Physical Principles of Electron Microscopy: An Introduction to TEM ...

50 Chapter 2
from the lens) for electrons of energy E0 as shown in Fig. 2-14. Because 1/f
increases as the electron energy decreases, electrons of energy E0 ��E0 will
have an image distance v ��v and arrive at the image plane a radial distance
ri from the optic axis. If the angle � of the arriving electrons is small,
ri = �v tan� ���v (2.15)
As in the case of spherical aberration, we need to know the x-displacement
of a second point object P' whose disk of confusion partially overlaps the
first, as shown in Fig. 2-14. As previously, we will take the required
displacement in the image plane to be equal to the disk radius ri , which will
correspond to a displacement in the object plane equal to rc = ri /M, where M
is
the image magnification given by:
M = v/u = tan�/tan��� / � (2.16)
From Eqs. (2.15) and (2.16), we have:
rc ���v/M ���v/M 2
(2.17)
Assuming a thin lens, 1/u + 1/v = 1/f and taking derivatives of this equation
(for a fixed object distance u) gives: 0 + (-2) v -2 �v = (�2) f - -2 �f , leading to:
�v = (v 2 /f 2 ) �f (2.18)
For M >> 1, the thin-lens equation, 1/u +1/(Mu) = 1/f , implies that u � f and
v � , so Eq. (2.18) becomes �v � M 2 Mf
�f and Eq. (4.7) gives:
rc � ��f (2.19)
r c
P
P'
�
x
u v
Figure 2-14. Ray diagram illustrating the change in focus and the disk of confusion resulting
from chromatic aberration. With two object points, the image disks overlap; the Rayleigh
criterion (about 15% reduction in intensity between the current-density maxima) is satisfied
when the separation PP’ in the object plane is given by Eq. (2.20).
��
�v
Q'
Q
r i