2013-12-20
Prof. Herbert Gross
Friedrich Schiller University Jena
Institute of Applied Physics
Albert-Einstein-Str 15
07745 Jena
Exercise
Lecture Imaging and aberration Theory – Part 5
Exercise 5-1: Triplet System
A classical triplet system consists of two positive lenses L1, L3 and one negative lens L2 in the
sequence (+) (-) (+) as shown in the figure.
f1
f3
f2
It is assumed that the lenses are thin and can be considered as close together. The outer positive
lenses are identical and of reversed orientation. They are made of a glass with high index n1 = n3 =
1.8 and the negative lens is made of a low index glass with n2=1.5. Calculate the individual focal
lengths of the lenses, if the total focal length of the system is f' = 100 mm and the field of view is
flattened according to the Petzval theorem.
What can be done to control and correct the spherical aberration of the system (qualitatively)? If the
negative lens is exactly in the middle position between the positive lenses, why can be seen coma
and distortion, although the setup is symmetrical? Why is the spherical aberration contribution of the
identical outer surfaces not equal?
Exercise 5-2: Galilei Telescope
A Galilean telescope consists of an afocal combination of a positive and a negative lens, having back
focal lengths f ' and f ' . The system transforms the diameter of a collimated incoming beam by a
magnification factor  which can be expressed by the focal lengths of the lenses as

f '
.
f '
1/3
2
What is the Petzval condition for a flattened field of view for this system? Derive the formula for the
magnification for a Galilei telescope with flat field! If the available glass materials have refractive
indices in the range n = 1.45...2.0, what is the achievable range of magnification?
Exercise 5-3: Condition of Achromatization
Assume an achromate consisting of two close thin lenses, made of glasses with refractive indices n 1
and n2, having radii of curvature R11, R12 and R21 and R22. Derive the condition of achromatization
without making any approximation for the behavior of the refractive index! Therefore, suppose that the
focal length is corrected for the F and C wavelengths.
Hint: use the lens maker’s equation, later resolve inverse R’s for the focal power. The Abbe number is
defined to be

n 1
.
nF  nC
Exercise 5-4: Achromate and field flattening
Design an achromate with the focal length f = 50 mm with the crown glass PK2 for the positive lens in
front and the flint glass SF6 for the negative lens. The cemented surface should be plane, the
wavelengths should be d (578nm), C (656nm) and F (486nm).
a) Calculate the focal length of both lenses and the radii of curvature of the outer surfaces!
b) Calculate the residual chromatic error (secondary spectrum) as a difference of the image location
between the green and the blue/red. Calculate the primary and secondary chromatic aberration of a
simple lens of K5 for comparison. What is the improvement of the chromatic difference in the
intersection length?
c) Compare the curvature of the image shell of Petzval for the achromate and the single lens. Which
is the system with the better performance? Why is it impossible to flatten the achromate in this setup?
The data of the materials are:
Glas
SF6
K5
PK2
486 nm
1.82775
1.52860
1.52372
587 nm
1.80518
1.52249
1.51821
656 nm
1.79609
1.51982
1.51576

25.4
59.5
65.05
Exercise 5-5: Derivation of the Coddington Equations
Derive the two Coddington equations for the astigmatic image locations (meaning the astigmatic focal
distances for (1) the sagittal and (2) tangential plane)!
Assume an aperture in front of the lens that allows for a circular light cone from a single off-axis object
point at some x=0, y≠0! Here, the tangential plane contains the optical axis and the object itself, the
sagittal plane is perpendicular to the latter and contains the chief ray. For understanding: the sagittal
3
points on the cone, parameterized with some angle α, have α=0° and 180°; the tangential ones have
α=90° and 270°, where the positive x-axis has α=0°.
Hint:
A full vectorial ray trace is to be avoided by virtue of rotational symmetries of the sphere.
Therefore, the sagittal case can be obtained by considering an auxiliary axis through the centre of
curvature of the surface and a subsequent ray trace (see Exercise: Imaging at a spherical surface.
This auxiliary axis may also cross the stop. The new y-axis is to be rotated to the sagittal points.)
Justification for that: the edge points for the sagittal line on the circle, seen from the object, have
equal distances to the auxiliary optical axis. Compare the expressions for the area of the triangle
Object – Interception point – Image, generated by generic angles at the interception point.
The tangential case is treated differently here: perform a ray trace for two rays with infinitesimal
distance in the height at the surface. The aim of that is to characterize the neighbourhood of the chief
ray.