2013-01-05
Prof. Herbert Gross
Friedrich Schiller University Jena
Institute of Applied Physics
Albert-Einstein-Str 15
07745 Jena
Exercise Solutions
Lecture Imaging and aberration Theory – Part 6
Exercise 6-1: Tele-photo System
Consider a tele-photo system (focal length: 100 mm). Both lenses be thin and have a distance of
30mm and are composed of glasses with refractive indices n=1.5 for the positive lens and n=1.8 for
the negative lens.
Compute the individual focal lengths of the system under the precondition that the image field is
flattened. Which is the second of the solutions to the ansatz?
Solution:
Condition for focal length:
Petzval condition:
1 1
1
d



f ' f '1 f '2 nf '1 f '2
1
1
1
1
 n'


0
rp
n1 f '1 n2 f '2
k nk  f k
Eliminate f'2 and solve for f'1:
f '1 
( n 2  n1 ) f ' ( n 2  n1 ) 2 f ' 2 4n1 n 2 df '
2n1
 50.83 mm
Inserting that, one obtains f'2 = -42.35 mm.
Die zweite Lösung der quadratischen Gleichung liefert mit den Brennweiten f' 1 = -70.83 mm und f'2 =
+59.01 mm ein Objektiv vom Retrofokustyp.
Exercise 6-2: Spherical Aberration of a Plane Parallel Plate
Derive the axial spherical aberration, which corresponds to the difference between the real and the
paraxial image location of a plane parallel plate with index n and thickness d. Calculate the lowest
order of the aberration as a function of a small value of sin u. If the diameter in the Gaussian image
plane should not be larger than 10 μm, calculate the greatest possible thickness of a plate in this
approximation with refractive index of n = 1.48 for a numerical aperture of sin u = 0.8.
1/7
2
Solution:
i
y1
image
plane
i'
y2
s'
n'
iu
The incidence angle is
If the ray height at the front side of the plate is given by y1, and the ray angle i' inside the plate is
n sin i '  sin i  sin u
sin u
sin i ' 
n
the height at the rear surface is
y2  y1  d  tan i'  y1  d 
sin u / n
1  sin u / n 
2
and the intersection length of the image plane behind the plate is given by
y1  d 
y
s'  2 
tan u
sin u / n
1  sin u / n 
sin u
2
1  sin 2 u
The paraxial approximation is given by
s0' 
y1
d

sin u n
Therefore the difference is the spherical aberration due to the plate
s '  

d 
1  sin 2 u

1
n  1  sin u / n 2 


By Taylor expansion we get the expression
1


1  sin 2 u


d
d  1
1
 

2

s'   
1    1  sin 2 u   1  2 sin 2 u 
1
n 1
n  2
  2n

sin 2 u 

2
 2n

d
1  d  (n 2  1)
 1
   sin 2 u     2  
 sin 2 u
3
n
2
2
n
2
n


The diameter of the lateral blurr spot is given by
D  2  s' sin u 
d  (n 2  1)
 sin 3 u
n3
Therefore we have for the allowed thickness of the plate
3
3
D
 n 
d 2

  0.053 mm
(n  1)  sin u 
Exercise 6-3: Achromate II
Consider an achromate made by a crown glass with index n1 = 1.573 and an Abbe number 1 = 57.4
and a flint glass with the corresponding data n2 = 1.689 and 2 = 31.2. The system should have the
focal length f = 100 mm and the crown lens should be shaped symmetrical biconvex corresponding to
the following figure.
r1
r2=-r1
r3
symmetrical
shape
crown
f1
flint
f2
a) Calculate the focal lengths of both thin lenses and the necessary radii of curvature.
b) Calculate the radii of curvature.
c) Why is it not possible to correct this system for spherical aberration at the boundary?
d) Consider a cemented 2-lens-component without achromatization which is used to focus collimated
light from an object point in infinity on axis. If the component is turned around, which of the primary
aberrations inclusive the two primary chromatic aberrations are changed, which are invariant and
which have the constant value of zero?
Solution:
a) Focal length condition
F  F1  F2 
Achromatic condition
F1
1

F2
2
1
f'
0
gives with the material data for crown
and flint glass
n1 = 1.573
n2 = 1689
the focal powers and focal lengths
F1 
b) crown lens symmetrical
1 = 57.4
2 = 31.2
F
 0.02191
1  2 /  1
F
F2 
 0.01191
1  1 / 2
1
2
 ( n1  1) 
f '1
r1
r2 = -52.30 mm
f'1 = 45.64 mm
f'2 = -83.97 mm
r1  2( n1  1) f '1  52.30 mm
4
flint lens :
resolved for r3 with r2=r1
 1 1
1
 ( n 2  1)  
f '2
 r2 r3 
1
r3 
 545.0 mm
1
1

r2 ( n 2  1) f ' 2
c) Since the crown lens should be symmetric, the degree of freedom of bending the component is not
available. This prevents the correction of spherical aberration.
d)
aberration type
spherical
coma
astigmatism
field curvature
distortion
axial chromatic
lateral chromatic
changed invariant constant = 0
x
x
x
x
x
x
x
explanation
Exercise 6-4: Stationary Phase Point
Consider a converging spherical wave with radius of curvature R = 100 mm. In the pupil plane, the
diameter of the lens is DExP = 20 mm. The wave is observed from a point A on the axis with a distance
z = 90 mm from the pupil.
Calculate the wave aberration as it is seen from the observation point as a function of the radial pupil
coordinate r in the pupil. Calculate the corresponding function in paraxial approximation. Establish in
addition an approximation for paraxial conditions and small values of the defocus z. Evaluate the
wave aberration at the rim of the pupil for all three approaches.
Now consider the case, where an additional contribution of fourth order is added to the wave
aberration with a coefficient a4. Calculate the values of a4 for all approaches in such a manner that in
the observation point A the point at the rim of the pupil radius fulfils the condition of a stationary
phase. What is the physical effect of this perturbation for the brightness of the focussed wave on
axis?
Solution:
r
DExP / 2
z
R
5
Wave aberration general along z-direction:

 

 

W1 (r )  R  1  1  (r / R) 2  z  1  1  (r / z ) 2 .
Paraxial approximation gives

W2 (r )  R  1  1  (r / R) 2  z  1  1  (r / z ) 2
 
 
r 2 
r 2 
 R  1  1  2   z  1  1  2 
  2 R 
  2 z 

r2 r2 r2 z  R
  
2R 2 z 2 z  R
If the value of the defocus has a small amount, we get
W3 (r ) 
r 2 z  R z  r 2


.
2 zR
2R 2
At the rim of the pupil with r = 10 mm we get:
W1 = - 0.0560246 mm
W2 = - 0.0555555 mm
W3 = - 0.05 mm
If now a fourth order contribution is added, the wave aberration reads in the various approximations

 

W1 (r )  R  1  1  (r / R) 2  z  1  1  (r / z ) 2  a4  r 4
W2 (r ) 
r2 z  R

 a4  r 4
2 zR
W3 (r ) 
z  r 2
 a4  r 4
2
2R
To get a point of stationary phase requires the vanishing of the derivative at the corresponding pupil
height of r=10 mm. This delivers
W1
r
r


 4a4  r 3  0
2
2
r R 1  (r / R)
z 1  (r / z )
a4 

1 
1
1




  0.000002825 mm 4
4  r 2  R 1  (r / R) 2 z 1  (r / z ) 2 


W2
Rz
r
 4 a4  r 3  0
r
zR
1 R  z
a4 

 0.000002778 mm3
4  r 2 Rz
6
W3
z
 r  2  4a4  r 3  0
r
R
 1 z
a4 
 2  0.000002500 mm3
2
4r R
A point of stationary phase would improve the brightness on axis considerably.
Exercise 6-4: Relation between Defocus and Zernike coefficient c4.
1
 NA2  z for a slight defocus z of an image and the Zernike
4  n
coefficient c4 = c2,0 for small numerical apertures NA. Use the relation W  c4  Z 4 (rp )  c4  2rp2  1
Derive the relationship c4 

with rp to be the radial pupil coordinate scaled in wave lengths.
Solution
The setup looks as follows:
rp
W
R
DExP / 2
R'
z
We have for a small defocussing two radii of the reference sphere R and R' with
z  R  R '
The sag difference at the edge of the pupil reads
z  R R D
2
2
Exp
/4
2
DExp
8R
, z' 
2
DExp
8R'
The defocus wave aberrations form this geometry
W  z ' z 
2
2
2
DExp
 1 1  DExp R  R' DExp z
   



8  R' R 
8
RR '
8 R2
The numerical aperture is used to eliminate DExp and R by
NA  n  sin u  n 
DExp
2R
and we have
Wedge 
2
2
DExp
z  2 NA  z NA2  z
 2 

 
8 R
2n 2
 n  8

7
On the other hand, the Zernike representation gives with a normalized radial pupil coordinate r p and c4
scaled in wavelengths.


W  c4  Z 4 (rp )  c4  2rp2 1
If the image domain is located in the medium with index n, the wavelength is /n. Therefore in
absolute units we get


W  c4  2rp2 1   / n
The largest difference is obtained between the edge and the center and is
Wedge  W (1)  W (0)  2c4   / n
By equating both expressions we get
NA2  z 2c4  

2n 2
n
2
NA  z
c4 
4n
Wedge 