2013-12-06
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 4
Exercise 4-1: Spherical Mirror
Derive the expression for the focal length of a spherical mirror in the paraxial limit and also
for finite height y of a collimated incoming beam (parallel to z axis). Calculate the formula of
the axial spherical aberration for this collimated incoming beam through third order in y!
Hint: use the reflection law to eliminate the appearing angles in favour of y and the radius of
curvature r!
Solution:
P
i
y
i'
r/2
u'
i
F
C
Q
a
s'
r
My solution:
i'  i ,
u '  2i,
y  a  tan u '  r  sin i.
Further holds ( z p being the surface sag)
s '  a  zp ,
i  arcsin( y / r ).
Rearranging the equation for a, namely
a
r sin i
y
r 2  2 y2


tan u ' tan(2*arcsin( y / r )) 2 r 2  y 2
Adding the surface sag
z p  r  r 2  y2
and subtracting that from the paraxial focal distance r/2, we get
1 
r
s ' r / 2  r 1 
2 
r 2  y2

y2
    O  y3  .

4r

Another solution by Prof. Gross:
According to the figure, we have the relations
i '  i , u '  2i
y  a  tan u '  r  sin i
in paraxial approximation we have
a tan u  a tan 2i  a sin 2i  2a sin i  r sin i
2a sin i  r sin i


f  s'  a  zP  r / 2  r  r 2  y 2 
r
2
With the geometric relations
sin i 
y
r/2
, cos i 
r
r  s'
we get
sin 2 i  cos 2 i  1
y2
r2 / 4

1
r 2 (r  s ' ) 2
r
s'  r 
2 1  ( y / r )2
The spherical aberration is given by
r 
1
s'  s'r / 2   1 

2 
1  ( y / r )2




In 3rd order the spherical aberrations is given by the easy expression

r 
1
y2


s'   1 

2  1  0.5 y 2 / r 2 
4r
Exercise 4-2: Spherical Aberration of a single lens
A thin plane-convex lens with focal length f' = 100 mm is used to focus a collimated beam
with diameter D = 10 mm and wavelength  = 1.06 m.
Draw a sketch of the lens in the optimal setup and explain why this orientation is
advantageous! Calculate the magnification parameter M and the bending parameter X of the
lens. The Quantity M is defined to be
M :
and
X :
uu'
,
u u'
r2  r1
,
r2  r1
where u and u’ are the angles of the ray with respect to the optical axis. The surface
contribution of the primary spherical wave aberration of a single lens can be expressed by
the formula
2
 3

2

2 n2  1
1
n
n2

 n  n  1 2 

As 

X

M

M



32n( n  1) f 3  n  1 n  1 
n2
n2





Calculate this coefficient for the refractive indices n = 1.4 and n = 2.0! Which choice of
refractive index is more advantageous? Also, calculate the transverse aberration y’ in the
image plane in the one-dimensional case with the pupil coordinate yp by using this formula in
the form WSPH y p  As  y 4p for both indices.


 
Compare the geometrical spot size and the diffraction Airy diameter!
Is the system diffraction limited?
Solution:
Sketch of the setup:
D
f'
In this orientation, the bending of the rays is distributed on two surfaces. This gives smaller
incidence angles and a smaller impact of the non-linearity of the law of refraction, which
generates aberrations.
The bending parameter is:
the magnification parameter is:
X = +1 (-1 in Kingslake!!!)
M = +1
By inserting the data for f, n, X and M we get the coefficients:
1. index n = 1.4 :
As = -4.598 10-7 mm-3
2. index n = 2.0 :
As = -1.25 10-7 mm-3
The transverse aberration is related by with the wave aberration by
y'   R 
W
  f '4 y 3p  As
 yp
with reference sphere radius R = f'. This gives for the two cases the geometrical spot sizes
n = 1.4
DGeo  2y'  46m
n = 2.0
DGeo  2y'  12.5m
The Airy diameter is given by
DAiry 
1.22   2.44    f '

 25.9m
sin u '
 in
Therefore, in the case of the higher index n=2, the system is diffraction limited, the smaller
index delivers a geometrical dominated broadening of the spot size.
Exercise 4-3: Thin Lens Formula for Spherical Aberration
The equation for the contribution of the spherical aberration of third order of a single
dielectric surface reads
n' (n'n)  1 1   n' n' n 
AS  
    

8n 2
s' 
 R s'   R
2
Use this formula to derive the corresponding formula for a single thin lens from Exercise 4-2:
( lens)
S
A
2
 n3

 n 2 (n  1)
1
n2 
2(n 2  1)
2








X


M


M

32n(n  1) f '3  n  1 n  1 
n2
n2



(Hint: elimination with the help of imaging and lens maker equation)
Solution:
From the expressions of the formula for the front surface
( lens)
S1
A
n(n  1)  1 1 
  

8  R1 s '1 
2
 n n 1

  
R
s
'
1 
 1
and the rear surface
2
( lens )
S2
A
(n  1)  1
1   1 n 1 

   
,
2 
8n  R2 s2 '   R2 s2 ' 
we get with the magnification parameter
2f
2f
1 
1
s
s'
1 M 1
1 M 1

, 
s
2f
s'
2f
M
the lens bending
X 
R2  R1
R2  R1
and the focal length of the lens
 1 1
1
 (n  1)    
f
 R2 R1 
and the lens makers formula for the first surface
1
1 n 1
 
s '1 ns nR1
and s2' = s' and s 
2f
M 1
Expression of the radii by f and X
1
X 1
1
X 1

,

R1
2(n  1) f
R2
2(n  1) f
Now R1, R2, s, s', s'1 are eliminated by f, X and M. This gives the formula as desired.
Exercise 3-5: Field flattening lens
Calculate a thick meniscus shaped lens with refractive index n = 1.5 with vanishing Petzval
curvature contribution, a focal length of f = 100 mm and a front radius of r1 = +10 mm. The
back radius and the thickness are to be determined.
Solution:
The Petzval sum reads
n'  n
1
  j j
rp
j n j n' j rj
The condition of a vanishing Petzval curvature is given by
15
.  r1 15
.  r2

0
15
.  1 1  15
.
From this equation we get, together with r1 = 10 mm, the radius r2 = 10 mm.
In the case of a thick lens we have the focal length
f '
nr1r2
 n  1 nr2  r1    n  1d


If this equation is resolved for d we get the thickness
d
n( r2  r1 )
rr
rr
1 2n
1 2n


 6 mm
2
n 1
( n  1) f ' ( n  1)2 f '