Now we consider the case of two slit
diffraction.

Notice
that the x axis has been drawn through the lower slit. Then the field at the
distant point is just the sum of the field from the two slits. Thus we can use
our solution to single slit diffraction for each slit and add them
together

E
=
ε
L
a
R
1
s
i
n
c
β
e
i
(
k
R
1
−
ω
t
)
+
ε
L
a
R
2
s
i
n
c
β
e
i
(
k
R
2
−
ω
t
)
E
=
ε
L
a
R
1
s
i
n
c
β
e
i
(
k
R
1
−
ω
t
)
+
ε
L
a
R
2
s
i
n
c
β
e
i
(
k
R
2
−
ω
t
)
Now we we will define
R
=
R
1
R
=
R
1
and use
R
2
=
R
−
d
sin
θ
R
2
=
R
−
d
sin
θ
E
=
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
ω
t
)
+
ε
L
a
R
−
d
sin
θ
s
i
n
c
β
e
i
(
k
R
−
k
d
sin
θ
−
ω
t
)
E
=
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
ω
t
)
+
ε
L
a
R
−
d
sin
θ
s
i
n
c
β
e
i
(
k
R
−
k
d
sin
θ
−
ω
t
)
Now we can ignore the
d
sin
θ
d
sin
θ
in the denominator, as it will not have a significant effect on that. However
in the exponent, we can not ignore it, since it could significantly affect the
phase of the harmonic function. Lets define
α
=
(
k
d
sin
θ
)
/
2
α
=
(
k
d
sin
θ
)
/
2
so now we can write:
E
=
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
ω
t
)
+
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
2
α
−
ω
t
)
E
=
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
ω
t
)
+
ε
L
a
R
s
i
n
c
β
e
i
(
k
R
−
2
α
−
ω
t
)
and start rearranging:
E
=
ε
L
a
R
s
i
n
c
β
[
2
cos
α
]
e
i
(
k
R
−
α
−
ω
t
)
E
=
ε
L
a
R
s
i
n
c
β
[
2
cos
α
]
e
i
(
k
R
−
α
−
ω
t
)
This is very similar to the case of single slit diffraction except that you
now get a factor
2
cos
α
2
cos
α
included and a phase shift in the harmonic function.

So we can see immediately the intensity
is
I
=
4
I
0
cos
2
α
s
i
n
c
2
β
I
=
4
I
0
cos
2
α
s
i
n
c
2
β
recall
α
=
(
k
d
sin
θ
)
/
2
α
=
(
k
d
sin
θ
)
/
2
and
β
=
(
k
a
sin
θ
)
/
2
β
=
(
k
a
sin
θ
)
/
2
If
d
d
goes to 0 then expression just becomes the expression for single slit
diffraction. If a goes to 0 then the expression just becomes that for Youngs
double slit. The double slit diffraction is just the product of these two
results. (Hey cool!)

Comments:"This book covers second year Physics at Rice University."