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90
Chapter 2. Geometrical optics
A physical aperture is located at the back focal plane of the first
lens, which coincides with the front focal plane of the second lens.
It is easy to verify from the figure that the Gaussian image plane
coincides with the back focal plane of the second lens. Both lenses
are assumed to be ideal in the equivalent system. Each ray emanating from a point object intersects the aperture plane at a unique
transverse position x A = (x A , y A ). Because the lenses are assumed
to be perfect, every ray has the same optical path length between
object and image. In the real system, the various rays have differing optical path length, owing to the aberrations.
The optical path difference for the primary aberration is given
by (2.223, 2.159)
z I
W 4 =
m 4 dz
(2.252)
z O
for the real system. We assume this to be known for each ray from
the preceding analysis. We now assume that all aberration of the
real system for a particular ray is concentrated in the aperture
plane of the equivalent system, manifest as an optical path difference W 4 from ideal.
In mathematical terms, we wish to find the intensity point spread
function (2.251) in the Gaussian image plane, given the optical
path difference (2.252) for each ray in the real system. For the
equivalent system we can express the momentum element in the
Gaussian image plane in terms of a unique area element in the
aperture plane as
∂P Ix ∂P Iy ∂P Ix ∂P Iy
p I
2
d
2 P I =
−
d
2 x A =
d
2 x A , (2.253)
∂x A ∂y A
∂y A ∂x A
f
where the large parenthesis is the Jacobian determinant, and where
we have made use of (2.196)
P(z) = p(z) [x O g
� (z) + x A h
� (z)]
P I = p I (x O g I
� + x A h
�
I )
(2.254)
in the paraxial approximation, where p(z) is the scalar kinetic momentum on axis. Also, h
�
I = 1/f where f is focal length of the final
�
� �
90
Chapter 2. Geometrical optics
A physical aperture is located at the back focal plane of the first
lens, which coincides with the front focal plane of the second lens.
It is easy to verify from the figure that the Gaussian image plane
coincides with the back focal plane of the second lens. Both lenses
are assumed to be ideal in the equivalent system. Each ray emanating from a point object intersects the aperture plane at a unique
transverse position x A = (x A , y A ). Because the lenses are assumed
to be perfect, every ray has the same optical path length between
object and image. In the real system, the various rays have differing optical path length, owing to the aberrations.
The optical path difference for the primary aberration is given
by (2.223, 2.159)
z I
W 4 =
m 4 dz
(2.252)
z O
for the real system. We assume this to be known for each ray from
the preceding analysis. We now assume that all aberration of the
real system for a particular ray is concentrated in the aperture
plane of the equivalent system, manifest as an optical path difference W 4 from ideal.
In mathematical terms, we wish to find the intensity point spread
function (2.251) in the Gaussian image plane, given the optical
path difference (2.252) for each ray in the real system. For the
equivalent system we can express the momentum element in the
Gaussian image plane in terms of a unique area element in the
aperture plane as
∂P Ix ∂P Iy ∂P Ix ∂P Iy
p I
2
d
2 P I =
−
d
2 x A =
d
2 x A , (2.253)
∂x A ∂y A
∂y A ∂x A
f
where the large parenthesis is the Jacobian determinant, and where
we have made use of (2.196)
P(z) = p(z) [x O g
� (z) + x A h
� (z)]
P I = p I (x O g I
� + x A h
�
I )
(2.254)
in the paraxial approximation, where p(z) is the scalar kinetic momentum on axis. Also, h
�
I = 1/f where f is focal length of the final