the cross-section in Aharonov-Bohm geometry.
(Choose the gauge properly).
The solution:
(1) Fraunhofer approximation:
ik
eikz e 2z (x
U (x, y) =
iλz
2 +y 2 )
ZZ
2π
0
0
u(x0 , y 0 )e−i λz (xx +yy ) dx0 dy 0
In optics, u(x’,y’) represents 2-dimensional ”mask”, or complex transmittance.
Born approximation for scattering:
~
Ψ(~r) = eik0 ·~r + f (Ω)
eik|~r|
m eik|~r|
~
= eik0 ·~r −
< k~Ω |V |k~0 >
|~r|
2π |~r|
Where:
k~0 is in z-direction (incident planar wave)
√
k = 2mE
k~Ω = k~nΩ = k(sin θ cos φ, sin θ sin φ, cos θ) = k(x/r, y/r, z/r)
Born approximation is similar to Fraunhofer approximation in case the scatterer is 2-dimensional:
V (x, y, z) → V (x, y)δ(z)
We are interested in ”far-field”, away from the scatterer, so in the denominator:
|~r| ≈ z
but in the exponent we need to be more careful:
|~r| =
p
x2 + y 2
x2 + y 2
z 2 + x2 + y 2 ≈ z(1 +
)
=
z
+
2z 2
2z
Now, let’s take a look on the last term:
< k~Ω |V |k~0 >=
Z
~
~
V (r~0 )e−i(kΩ −k0 )·r~0 dr~0
1
Where r~0 = (x0 , y 0 , 0) in the plane of the scatterer, and so:
k~0 · r~0 = 0
k
k
k~Ω · r~0 = (xx0 + yy 0 ) ≈ (xx0 + yy 0 )
r
z
And so we get:
ik
Ψ(r) = eikz −
m eikz e 2z (x
2π
z
2 +y 2 )
Z
k
0
0
V (x0 , y 0 )e−i z (xx +yy ) dx0 dy 0
If we take:
−
m
1
V (x0 , y 0 ) = u(x0 , y 0 )
2π
iλ
then the Born approximation is the same as Fraunhofer approximation.
(2) The transmittance of a glass half-plane is
iφ 0
e ,x > 0
0
u(x ) =
1, x0 < 0
The calculation in this part is in 2D and not 3D!
U (x) =
=
√
√
ik
eikz e 2z x
√
2π
iλ z
ik
2
Z
0
e
−i kz xx0
−∞
0
∞
Z
dx +
iφ −i kz xx0
e e
dx
0
=
0
ik
2
2
√ eikz e 2z x z
eikz e 2z x iz
√
√
2π
[1 − eiφ ] = 2π
(−2i)eiφ/2 sin(φ/2) =
iλ z kx
λ z kx
ik
2
eikz e 2z x eiφ/2 2z
√
sin(φ/2)
= √
z i 2π x
In our approximation,
x
2z
≈
θ
2
≈ sin( 2θ ) and therefore we get:
eikz
U (x) = √ f (θ)
z
ik
2
e 2z x eiφ/2 sin(φ/2)
√
f (θ) =
i 2π sin(θ/2)
(3) The exact calculation as published by Anokhov 1 :
Z U
1 + eiφ
1−i
2
E(x, z) = eikz
+ [1 − eiφ ] √
×
eiµ dµ
2
2π
0
q p
U = ± k[ x2 + z 2 − z]
µ2 = k(r − z)
2
(4) To use the result of (2) in Aharonov-Bohm Geometry, we need to define the phase-shift due to
the magnetic flux. We choose the vector-potential (gauge):
A = Φδ(θ − π/2)
The phase-shift for the upper half is:
φ=
eΦ
c~
The notations in Aharonov-Bohm article are slightly different:
θ0 = π − θ =⇒ sin(θ/2) = sin((π − θ0 )/2) = cos(θ0 /2)
eΦ
=⇒ sin(φ/2) = sin(πα)
ch
Substitution of these notations in our previous result gives the following cross-section:
α=
σ(θ0 ) = |f (θ0 )|2 =
1
sin2 (πα)
2π cos2 (θ0 /2)
S. Anokhov, ”Plane wave diffraction by a perfectly transparent half-plane”, J. Opt. Soc. Am. A 24 (2007)
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