Diffraction
S.M.Lea
Fall 1998
Diffraction occurs when EM waves approach an aperture (or an obstacle) with dimension
d > ¸: We shall refer to the region containing the source of the waves as region I and the
region containing the diffracted ®elds as region II. The total volume comprises both regions
I and II. Region II is bounded by a surface S = S1 + S2 : The surface S2 is at |in®nity},
while S1 is a screen containing apertures, or else a set of obstacles.
1 Kirchhoffzs method
We may write all the ®elds in the form
à / e¡i!t
and then à satis®es the Helmholtz equation
¡ 2
¢
r + k2 Ã = 0
1
~ or B,
~ or another variable describing the ®eld,
where k = !=c and à is a component of E
~ We can solve the Helmholtz equation with the outgoing wave
such as a component of A:
Greenzs function. Since there are no sources in region II, the volume integral is zero.
Z h
i
~ 0 G ¡ G^
~ 0 Ã (~x 0 ) da0
à (~x) =
à (~x 0 ) ^
n¢r
n¢r
(1)
S
where
1 eikR
4¼ R
·
µ
¶
¸
0
eikR
i
0
0
~
^
à (~x ) ik 1 +
n
^ ¢R +^
n ¢ r à (~x ) da0
R
kR
G=
Thus
Z
1
à (~x ) = ¡
4¼
S 1+S 2
(2)
The integral reduces to an integral over S1 ; since à » 1=R at 1:
We can evaluate the integral using the Kirchhoff approximation:
1. Ã and
@Ã
@n
vanish on S1 except in the openings, and
2. in the openings, Ã and
@Ã
@n
are equal to the values in the incident ®eld.
Strictly speaking, this is not a mathematically sound procedure, since if both à and @Ã
@n
vanish on any ®nite surface, then the solution à must be zero. Also the |solution} we obtain
does not yield the assumed values of à on S1 : Yet the solution does a pretty good job of
reproducing experimental results.
We can partially ®x things up by using a better Greenzs function. If à is known (or
approximated) on S1 ; we should use a Dirichlet Greenzs function, for which G D (~x; ~x 0) = 0
for ~x0 on S1 : Then the solution is:
Z
~ 0 GD da0
à (~x ) =
à (~x 0) ^
n¢r
(3)
S1
@Ã
@n
Similarly, if
is known or approximated on S1 ; we should use the Neumann Greenzs
N
function, @G
=
0 on S1 : Then
@n
Z
~ 0 Ã (~x0 ) da0
à (~x) = ¡
G^
n ¢r
(4)
S1
If the screen S1 is plane we can easily ®nd the appropriate Greenzs functions using the
method of images:
Ã
!
0
1
eikR
eikR
GD;N =
¨
4¼
R
R0
~ = ~x ¡ ~x 0 and R
~ 0 = ~x ¡ ~x 00 and ~x 00 describes the point that is the mirror image of ~x 0
where R
in S 1; and we let both points approach the surface S1 : Then
GD ! 0 as z 0 ! 0
and
@G N
@n0
=
!
·
¸µ
¶
@G N
1 ¡ (z ¡ z0 )
(z + z 0)
1 eikR
=
+
ik ¡
@z 0
4¼
R
R
R
R
µ
¶
2z 0 eikR 1
¡ ik as z 0 ! 0
4¼R R
R
2
and thus equation (3) becomes:
à (~x)
=
=
Z
¶
1
¡ ik da0
R
S1
Z
µ
¶
ikR
1
1
0 z e
à (~x )
¡ ik da0
2¼ S1
R R
R
2z eikR
4¼R R
à (~x0 )
µ
with z=R = cos µ; while equation (4) becomes:
Z
1
e ikR @
à (~x) = ¡
à (~x 0 ) da0
2¼ S 1 R @z 0
(5)
(6)
Now to estimate the differences between these expressions, we choose a point source in
region I with coordinate ~r0 so that the wave from the source travels in direction ¡^
r 0 to
reach the center of a small aperture at the origin. Then we can write
0
0
~ ~
eik¢R
e ¡ik (r0+^r0¢~x )
e¡ikr0 e¡ik^r0¢~x
à inc = à 0
= Ã0
= Ã0
R
r0
r0
and
0
@Ã inc
e¡ikr0 e¡ik^r0¢~x
= Ã0
(¡ik~r0 ¢ n
^)
@n
r0
where ~r0 ¢ ^
n = cos (¼ ¡ µ 0 ) = ¡ cos µ 0 : Then the 3 expressions for à may be written:
outgoing wave Greens function:
·
µ
¶
¸
Z
1
eikR
i
@
0
0
^
à (~x ) = ¡
à (~x ) ik 1 +
n ¢ R + 0 Ã (~x ) da0
^
4¼ S 1+S 2 R
kR
@z
Z
i
ikr ¡ikr0 ¡ik^
r0¢~
x0 h
Ã
e e
e
^ + (¡ik~r0 ¢ n
' ¡ 0
ik^
n ¢R
^ ) da0
4¼ S1 +S2 r
r0
Z
ikr ¡ikr0
Ã
e e
' ¡ik 0
[cos µ 0 + cos µ] da0
4¼ S1+ S2 r
r0
Dirichlet Greenzs function:
ik
Ã
2¼ 0
Z
eikr e¡ ikr0
cos µda0
r
r0
ik
à (~x) = ¡ à 0
2¼
Z
eikr e¡ ikr0
cos µ0 da0
r
r0
à (~x) = ¡
S1
and
Neumann Greenzs function:
S1
If the aperture is small compared with the distances r and r0 ; then the factors cosµ and
cos µ0 are essentially constant throughout the range of integration. Then all 3 expressions
give the same diffraction pattern: the overall intensity smplitude differs by small factors of
order 1. Thus we can use whichever of the expressions is most convenient.
3
2
Fraunhoffer and Fresnel diffraction
Thee exponential that appears in equation (2) may be approximated:
p
kR = k j~x ¡ ~x 0 j = k r2 + r 02 ¡ 2~x ¢ ~x0
r
r02
r ¢ ~x 0
^
= kr 1 + 2 ¡ 2
r
r
³
´
k
2
= kr ¡ k^r ¢ ~x0 ¡
r 02 ¡ (^
r ¢ ~x 0) + ¢ ¢ ¢
2r
Thus
µ
´¶
k ³ 02
ikR
ikr ¡ik^
r¢~
x0
0 2
e
=e e
exp ¡i
r ¡ (^r ¢ ~x )
2r
The the exponent in the second term is of order:
d
k^r ¢ ~x0 =
¸
where d is the dimension of the aperture, and its ratio to the ®rst term is of order d=r: . The
exponent in the third term is of order
d2
¸r
In Fraunhauffer diffraction we ignore the third term. This means that we need
d2
¿1
¸r
and
third term
d
= ¿1
second term
r
So we need the observation point to be a long way from the aperture.
If d=¸ > 1; the third term may become important. This is the regime of Fresnel
diffraction. Fresnel diffraction occurs for
d
d &rÀd
¸
Typically we ®nd that Fresnel diffraction patterns are more complex than Fraunhoffer
patterns. Most of the problems in the text refer to the Fraunhoffer regime.
3 Vectorial diffraction theory
A more careful analysis requires that we consider the vector nature of the ®elds. Thus
we replace equation (2) with the vector relation:
Z h ³
´
³
´ i
0
0
~ (~x ) =
~ ^
~ G ¡G ^
~ E
~ da0
E
E
n0 ¢ r
n0 ¢ r
S
First we rewrite the integrand as:
³
´
³
´³
´
~ n
~0 G¡ ^
~ 0 GE
~
2E
^0 ¢ r
n0 ¢ r
4
And then we convert the surface integral of the second term to a volume integral. Recall that
the normal n
^ 0 is inward:
Z
Z
³
´³
´
³
´
~ 0 GE
~ da0 =
~ dV 0
¡ n
^0 ¢ r
r2 G E
s
ZV h ³
´
³
´i
~ r
~ ¢ GE
~ ¡r
~ £ r
~ £ GE
~ dV
=
r
V
Z h ³
´
³
´i
~ ¢ GE
~ ¡^
~ £ GE
~ da0
= ¡
n
^0 r
n0 £ r
S
where
~ ¢ GE
~ = Gr
~ ¢E
~+E
~ ¢ rG
~ = 0+E
~ ¢ rG
~
r
since there is no charge density on S;and
~ £ GE
~ = rG
~ £E
~ + Gr
~ £E
~
r
³
´
~ £E
~ +G ikB
~
= rG
Thus we have:
Z h ³
´
³
´
³ 0
³
´´i
~ (~x) =
~ n
~ 0 G ¡ ^n0 E
~ ¢r
~ 0G + n
~ G£E
~ +G ikB
~
E
2E
^0 ¢ r
^0 £ r
da0
ZS h ³
´
³
´
³
´
³
´
i
~ n
~ 0 G ¡ ^n0 E
~ ¢r
~ 0G + r
~ 0G n
~ ¡E
~ ^
~ 0 G + Gik^n0 £ B
~ da0
=
2E
^0 ¢ r
^0 ¢ E
n0 ¢ r
ZS h ³
´
³
´
³
´
i
~ ^
~0 G ¡n
~ ¢r
~ 0G + r
~ 0G ^
~ + ik^
~
=
E
n0 ¢ r
^0 E
n0 ¢ E
n0 £ BG
da0
S
Z h³
´
³
´ 0
i
~ £r
~ 0G + n
~ r
~ G + ik^
~
=
n0 £ E
^
^0 ¢ E
n 0 £ BG
da0
S
Now we can take G = eikR =R and proceed as with the scalar theory. We can make the same
Kirchhoff approximations, and the same inconsistencies remain. For r0 very large, we have
0
0
eikr ¡ik^r0 ¢~x
eikr
G=
e
'
r0
r0
Now if we choose S2 to be a large hemisphere of radius r 0 ! 1; then ^r0 = ¡^
n 0; and
~ 0 G = ¡ik^
r
n0 G
and the integral over S2 becomes:
Z h³
´
³
´
i
~ £^
~ ^
~
¡ik
n0 £ E
^
n0 G + n
^0 ¢ E
n0 G¡^
n0 £ BG
da0
ZS 2 h
i
~ ¡^
~ Gda0 = 0
= ¡ik
E
n0 £ B
S2
since
for outgoing waves, and thus:
~ = ikB
~
i~k £ E
~ = B=
~ ¡^
~
^r0 £ E
n0 £ E
5
³
´
~ = ¡^
~ =E
~
^0 £ B
n
n0 £ n
^0 £ E
and so
on S2 : Thus the integral reduces to an integral over S1 :
Now if our observation point P is very ³
distant from
´ the sources on S 1;then as in the
0
eikR
eikr
~¢~
scalar theory we ®nd G =
'
exp ¡ik
x and we expect the diffracted ®elds to
4¼R
4¼ r
have a similar form
³
´
ikr
~ d = e F~ ~k; ~k0
E
r
Then the amplitude of the diffracted ®eld may be written:
Z h³
³
´
´ ³ ´ ³
´³ ´
i
i
~
~
~
~ £ ¡~k + ^
~
~ da0
F k; k0
=
n
^0 £ E
n0 ¢ E
¡~k +k^
n0 £ B
4¼ S
Z h
³
´
³
´
i
i
~k £ ^
~ ¡ ~k ^
~ + k^
~ da0
=
n0 £ E
n0 ¢ E
n0 £ B
4¼ S
where the ®elds in the integral are also the diffracted ®elds. We know that F~ must be
perpendicular to ~k: In fact we can write the last two terms in the integrand, like the ®rst, in
the form ~k £ ~v for some vector ~v : First note that
h
³
´i
h ³
´i
³
´
~k £ ~k £ ^
~
~
~
n0 £ B
= ~k ~k ¢ ^
n0 £ B
¡ k2 ^
n0 £ B
and from Amperezs law:
giving:
³
´
³
´
~k ¢ n
~ = ¡^
~ =n
~
^0 £ B
n 0 ¢ ~k £ B
^ 0 ¢ kE
³
´
~
k n
^0 £ B
h ³
´i
h
³
´i
^ ~k ¢ ^
~
^ ~k £ ^
~
= k
n0 £ B
¡ k£
n0 £ B
³
´
h
³
´i
~ ¡ k£
^ ~k £ ^
~
= ~k ^
n0 ¢ E
n0 £ B
so the terms in the integrand are:
³
´
h
³
´i
~ ¡ ~k n
~ = ¡ 1 ~k £ ~k £ n
~
k^
n0 £ B
^0 ¢ E
^0 £ B
k
and thus
Z h
³
´
³
´
i
¡i ~
~k £ n
~ ¡^
~ da0
F~ ~k; ~k0 =
k£
^0 £ B
n0 £ E
4¼k
S1
The integrand also depends on ~k0 if we use the incident ®elds in the apertures.
4 Special case of plane conducting screen
Let the conducting screen be the z = 0 plane, with the incident ®elds propagating
toward the screen from negative z: We can decompose the total electric ®eld into three parts:
~ 0 ; the ®eld due to re-ection by a solid conducting sheet, E
~ r ; and the
the incident ®eld E
~
diffracted ®eld E d : Then
~ =E
~0 + E
~r + E
~d = E
~0 + E
~1
E
6
~0 + E
~ r = 0:
Then in Region II (z > 0) E
~r + E
~d = E
~ 1 are produced by currents in the conducting sheet:
The ®elds E
~j = jx ^
x+j y ^
y
The resulting vector potential is given by:
¡ 2
¢
~ = 4¼ ~j
r + k2 A
c
and consequently Az = 0: Also, since the operator on the LHS is even in z; and the right
~ is even in z: Then
hand side is zero except for z = 0; then A
³
´
@Ax
~ £A
~
By = r
=
y
@z
is odd in z: Similarly,
@Ax
Ex = ¡
@t
is even in z: Thus we have
E x ; E y ; and Bz are even in z
and
Ez ; Bx ; and By are odd in z
The ®elds that are odd in z need not be zero at z = 0 where the conducting screen exists,
because of the surface charge density and currents in the screen. But in the apertures, these
~ tangential and E normal are the incident ®elds
odd ®elds must be zero. Thus in the apertures, B
alone.
~ is a Neumann problem, because
Now the mathematical problem that determines A
~
@A
~ ta ngential is determined (through Amperezs law) by the currents in the screen. Thus
=B
@z
we must use the Neumann Greenzs function:
Z
~
1
e ikR @ A
~
A= ¡
da0
2¼ screen R @z 0
Z
Z
1
eikR @Ax 0
1
eikR
Ax = ¡
da
=
¡
By da0
2¼ screen R @z 0
2¼ screen R
Z
´
1
eikR ³
~ 1 da0
=
z£B
^
2¼ screen R
x
A similar expression holds for Ay ; and thus
Z
´
e ikR ³
~1 = r
~ £A
~=r
~ £ 1
~ 1 da0
B
^z £ B
(7)
2¼ screen R
~ 1 = 0 in the aperture, the integral is over the conducting material only. We can
Since ^z £ B
evaluate the integral on either side of the screen. For z < 0; the normal becomes ¡^z instead
~ tan :
of ^
z; thus con®rming our expectations about the oddness of B
~ ! E;
~
Now the source-free Maxwell equations are invariant under the transformation B
7
~ ! ¡B;
~ so we can write a similar expression for E
~1 :
E
Z
´
eikR ³
~1 = r
~ £ 1
~ 1 da0
E
z£E
^
2¼ scr een R
Now if we want the electric ®eld for z < 0; we must change the sign in front of the integral
as well as the sign in front of ^
z in order to obtain the correct symmetry. Since E 1z is zero in
the aperture, we cannot simplify this integral as it stands. However:
Z
³
´´
eikR ³
~1 = r
~ £ 1
~1 + E
~0 ¡ E
~ 0 da0
E
z£ E
^
2¼ screen R
Z
Z
´
´
eikR ³
eikR ³
~ £ 1
~ da0 ¡ r
~ £ 1
~ 0 da0
= r
z£E
^
z£E
^
2¼ screen R
2¼ scr een R
~
where E is the total electric ®eld, whose tangential component vanishes on the conducting
surface. Thus the ®rst integral is an integral over the holes alone. The second term
Z
´
1
eikR ³
~
~ 0 da0 = E
~ 0 for z > 0
r£
z£E
^
2¼ scr een R
Thus
Z
´
e ikR ³
~ =E
~1 + E
~0 = E
~d = r
~ £ 1
~ da0
E
^z £ E
(8)
2¼ holes R
where the ®eld in the integrand is the total electric ®eld in the aperture. In practical
applications we can approximate by using the incident ®elds as we have indicated previously.
Equation (8) is the vector Smythe-Kirchoff relation.
5 Babinetzs principle
Babinetzs principle relates the diffraction patterns due to 2 complementary screens S and
Sc . The complementary screen Sc has apertures where S is solid and vice-versa. Thus the
sum S + Sc is a solid surface.
5.1
Scalar principle
For any complete closed surface we can write a mathematical identity for any scalar function
à : equation (1) or, using the outgoing wave Greenzs function, equation (2) Thus:
à (~x)
·
µ
¶
¸
0
eikR
i
0
0
^
~
à (~x ) ik 1 +
n
^ ¢R +n
^ ¢ r à (~x ) da0
kR
S+S c +S 2 R
Z
·
µ
¶
¸
0
1
e ikR
i
0
0
^
~
= ¡
à (~x ) ik 1 +
n
^¢R+n
^ ¢r à (~x ) da0
4¼ S+S c R
kR
1
= ¡
4¼
Z
where as usual, the integral over S2 vanishes. This result is exact. Now, using the Kirchhoff
approximation, the integral over S c (the apertures in S) gives the diffracted ®eld à a in
region II due to the screen S: Similarly the integral over S gives the diffracted ®eld à b due
8
to the complementary screen Sc : Thus
à = à a + Ãb
Now in directions where à = 0;we have à a = ¡Ã b ; and thus the diffraction pattern, which
depends on à 2 ; is the same. Thus for example the diffraction pattern due to a hole of radius
a and due to a disk of radius a are the same except }straight ahead}, where the incident ®eld
is non-zero.
5.2
Vector principle
A more careful statement can be made for a plane conducting screen. First we de®ne the
two situations that are complementary:
~ 0; B
~ 0 and the screen is S0 :
1. The incident ®elds are E
~c = B
~ 0; B
~ c = ¡E
~ 0 ; and the complementary screen Sc :
2. The incident ®elds are E
Then we can use equation (8) to ®nd the diffracted ®eld in problem 1:
Z
´
1
eikR ³
~
~
~ 1 da0
E1 = r £
z£E
^
2¼ holes in S0 R
For problem 2, we use equation (7):
Z
´
e ikR ³
~2 = r
~ £ 1
~ 2 da0
B
^z £ B
2¼ conducting part of Sc R
These two integrals are taken over the same surface. Since these two expressions are
mathematically identical, then
~1 = B
~2
E
~
in region II. However the expression for B2 gives the ®eld due to the currents in the sheet.
The total magnetic ®eld in region 2 is
~ 2,tot = B
~c + B
~ 2 = ¡E
~0 + E
~ 1:
B
¯ ¯2
¯ ¯2
¯~ ¯
¯~ ¯
The intensity of the radiation is given by ¯E
¯ or equivalently ¯B
¯ : Thus the diffraction
~ 0 is zero.
pattern is the same for both screens wherever E
A similar analysis shows that
Z
´
1
eikR ³
~
~
~ 2 da0 = ¡ B
~1
E2 = r £
z£E
^
2¼ holes in Sc R
~ 1 is the ®eld due to the sheet alone, and the minus sign appears so that
where here again B
both waves are outgoing. Thus
~ 1,tot = B
~0 + B
~1 = B
~0 ¡ E
~2
B
~0 ¡ E
~2
= ^k0 £ E
which leads to the same prediction with regard to the diffraction pattern.
Babinetzs principle is used in the design of microwave antennae. A slot in a conducting
plate radiates the same pattern as a metal strip. Thus slots in wave-guides can serve as
9
effective microwave radiatiors.
10