Consider the contribution to the field
E
⃗
E
⃗
at a P due to a small element of the slit
ⅆ
y
ⅆ
y
at
y
y
.
It is a distance
r
r
from P.
R
R
is the distance from the center of the slit to P.
lets define
ε
L
ε
L
which is the source strength per unit length, which is a constant.
then
d
E
=
ε
L
r
ⅆ
y
e
i
(
k
r
−
ω
t
)
d
E
=
ε
L
r
ⅆ
y
e
i
(
k
r
−
ω
t
)
Now from the drawing
r
2
=
(
R
−
y
sin
θ
)
2
+
(
y
cos
θ
)
2
=
R
2
+
y
2
sin
2
θ
−
2
R
y
sin
θ
+
y
2
cos
2
θ
=
R
2
+
y
2
−
2
R
y
sin
θ
=
R
2
[
1
−
2
y
R
sin
θ
+
y
2
R
2
]
r
2
=
(
R
−
y
sin
θ
)
2
+
(
y
cos
θ
)
2
=
R
2
+
y
2
sin
2
θ
−
2
R
y
sin
θ
+
y
2
cos
2
θ
=
R
2
+
y
2
−
2
R
y
sin
θ
=
R
2
[
1
−
2
y
R
sin
θ
+
y
2
R
2
]
Now assume that
y
<
<
R
y
<
<
R
(which gives us the Franhaufer condition) and
r
=
R
[
1
−
2
y
R
sin
θ
]
1
2
r
=
R
[
1
−
2
y
R
sin
θ
]
1
2
now expand the square root
r
=
R
[
1
−
y
R
sin
θ
+
…
]
r
=
R
[
1
−
y
R
sin
θ
+
…
]
and neglect higher terms so that
r
=
R
−
y
sin
θ
r
=
R
−
y
sin
θ
thus
d
E
=
ε
L
R
e
i
(
k
(
R
−
y
sin
θ
)
−
ω
t
)
ⅆ
y
d
E
=
ε
L
R
e
i
(
k
(
R
−
y
sin
θ
)
−
ω
t
)
ⅆ
y
where now we have used R in the denominator since it is much bigger than y
d
E
=
ε
L
R
e
i
(
k
R
−
ω
t
)
e
−
i
k
y
sin
θ
ⅆ
y
d
E
=
ε
L
R
e
i
(
k
R
−
ω
t
)
e
−
i
k
y
sin
θ
ⅆ
y
now integrate assuming that
θ
θ
is a constant over the slit
E
=
ε
L
R
e
i
(
k
R
−
ω
t
)
∫
−
a
/
2
a
/
2
e
−
i
k
y
sin
θ
ⅆ
y
=
ε
L
R
e
i
(
k
R
−
ω
t
)
e
−
i
k
y
sin
θ
−
i
k
sin
θ

−
a
/
2
a
/
2
=
ε
L
R
e
i
(
k
R
−
ω
t
)
e
−
i
k
a
2
sin
θ
−
e
i
k
a
2
sin
θ
−
i
k
sin
θ
=
ε
L
R
e
i
(
k
R
−
ω
t
)
−
2
i
sin
(
k
a
2
sin
θ
)
−
i
k
sin
θ
=
ε
L
R
e
i
(
k
R
−
ω
t
)
2
sin
(
k
a
2
sin
θ
)
k
sin
θ
=
ε
L
a
R
e
i
(
k
R
−
ω
t
)
sin
(
k
a
2
sin
θ
)
k
a
2
sin
θ
E
=
ε
L
R
e
i
(
k
R
−
ω
t
)
∫
−
a
/
2
a
/
2
e
−
i
k
y
sin
θ
ⅆ
y
=
ε
L
R
e
i
(
k
R
−
ω
t
)
e
−
i
k
y
sin
θ
−
i
k
sin
θ

−
a
/
2
a
/
2
=
ε
L
R
e
i
(
k
R
−
ω
t
)
e
−
i
k
a
2
sin
θ
−
e
i
k
a
2
sin
θ
−
i
k
sin
θ
=
ε
L
R
e
i
(
k
R
−
ω
t
)
−
2
i
sin
(
k
a
2
sin
θ
)
−
i
k
sin
θ
=
ε
L
R
e
i
(
k
R
−
ω
t
)
2
sin
(
k
a
2
sin
θ
)
k
sin
θ
=
ε
L
a
R
e
i
(
k
R
−
ω
t
)
sin
(
k
a
2
sin
θ
)
k
a
2
sin
θ
now we define
β
=
k
a
2
sin
θ
β
=
k
a
2
sin
θ
and see that we can rewrite our expression as
E
=
ε
L
a
R
sin
β
β
e
i
(
k
R
−
ω
t
)
E
=
ε
L
a
R
sin
β
β
e
i
(
k
R
−
ω
t
)
or equivalently
E
=
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
ω
t
)
E
=
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
ω
t
)
The intensity will go like the square of this so
I
=
I
0
s
i
n
c
2
β
I
=
I
0
s
i
n
c
2
β
Plot of
sin
2
β
β
2
sin
2
β
β
2
The Intensity has a maximum at
β
=
0
β
=
0
or
θ
=
0
θ
=
0
.
there are minima when
sin
β
=
0
sin
β
=
0
or
β
=
k
a
2
sin
θ
=
n
π
β
=
k
a
2
sin
θ
=
n
π
2
π
λ
a
2
sin
θ
=
n
π
2
π
λ
a
2
sin
θ
=
n
π
sin
θ
=
n
λ
a
sin
θ
=
n
λ
a
in the case of small
θ
θ
we see that
Δ
θ
=
λ
a
Δ
θ
=
λ
a
is the distance between adjacent minima.
As
a
a
becomes large, we see that the minima will merge together. This is consistent
with what we said at the beginning, that if
a
>
>
λ
a
>
>
λ
then you just get shadowing but not diffraction.
Finding the secondary maxima is more difficult. (Take the derivative of I and
then look for zeros.) This can not be done analytically.
Note that wee have been considering only one dimension. If the length of the
slit is
L
L
then we have only considered the case that
L
>
>
λ
L
>
>
λ
and so diffraction occurs only in the other dimension.
"This book covers second year Physics at Rice University."