Consider an array of rectangular slits of widths a and b with spacing d in the
y direction, illuminated by a plane wave. We want to derive the Fraunhofer
diffracted field
E
(
k
x
,
k
y
)
E
(
k
x
,
k
y
)
.
The simplest way to do this problem is to use Fourier optics: The array is a
convolution of a single rectangular slit with an array of Dirac delta
functions.
That is the aperture function can be written
A
=
A
0
(
f
⊗
h
)
A
=
A
0
(
f
⊗
h
)
where the symbol
⊗
⊗
represents convolution and
f
=
1
w
h
e
n

x

≤
a
/
2
a
n
d

y

≤
b
/
2
=
0
o
t
h
e
r
w
i
s
e
f
=
1
w
h
e
n

x

≤
a
/
2
a
n
d

y

≤
b
/
2
=
0
o
t
h
e
r
w
i
s
e
h
=
∑
n
y
=
0
N
−
1
δ
(
y
−
n
y
d
)
.
h
=
∑
n
y
=
0
N
−
1
δ
(
y
−
n
y
d
)
.
We now have that
E
=
ɛ
A
R
e
−
i
(
k
R
−
ω
t
)
Ϝ
{
f
}
Ϝ
{
h
}
E
=
ɛ
A
R
e
−
i
(
k
R
−
ω
t
)
Ϝ
{
f
}
Ϝ
{
h
}
where
R
R
is the distance from the origin to the point of measurement. and
F
{
}
F
{
}
represents a Fourier transform. Now we have
Ϝ
{
f
}
=
∫
−
b
/
2
b
/
2
∫
−
a
/
2
a
/
2
e
i
(
k
x
x
′
+
k
y
y
′
)
ⅆ
x
′
ⅆ
y
′
Ϝ
{
f
}
=
∫
−
b
/
2
b
/
2
∫
−
a
/
2
a
/
2
e
i
(
k
x
x
′
+
k
y
y
′
)
ⅆ
x
′
ⅆ
y
′
or rearranging,
Ϝ
{
f
}
=
∫
−
b
/
2
b
/
2
ⅆ
y
′
e
i
k
y
y
′
∫
−
a
/
2
a
/
2
ⅆ
x
′
e
i
k
x
x
′
.
Ϝ
{
f
}
=
∫
−
b
/
2
b
/
2
ⅆ
y
′
e
i
k
y
y
′
∫
−
a
/
2
a
/
2
ⅆ
x
′
e
i
k
x
x
′
.
These are integrals we have done a number of times already:
Ϝ
{
f
}
=
b
s
i
n
c
(
k
y
b
/
2
)
a
s
i
n
c
(
k
x
a
/
2
)
Ϝ
{
f
}
=
b
s
i
n
c
(
k
y
b
/
2
)
a
s
i
n
c
(
k
x
a
/
2
)
Next we transform the sums of the delta functions
Ϝ
{
h
}
=
∑
n
y
=
0
N
−
1
∫
−
∞
∞
e
i
k
y
y
δ
(
y
−
n
y
d
)
ⅆ
y
,
Ϝ
{
h
}
=
∑
n
y
=
0
N
−
1
∫
−
∞
∞
e
i
k
y
y
δ
(
y
−
n
y
d
)
ⅆ
y
,
which upon integration becomes
Ϝ
{
h
}
=
∑
n
y
=
0
N
−
1
e
i
k
y
n
y
d
.
Ϝ
{
h
}
=
∑
n
y
=
0
N
−
1
e
i
k
y
n
y
d
.
We have also performed this sum before and it gives
F
{
h
}
=
e
i
k
y
(
N
−
1
)
d
/
2
sin
(
N
k
y
d
/
2
)
sin
(
k
y
d
/
2
)
.
F
{
h
}
=
e
i
k
y
(
N
−
1
)
d
/
2
sin
(
N
k
y
d
/
2
)
sin
(
k
y
d
/
2
)
.
So we have
E
=
ɛ
A
R
e
−
i
(
k
R
−
ω
t
)
e
i
k
y
(
N
−
1
)
d
/
2
sin
(
N
k
y
d
/
2
)
sin
(
k
y
d
/
2
)
b
s
i
n
c
(
k
y
b
/
2
)
a
s
i
n
c
(
k
x
a
/
2
)
E
=
ɛ
A
R
e
−
i
(
k
R
−
ω
t
)
e
i
k
y
(
N
−
1
)
d
/
2
sin
(
N
k
y
d
/
2
)
sin
(
k
y
d
/
2
)
b
s
i
n
c
(
k
y
b
/
2
)
a
s
i
n
c
(
k
x
a
/
2
)
or if we define
k
R
c
=
k
R
−
(
N
−
1
)
d
k
y
/
2
k
R
c
=
k
R
−
(
N
−
1
)
d
k
y
/
2
E
=
ɛ
A
a
b
R
e
−
i
(
k
R
c
−
ω
t
)
sin
(
N
k
y
d
/
2
)
sin
(
k
y
d
/
2
)
s
i
n
c
(
k
y
b
/
2
)
s
i
n
c
(
k
x
a
/
2
)
.
E
=
ɛ
A
a
b
R
e
−
i
(
k
R
c
−
ω
t
)
sin
(
N
k
y
d
/
2
)
sin
(
k
y
d
/
2
)
s
i
n
c
(
k
y
b
/
2
)
s
i
n
c
(
k
x
a
/
2
)
.
"This book covers second year Physics at Rice University."