� �
�
�
�
�
�
�
�
�
�
�
�
�
�
�
and k = 2π/λ is the wave number of the particle. We define a
two-vector position r by
f 2
k
2
r = − K,
d
2 K =
d
2 r.
(3.317)
k
f 2
It follows that
J ˜ (K) =
k
2
d
2 r
� P (r
� ) P (r
� − r).
(3.318)
2πf 2
Geometrically, the integral on the right is the common area of the
pupil with itself displaced by r.
An important special case is a round aperture, for which the pupil
function is
P (r) = 1
(3.319)
for 0 ≤ r ≤ a, and zero for r > a. It follows that
⎡
⎤
⎦
2
2
k
r
r
r
J ˜ (K) =
· 2a
2
⎣ cos
−1
−
1 −
.

2πf 2
2a
2a
2a
(3.320)
This depends only on the magnitude r = |r|, because of the radial
symmetry of the pupil function P (r). Substituting r = f 2 K/k, we
obtain
⎤
⎡
2
2
ka

f 2 K
f 2 K 1 −

f 2 K
⎢
⎣ cos

−1
⎥
⎦
˜
J(K) = 2
−

·

,

2πf 2
2ka
2ka
2ka
(3.321)
where this is a function of the magnitude K. The optical transfer
function is
⎤
⎡
2
2

f 2 K
f 2 K 1 −

f 2 K
⎢
⎣ cos

−1
⎥
⎦
O(K) =

−

.
(3.322)

π
2ka
2ka
2ka

This is unity at K = 0 as required, and zero for
2ka
K c =
,
(3.323)
f 2
221
3.3. Diffraction