Fresnel diffraction by a screen

1
Fresnel diffraction by a screen

• Pasi Manninen
• pasi.manninen_at_tkk.fi
• Metrology Research Institute / TKK

2
Outline

• Fresnel-Kirchhoff diffraction integral
• Fresnel approximation
• for F-K diffraction integral
• Fresnels integrals
• Fresnel diffraction behind a half plane
• Homework

3
Fresnel-Kirchhoff diffraction integral
(1)
x
P0(x0,y0,z0)
r
Q(?,?,0)
r
s
O
y
s
P(x,y,z)
z
4
Fresnel approximation for diffraction integral
1/4
When the distances of P0 and P from the screen
are large compared to the size of the aperture
(2)
(3)
Now the equation (1) reduces to
(4)
5
Fresnel approximation for diffraction integral
2/4
The quantities r, s, rand s can be expressed by
the coordinates of points P0, P and Q
(5)
(6)
(7)
6
Fresnel approximation for diffraction integral
3/4
The aperture is small compared to both r and s
(8)
We denote the new variables l, m, l0 and m0
(9)
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Fresnel approximation for diffraction integral
4/4
Now, we have approximation which is known as
Fresnel diffraction
(10)
where
(11)
Fraunhofer diffraction is obtained when both the
source and the point of observation are at large
distances from the aperture. In this case the
quadratic and higher order terms in ? and ? can
be neglect.
8
Fresnels integrals 1/7
The integral equation (10) can be written in the
form
(12)
(intensity I (P )B 2(C 2S 2))
where
(13)
(14)
As the x direction the projection of the line P0P
to the plane of the aperture
(15)
9
Fresnels integrals 2/7
Now, the expression (11) reduces to
(16)
We neglect terms of third and higher order in ?
and ?. This (16) is put to the equation (14) and
chosen new variables of integration of u, v
(17)
Then
(18)
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Fresnels integrals 3/7
Now, we have the integral (14) in the form
(19)
where
(20)
The relations for the cosine and sine functions
(21)
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Fresnels integrals 4/7
To evaluate (19) we must consider Fresnels
integrals
(22)
The cosine and sine into power series and
integration term by term
(23)
(23) is for small w. For large w evaluation is
done by series in inverse powers of w and
integrating by parts
(24)
12
Fresnels integrals 5/7
Fresnels cosine and sine integrals
13
Fresnels integrals 6/7
Finally, we obtain
(25)
where
(26)
We combine (22) into a comblex integral
(27)
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Fresnels integrals 7/7
1.5
N
0.7 0.6 0.5 0.4 0.3 0.2 0.1
From Cauchys residue theorem integral taken
along oblique line through the origin is equal to
the value of the integral taken along the real
axis gt
P
1.0
(28)
0.5
M
-0.7 -0.6 -0.5 -0.4 -0.3 0.2 -0.1
0.1 0.2 0.3 0.4 0.5 0.6 0.7
-0.5
-0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7
Thus
w
(29)
-1.0
P-
Cornus spiral
Cornus spiral
-1.5
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Fresnel diffraction behind a half plane 1/4
The integration extends troughout the region
x
or, in terms of u and v,
P
where
d
x0
(30)
O
y
P0
z
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Fresnel diffraction behind a half plane 2/4
The diffraction integrals (19) become
(31)
From the relations (29), M(-w)-M(w) and
N(-w)-N(w) we have
(32)
(33)
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Fresnel diffraction behind a half plane 3/4
Hence (31) becomes
(34)
Substitution into the parenthesis expression of
(12) gives for the intentsity
(35)
(36)
On the edge of the shadow (w 0) I /I (0) ¼.
This can be see at the figure of the next slide.
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Fresnel diffraction behind a half plane 4/4
w
x
x0
x0 5 (S.F.)
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Homework

• 1. Plane wave propagates towards a circular
  obstacle. The normal to the obstacle is parallel
  with the direction of propagation of plane wave.
  Show that in Fresnels approximation intensity is
  the same with and without the obstacle on a
  straight line which follows via the centre of the
  circle and is parallel with the direction of
  propagation of plane wave.
• ( )
• 2. Define the following intensities by Cornus
  spiral
• I (1.5)I (-0.5)/I (0)
• I (1.0)/I (0)

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References

• M. Born, Principles of optics, 1997.
• M.V. Klein, T.E. Furtak, Optics, 1986.