Réponse impulsionnelle Modifier La répartition du champ dans le plan image peut être exprimée
U
i
(
x
,
y
)
=
∫
−
∞
∞
∫
−
∞
∞
U
o
(
a
,
b
)
h
(
x
,
y
,
a
,
b
)
d
a
d
b
{\displaystyle U_{i}(x,y)=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }U_{o}(a,b)\,h(x,y,a,b)\ \mathrm {d} a\ \mathrm {d} b}
Entre l'objet et la lentille Modifier Le champ au niveau de la face d'entrée de la lentille en fonction du champ dans le plan objet est :
U
l
(
X
,
Y
,
a
,
b
)
=
e
i
k
D
o
i
λ
D
o
∫
−
∞
∞
∫
−
∞
∞
U
o
e
i
k
2
D
o
[
(
X
−
a
)
2
+
(
Y
−
b
)
2
]
d
X
d
Y
{\displaystyle U_{l}(X,Y,a,b)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{o}}}{\mathrm {i} \,\lambda D_{o}}}\ \!\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }U_{o}\,\mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left[(X-a)^{2}+(Y-b)^{2}\right]}\ \mathrm {d} X\ \mathrm {d} Y}
Si on considère un seul point source , afin de trouver l'expression de la réponse impulsionnelle en amplitude, le champ
U
o
{\displaystyle U_{o}}
δ
{\displaystyle \delta }
(
a
,
b
)
{\displaystyle (a,b)}
U
l
{\displaystyle U_{l}}
U
l
(
X
,
Y
,
a
,
b
)
=
e
i
k
D
o
i
λ
D
o
e
i
k
2
D
o
[
(
X
−
a
)
2
+
(
Y
−
b
)
2
]
{\displaystyle U_{l}(X,Y,a,b)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{o}}}{\mathrm {i} \,\lambda D_{o}}}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left[(X-a)^{2}+(Y-b)^{2}\right]}}
Au niveau de la lentille Modifier p.97
L'épaisseur de la lentille à traverser dépend de la position
(
X
,
Y
)
{\displaystyle (X,Y)}
Δ
=
Δ
0
−
X
2
+
Y
2
2
(
1
R
1
−
1
R
2
)
{\displaystyle \Delta =\Delta _{0}-{\frac {X^{2}+Y^{2}}{2}}\left({\frac {1}{R_{1}}}-{\frac {1}{R_{2}}}\right)}
qui se traduit dans l'équation de la propagation par la multiplication par un facteur de module unité :
e
i
k
n
Δ
0
e
−
i
k
(
n
−
1
)
X
2
+
Y
2
2
(
1
R
1
−
1
R
2
)
{\displaystyle \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}\mathrm {e} ^{-\mathrm {i} \,k(n-1){\frac {X^{2}+Y^{2}}{2}}\left({\frac {1}{R_{1}}}-{\frac {1}{R_{2}}}\right)}}
où on reconnait l'expression de la vergence de la lentille mince si bien que :
e
i
k
n
Δ
0
e
−
i
k
2
f
(
X
2
+
Y
2
)
{\displaystyle \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}\mathrm {e} ^{-{\frac {\mathrm {i} \,k}{2f}}\left(X^{2}+Y^{2}\right)}}
Après le passage de la lentille, on trouve :
U
l
′
(
X
,
Y
)
=
t
(
X
,
Y
)
e
i
k
n
Δ
0
e
−
i
k
2
f
(
X
2
+
Y
2
)
U
l
(
X
,
Y
)
{\displaystyle U_{l}^{'}(X,Y)=t(X,Y)\ \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}\ \mathrm {e} ^{-{\frac {\mathrm {i} \,k}{2f}}\left(X^{2}+Y^{2}\right)}U_{l}(X,Y)}
Entre la lentille et l'image Modifier Ce qui amène l'expression
U
i
(
x
,
y
)
=
e
i
k
D
i
i
λ
D
i
∫
−
∞
∞
∫
−
∞
∞
U
l
′
e
i
k
2
D
i
[
(
x
−
Y
)
2
+
(
y
−
Y
)
2
]
d
X
d
Y
{\displaystyle U_{i}(x,y)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}}{\mathrm {i} \,\lambda D_{i}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }U_{l}^{'}\,\mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left[(x-Y)^{2}+(y-Y)^{2}\right]}\ \mathrm {d} X\ \mathrm {d} Y}
Entre le plan objet et le plan image Modifier En combinant les trois résultats précédents :
U
i
(
x
,
y
,
a
,
b
)
=
h
(
x
,
y
,
a
,
b
)
=
e
i
k
D
i
i
λ
D
i
e
i
k
D
o
i
λ
D
o
e
i
k
n
Δ
0
∫
−
∞
∞
∫
−
∞
∞
t
(
X
,
Y
)
e
−
i
k
2
f
(
X
2
+
Y
2
)
e
i
k
2
D
o
[
(
X
−
a
)
2
+
(
Y
−
b
)
2
]
e
i
k
2
D
i
[
(
x
−
Y
)
2
+
(
y
−
Y
)
2
]
d
X
d
Y
{\displaystyle U_{i}(x,y,a,b)=h(x,y,a,b)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}}{\mathrm {i} \,\lambda D_{i}}}\ {\frac {\mathrm {e} ^{\mathrm {i} \,kD_{o}}}{\mathrm {i} \,\lambda D_{o}}}\ \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{-{\frac {\mathrm {i} \,k}{2f}}\left(X^{2}+Y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left[(X-a)^{2}+(Y-b)^{2}\right]}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left[(x-Y)^{2}+(y-Y)^{2}\right]}\ \mathrm {d} X\ \mathrm {d} Y}
h
(
x
,
y
,
a
,
b
)
=
−
e
i
k
D
i
e
i
k
D
o
e
i
k
n
Δ
0
λ
2
D
i
D
o
∫
−
∞
∞
∫
−
∞
∞
t
(
X
,
Y
)
e
i
k
2
(
1
D
i
+
1
D
o
−
1
f
)
(
X
2
+
Y
2
)
e
i
k
2
D
i
(
x
2
+
y
2
)
e
i
k
2
D
o
(
a
2
+
b
2
)
e
−
i
k
[
(
a
D
o
+
x
D
i
)
X
+
(
b
D
o
+
y
D
i
)
Y
]
d
X
d
Y
{\displaystyle h(x,y,a,b)=-{\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}\ \mathrm {e} ^{\mathrm {i} \,kD_{o}}\ \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}}{\lambda ^{2}D_{i}D_{o}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2}}\left({\frac {1}{D_{i}}}+{\frac {1}{D_{o}}}-{\frac {1}{f}}\right)\left(X^{2}+Y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left(x^{2}+y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left(a^{2}+b^{2}\right)}\ \mathrm {e} ^{-\mathrm {i} \,k\left[\left({\frac {a}{D_{o}}}+{\frac {x}{D_{i}}}\right)X+\left({\frac {b}{D_{o}}}+{\frac {y}{D_{i}}}\right)Y\right]}\ \mathrm {d} X\ \mathrm {d} Y}
D'une part, la relation de conjugaison apporte
1
D
i
+
1
D
o
−
1
f
=
0
{\displaystyle {\frac {1}{D_{i}}}+{\frac {1}{D_{o}}}-{\frac {1}{f}}=0}
γ
t
=
−
D
i
D
o
{\displaystyle \gamma _{t}=-{\frac {D_{i}}{D_{o}}}}
D'autre part, si on considère que la lentille donne une image convenable (pas trop floue), on peut considérer que les points
(
a
,
b
)
{\displaystyle (a,b)}
k
2
D
o
(
a
2
+
b
2
)
{\displaystyle {\frac {k}{2D_{o}}}\left(a^{2}+b^{2}\right)}
(
x
γ
t
,
y
γ
t
)
{\displaystyle \left({\frac {x}{\gamma _{t}}},{\frac {y}{\gamma _{t}}}\right)}
a
≃
x
γ
t
{\displaystyle a\simeq {\frac {x}{\gamma _{t}}}}
b
≃
y
γ
t
{\displaystyle b\simeq {\frac {y}{\gamma _{t}}}}
h
(
x
,
y
,
a
,
b
)
=
−
e
i
k
D
i
e
i
k
D
o
e
i
k
n
Δ
0
e
i
k
2
D
i
(
x
2
+
y
2
)
e
i
k
2
D
o
x
2
+
y
2
γ
t
2
λ
2
D
i
D
o
∫
−
∞
∞
∫
−
∞
∞
t
(
X
,
Y
)
e
−
i
k
[
(
−
γ
t
a
D
i
+
x
D
i
)
X
+
(
−
γ
t
b
D
i
+
y
D
i
)
Y
]
d
X
d
Y
{\displaystyle h(x,y,a,b)=-{\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}\ \mathrm {e} ^{\mathrm {i} \,kD_{o}}\ \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left(x^{2}+y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}{\frac {x^{2}+y^{2}}{\gamma _{t}^{2}}}}}{\lambda ^{2}D_{i}D_{o}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{-\mathrm {i} \,k\left[\left({\frac {-\gamma _{t}a}{D_{i}}}+{\frac {x}{D_{i}}}\right)X+\left({\frac {-\gamma _{t}b}{D_{i}}}+{\frac {y}{D_{i}}}\right)Y\right]}\ \mathrm {d} X\ \mathrm {d} Y}
Si on ne tient pas compte des divers déphasages la relation devient :
h
(
x
−
γ
t
a
,
y
−
γ
t
b
)
=
−
1
λ
2
D
i
D
o
∫
−
∞
∞
∫
−
∞
∞
t
(
X
,
Y
)
e
−
i
k
D
i
[
(
x
−
γ
t
a
)
X
+
(
y
−
γ
t
b
)
Y
]
d
X
d
Y
{\displaystyle h(x-\gamma _{t}a\,,\,y-\gamma _{t}b)=-{\frac {1}{\lambda ^{2}D_{i}D_{o}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{{\frac {-\mathrm {i} \,k}{D_{i}}}\left[\left(x-\gamma _{t}a\right)X+\left(y-\gamma _{t}b\right)Y\right]}\ \mathrm {d} X\ \mathrm {d} Y}
h
(
x
,
y
)
=
−
1
λ
2
D
i
D
o
∫
−
∞
∞
∫
−
∞
∞
t
(
X
,
Y
)
e
−
i
2
π
λ
D
i
(
x
X
+
y
Y
)
d
X
d
Y
{\displaystyle h(x,y)=-{\frac {1}{\lambda ^{2}D_{i}D_{o}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{{\frac {-\mathrm {i} \,2\pi }{\lambda D_{i}}}\left(xX+yY\right)}\ \mathrm {d} X\ \mathrm {d} Y}
Fort de ces approximations, la réponse impulsionnelle en amplitude d'une lentille, dans le plan image, et équivalente à la figure de diffraction de Fraunhofer d'une ouverture étudiée en champ lointain.
Fonction de transfert en amplitude Modifier En introduisant les variable réduites
X
′
=
−
X
λ
D
i
{\displaystyle X'={\frac {-X}{\lambda D_{i}}}}
Y
′
=
−
Y
λ
D
i
{\displaystyle Y'={\frac {-Y}{\lambda D_{i}}}}
d
X
=
−
λ
D
i
X
′
{\displaystyle \mathrm {d} X=-\lambda D_{i}X'}
h
(
x
,
y
)
=
γ
t
∫
−
∞
∞
∫
−
∞
∞
t
(
−
λ
D
i
X
′
,
−
λ
D
i
Y
′
)
e
i
2
π
(
x
X
′
+
y
Y
′
)
d
X
′
d
Y
′
{\displaystyle h(x,y)=\gamma _{t}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t\left(-\lambda D_{i}X',-\lambda D_{i}Y'\right)\ \mathrm {e} ^{\mathrm {i} \,2\pi \left(xX'+yY'\right)}\ \mathrm {d} X'\ \mathrm {d} Y'}
h
(
x
,
y
)
=
γ
t
F
−
1
{
t
(
−
λ
D
i
X
′
,
−
λ
D
i
Y
′
)
}
{\displaystyle h(x,y)=\gamma _{t}\ {\mathcal {F}}^{-1}\left\{t(-\lambda D_{i}X',-\lambda D_{i}Y')\right\}}
La fonction de transfert en amplitude
h
^
(
ν
x
,
ν
y
)
=
F
{
h
(
x
,
y
)
}
{\displaystyle {\hat {h}}(\nu _{x},\nu _{y})={\mathcal {F}}\left\{h(x,y)\right\}}
h
^
(
ν
x
,
ν
y
)
=
∫
−
∞
∞
∫
−
∞
∞
h
(
x
,
y
)
e
−
i
2
π
(
x
ν
x
+
y
ν
y
)
d
x
d
y
{\displaystyle {\hat {h}}(\nu _{x},\nu _{y})=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }h\left(x,y\right)\ \mathrm {e} ^{-\mathrm {i} \,2\pi \left(x\nu _{x}+y\nu _{y}\right)}\ \mathrm {d} x\ \mathrm {d} y}
h
^
(
ν
x
,
ν
y
)
=
F
{
h
(
x
,
y
)
}
=
F
{
γ
t
F
−
1
{
t
(
−
λ
D
i
X
′
,
−
λ
D
i
Y
′
)
}
}
{\displaystyle {\hat {h}}(\nu _{x},\nu _{y})={\mathcal {F}}\left\{h(x,y)\right\}={\mathcal {F}}\left\{\gamma _{t}\ {\mathcal {F}}^{-1}\left\{t(-\lambda D_{i}X',-\lambda D_{i}Y')\right\}\right\}}
h
^
(
ν
x
,
ν
y
)
=
γ
t
t
(
−
λ
D
i
ν
x
,
−
λ
D
i
ν
y
)
{\displaystyle {\hat {h}}(\nu _{x},\nu _{y})=\gamma _{t}\ t(-\lambda D_{i}\nu _{x},-\lambda D_{i}\nu _{y})}
Compte-tenu de la symétrie des systèmes étudiés, les signes – peuvent être supprimés (toutes les fonctions sont paires).
h
^
(
ν
x
,
ν
y
)
=
γ
t
t
(
λ
D
i
ν
x
,
λ
D
i
ν
y
)
{\displaystyle {\hat {h}}(\nu _{x},\nu _{y})=\gamma _{t}\ t(\lambda D_{i}\nu _{x},\lambda D_{i}\nu _{y})}
Fonction de transfert optique Modifier
I
i
(
x
,
y
)
=
∫
−
∞
∞
∫
−
∞
∞
I
o
(
a
,
b
)
H
(
x
,
y
,
a
,
b
)
d
a
d
b
{\displaystyle I_{i}(x,y)=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }I_{o}(a,b)\,{\mathcal {H}}(x,y,a,b)\ \mathrm {d} a\ \mathrm {d} b}
I
i
(
x
,
y
)
=
∫
−
∞
∞
∫
−
∞
∞
I
o
(
a
,
b
)
|
h
(
x
,
y
,
a
,
b
)
|
2
d
a
d
b
{\displaystyle I_{i}(x,y)=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }I_{o}(a,b)\,|h(x,y,a,b)|^{2}\ \mathrm {d} a\ \mathrm {d} b}
I
i
(
x
,
y
)
=
∫
−
∞
∞
∫
−
∞
∞
I
o
(
a
,
b
)
|
h
(
x
−
γ
t
a
,
y
−
γ
t
b
)
|
2
d
a
d
b
{\displaystyle I_{i}(x,y)=\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }I_{o}(a,b)\,|h(x-\gamma _{t}a,y-\gamma _{t}b)|^{2}\ \mathrm {d} a\ \mathrm {d} b}
La fonction de transfert optique est :
H
^
(
ν
x
,
ν
y
)
=
F
{
|
h
(
x
,
y
)
|
2
}
=
F
{
h
(
x
,
y
)
}
∗
F
{
h
(
x
,
y
)
}
=
h
^
(
ν
x
,
ν
y
)
∗
h
^
(
ν
x
,
ν
y
)
{\displaystyle {\hat {\mathcal {H}}}(\nu _{x},\nu _{y})={\mathcal {F}}\left\{|h(x,y)|^{2}\right\}={\mathcal {F}}\left\{h(x,y)\right\}*{\mathcal {F}}\left\{h(x,y)\right\}={\hat {h}}(\nu _{x},\nu _{y})*{\hat {h}}(\nu _{x},\nu _{y})}
H
^
(
ν
x
,
ν
y
)
=
γ
t
2
t
(
−
λ
D
i
ν
x
,
−
λ
D
i
ν
y
)
∗
t
(
−
λ
D
i
ν
x
,
−
λ
D
i
ν
y
)
{\displaystyle {\hat {\mathcal {H}}}(\nu _{x},\nu _{y})=\gamma _{t}^{2}\ t(-\lambda D_{i}\nu _{x},-\lambda D_{i}\nu _{y})*t(-\lambda D_{i}\nu _{x},-\lambda D_{i}\nu _{y})}
La fonction de transfert optique normalisée est :
H
^
1
(
ν
x
,
ν
y
)
=
F
{
|
h
(
x
,
y
)
|
2
}
∫
∫
−
∞
∞
|
h
(
x
,
y
)
|
2
d
x
d
y
{\displaystyle {\hat {\mathcal {H}}}_{1}(\nu _{x},\nu _{y})={\frac {{\mathcal {F}}\left\{|h(x,y)|^{2}\right\}}{\int \!\int _{-\infty }^{\infty }|h(x,y)|^{2}\mathrm {d} x\mathrm {d} y}}}
Ouverture circulaire Modifier
t
(
X
,
Y
)
=
{
1
,
si
X
2
+
Y
2
≤
d
2
0
,
sinon
{\displaystyle t(X,Y)={\begin{cases}1,&{\text{si }}{\sqrt {X^{2}+Y^{2}}}\leq {\frac {d}{2}}\\0,&{\text{sinon }}\end{cases}}}
On observe que la fonction de transfert sera nulle si
ν
x
2
+
ν
y
2
≤
ν
c
=
d
λ
f
=
1
λ
N
{\displaystyle {\sqrt {\nu _{x}^{2}+\nu _{y}^{2}}}\leq \nu _{c}={\frac {d}{\lambda f}}={\frac {1}{\lambda N}}}
Autoconvolution d'un disque L’auto-convolution peut être calculée en déterminant la surface d'intersection de deux disques de rayons
ν
c
2
=
d
λ
f
=
1
2
λ
N
{\displaystyle {\frac {\nu _{c}}{2}}={\frac {d}{\lambda f}}={\frac {1}{2\lambda N}}}
ν
c
{\displaystyle \nu _{c}}
A
=
2
(
θ
ν
c
2
2
−
ν
c
cos
θ
ν
c
sin
θ
)
{\displaystyle A=2\left({\frac {\theta \nu _{c}^{2}}{2}}-\nu _{c}\cos \theta \,\nu _{c}\sin \theta \right)}
A
=
2
ν
c
2
(
θ
2
−
sin
2
θ
2
)
{\displaystyle A=2\,\nu _{c}^{2}\left({\frac {\theta }{2}}-{\frac {\sin 2\theta }{2}}\right)}
A
=
2
ν
c
2
(
θ
2
−
cos
θ
sin
θ
)
{\displaystyle A=2\,\nu _{c}^{2}\left({\frac {\theta }{2}}-{\cos \theta \sin \theta }\right)}
Au maximum, l'aire vaut
A
m
a
x
=
π
ν
c
2
{\displaystyle A_{\mathrm {max} }=\pi \,\nu _{c}^{2}}
ν
ν
c
=
cos
θ
{\displaystyle {\frac {\nu }{\nu _{c}}}=\cos \theta }
sin
2
θ
=
1
−
(
ν
ν
c
)
2
{\displaystyle \sin ^{2}\theta =1-\left({\frac {\nu }{\nu _{c}}}\right)^{2}}
A
=
2
ν
c
2
(
arccos
(
ν
/
ν
c
)
2
−
ν
ν
c
1
−
(
ν
ν
c
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2
)
{\displaystyle A=2\,\nu _{c}^{2}\left({\frac {\arccos(\nu /\nu _{c})}{2}}-{{\frac {\nu }{\nu _{c}}}{\sqrt {1-\left({\frac {\nu }{\nu _{c}}}\right)^{2}}}}\right)}
En divisant par
A
m
a
x
{\displaystyle A_{\mathrm {max} }}
H
^
1
(
ν
)
=
2
π
(
arccos
(
|
ν
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/
ν
c
)
2
−
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ν
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ν
c
1
−
(
ν
ν
c
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2
)
{\displaystyle {\hat {\mathcal {H}}}_{1}(\nu )={\frac {2}{\pi }}\left({\frac {\arccos(|\nu |/\nu _{c})}{2}}-{{\frac {|\nu |}{\nu _{c}}}{\sqrt {1-\left({\frac {\nu }{\nu _{c}}}\right)^{2}}}}\right)}
Ouverture rectangulaire Modifier
t
(
X
,
Y
)
=
Π
L
/
2
,
H
/
2
(
X
,
Y
)
=
{
1
,
si
−
L
2
≤
X
≤
L
2
et si
−
H
2
≤
Y
≤
H
2
0
,
sinon
{\displaystyle t(X,Y)=\Pi _{{L}/2,H/2}(X,Y)={\begin{cases}1,&{\text{si }}-{\frac {L}{2}}\leq X\leq {\frac {L}{2}}{\text{ et si }}-{\frac {H}{2}}\leq Y\leq {\frac {H}{2}}\\0,&{\text{sinon }}\end{cases}}}
Brouillon 1
Brouillon 2
Brouillon 3
Brouillon 4
Brouillon 5
Brouillon 6
Brouillon 9
![{\displaystyle U_{l}(X,Y,a,b)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{o}}}{\mathrm {i} \,\lambda D_{o}}}\ \!\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }U_{o}\,\mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left[(X-a)^{2}+(Y-b)^{2}\right]}\ \mathrm {d} X\ \mathrm {d} Y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25c026bd4c416622d728528aa24a14400e4219da)
![{\displaystyle U_{l}(X,Y,a,b)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{o}}}{\mathrm {i} \,\lambda D_{o}}}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left[(X-a)^{2}+(Y-b)^{2}\right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7234b96f7eea209a80eafd3c6ac537d15b14669c)
![{\displaystyle U_{i}(x,y)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}}{\mathrm {i} \,\lambda D_{i}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }U_{l}^{'}\,\mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left[(x-Y)^{2}+(y-Y)^{2}\right]}\ \mathrm {d} X\ \mathrm {d} Y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/765c30cc39803bd9e2453230deb7abbbf8ecd156)
![{\displaystyle U_{i}(x,y,a,b)=h(x,y,a,b)={\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}}{\mathrm {i} \,\lambda D_{i}}}\ {\frac {\mathrm {e} ^{\mathrm {i} \,kD_{o}}}{\mathrm {i} \,\lambda D_{o}}}\ \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{-{\frac {\mathrm {i} \,k}{2f}}\left(X^{2}+Y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left[(X-a)^{2}+(Y-b)^{2}\right]}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left[(x-Y)^{2}+(y-Y)^{2}\right]}\ \mathrm {d} X\ \mathrm {d} Y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e57a20ecfd71b8c5a01fb9a90c305415f7d21bc)
![{\displaystyle h(x,y,a,b)=-{\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}\ \mathrm {e} ^{\mathrm {i} \,kD_{o}}\ \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}}{\lambda ^{2}D_{i}D_{o}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2}}\left({\frac {1}{D_{i}}}+{\frac {1}{D_{o}}}-{\frac {1}{f}}\right)\left(X^{2}+Y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left(x^{2}+y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}\left(a^{2}+b^{2}\right)}\ \mathrm {e} ^{-\mathrm {i} \,k\left[\left({\frac {a}{D_{o}}}+{\frac {x}{D_{i}}}\right)X+\left({\frac {b}{D_{o}}}+{\frac {y}{D_{i}}}\right)Y\right]}\ \mathrm {d} X\ \mathrm {d} Y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ba4f32446686519ba8f5974cb2163ed501525e8)
![{\displaystyle h(x,y,a,b)=-{\frac {\mathrm {e} ^{\mathrm {i} \,kD_{i}}\ \mathrm {e} ^{\mathrm {i} \,kD_{o}}\ \mathrm {e} ^{\mathrm {i} \,kn\Delta _{0}}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{i}}}\left(x^{2}+y^{2}\right)}\ \mathrm {e} ^{{\frac {\mathrm {i} \,k}{2D_{o}}}{\frac {x^{2}+y^{2}}{\gamma _{t}^{2}}}}}{\lambda ^{2}D_{i}D_{o}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{-\mathrm {i} \,k\left[\left({\frac {-\gamma _{t}a}{D_{i}}}+{\frac {x}{D_{i}}}\right)X+\left({\frac {-\gamma _{t}b}{D_{i}}}+{\frac {y}{D_{i}}}\right)Y\right]}\ \mathrm {d} X\ \mathrm {d} Y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1195da195e9f66345ee473d0d97bece2f64b9de7)
![{\displaystyle h(x-\gamma _{t}a\,,\,y-\gamma _{t}b)=-{\frac {1}{\lambda ^{2}D_{i}D_{o}}}\int _{-\infty }^{\infty }\!\int _{-\infty }^{\infty }t(X,Y)\ \mathrm {e} ^{{\frac {-\mathrm {i} \,k}{D_{i}}}\left[\left(x-\gamma _{t}a\right)X+\left(y-\gamma _{t}b\right)Y\right]}\ \mathrm {d} X\ \mathrm {d} Y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c4c0689f425fc9836e8c77c0302525fa298ce9d)
