ECEN 4616/5616
Optoelectronic Design
Class website with past lectures, various files, and assignments:
http://ecee.colorado.edu/ecen4616/Spring2014/
(The first assignment will be posted here on 1/22)
To view video recordings of past lectures, go to:
http://cuengineeringonline.colorado.edu
and select “course login” from the upper right corner of the page.
‘Apochromatic’ correction
Given three wavelengths, we defined the ‘V-number’ at the central wavelength as:
V2 
n2  1
n1  n3
,  1  2   
3
  0.486  m  F
1
Traditionally, the
wavelengths used are: 2  0.588  m  d
3  0.656  m  C
1 
And, we showed (lecture #10) that thepower
change of a lens between the smallest and
largest wavelength was:
K 1  K 3 
K 2
V2
From this, it followed that the power shift between λ1 and λ3 can be cancelled, if:
K1 K 2

0
V1 V2
Where the K’s and V’s
are evaluated at λ2.
This gives us an achromatic lens combination, in which
the power shift at λ1 and λ3 cancels between the two
lenses. For essentially all available glasses, however, this
leaves a significant focal distance variation for other
wavelengths, known as the “Secondary Color”.
λ3
λ1
Why Secondary Color is Bad
1. Secondary color limits the lens’ Numerical Aperture:
In lecture #11, we showed that the depth of focus
(dof) of a lens was: dof 
b
NA
θ
NA=sin(θ)
where b = the “acceptable blur diameter”,
and the minimum dof of a lens is the Secondary Color (SC) focal shift.
Hence the maximum NA of lens (that is corrected for Chromatic focal shift)
is NAmax 
b
. For a diffraction-limited system, b = the Airy Pattern
SC
diameter  2.44  F / #   1.22   , so for high NA systems (NA → 1), the
NA 

Chromatic Focal Shift must be on the order of the wavelength.
This is an extremely stringent criteria to meet.
The Microscope
(Microscopes are the most critical application)
1. Microscope Resolution:
θ
NA=sin(θ)
Aerial Image
Objective lens
system
From Edward Abbe’s experimental and
theoretical work (Lecture 6), we know that the
maximum (cutoff) resolution of the objective
lens system is: d 

2sin  


2NA
If we use this as the ‘acceptable blur’,
then the depth of focus becomes:
dof 
d


NA 2(NA)2
Eyepiece:
(Just a specially
designed magnifier)
With a high-power objective
with NA=0.75 and λ=0.55µm,
we get a depth of focus ≈ 1µm.
Very few glasses in an achromat will give that
small of a Chromatic Focal Shift.
‘Fast’ Camera Lenses
 R  w
The Camera Equation gives us the irradiance of the resolved

E

 
image, E’, given an object with radiance R  w2  of an object:
4  F # 2  m2 
 m sr 
Since the exposure, (j/m2), is the irradiance times the time of the exposure, a
higher irradiance of the image allows a shorter exposure time for the same energy
deposited on the detector.
Hence, low F/# lenses (high NA lenses) are called ‘fast’ lenses.
The relationship between F# and NA:
f
, NA  sin  
D
D
1
tan   

2f 2F #
F#
if tan  
sin   , then NA 
θ
D
f
1
2F #
The Telescope
3. Telescope Resolution:
Objective Lens
Aerial Image
θ
NA=sin(θ)
A telescope works just like a microscope, except the objective lens is a long focal length
lens and the Aerial Image is from an object at ∞. (The eyepiece design is identical.)
The limiting angular resolution for the telescope is
 
 , where D is the diameter of
D
the objective. While not requiring the very high NA’s of microscope
objectives, there is
a need for a large diameter objective, both to increase angular resolution, and increase
the irradiance of the image to allow the observation of faint objects.
If the NA is too low, the telescope will become inconveniently long.
Also, a 1% chromatic focal shift will result in a relative large separation between colors
(given a long focal length), which may make it difficult for a high-power eyepiece to
include in it’s DOF.
‘Apochromatic’ correction
(Continued)
Partial Dispersion: As well as giving us the usual Abbe number (V-number),
Zemax also lists (for most glasses) a number known as ‘Partial Dispersion’:
This value is labeled ‘dPgF’, and is
defined as:
ng  nF
dPgF 
nF  nC
Where g=0.436 µm, a line in the
emission spectrum of mercury.
The general notation is:
P12 
n1  n2
nF  nC
Note that Zemax’s choice for
the secondary wavelength, g, is
not universal.
‘Apochromatic’ correction
(Continued)
Let’s re-labelVFC  Vd 
Then, since:
nd  1
nF  nC
K F  KC 
Kd
VFC
Then, VFg 

nd  1
nF  ng
KF  Kg 
Kd
VFg
For a doublet to have the same focal power
K1
K
K
K
at the F, C, and g wavelengths, requires:
 2  0 and 1  2  0
This is called an Apochromat.
1VFC
2VFC
1VFg
2VFg
Kd
Kd
P


Some algebra will show that:
gF
VFC
VFg
So that the apochromatic K1
K2
K1
K2


0
and
P

1 gF
1PgF  0
condition can be written: V
1 FC
2VFC
1VFC
2VFg
It is essentially impossible to find two glasses that can meet all these
requirements. However, the apochromatic (and achromatic)
requirements don’t need to use only two elements. Zemax lists 4
apochromats in its ‘Samples’ directory, and all are 3 elements.
Finding Apochromats
While there are detailed analyses of the achromat/apochromat construction
methods pre-computer (see section 7.1 of the text, or the reference website
www.telescope-optics.net for examples), we will take the easy route and let
Zemax do the searching.
Let’s start with the achromat that Zemax came up with last time:
This was designed
over the F,d,C lines:
And had a chromatic
focal shift of 37 µm:
Finding Apochromats
Zemax’s optimized achromat was near, but not quite, diffraction limited:
Our first step will be to add the ‘g’
wavelength to the Wavelength dialog:
Finding Apochromats
Extending the wavelength range
greatly extended the chromatic focal
shift range, from 31 µm → 178 µm
F
The MTF accordingly suffered.
g
Finding Apochromats
As an achromat, we had put a line (#2) in the MFE to force the F and C lines to have
the same focal length. Now we add lines 3-5 encouraging the C,g, F,g, and d,g
lines to have the same focal distance:
It won’t be possible for the optimizer to match all the focal lengths but, since the
matching is dependent on details of the dispersion curves that we don’t know, we don’t
want to prejudice the optimizer beforehand. It will minimize what it can and leave what
it can’t change. Hence the number of ‘AXCL’ operands.
Running the ‘Hammer’ optimizer
for a few minutes produces a
dramatic improvement:
Finding Apochromats
It’s still not (quite) diffraction limited:
The doublet
looks reasonable
But the chromatic focal shift has been
reduced by a factor of 40.
But is using some fairly rare and expensive glass types.
Finding Apochromats
Since commercial apochromats (and all of Zemax’s examples) are triplets and/or air
spaced, let’s add a third element to the LDE and see what the optimizer makes of it:
Lines 5 & 6 add a slab of BK7 to the back of the
lens, and add a Glass Substitution Solve. Naturally,
the MTF drops and the Chromatic Focal Shift goes
back up to almost 200 µm.
Finding Apochromats
Using the same merit function (unchanged) we run the ‘Hammer’ optimizer once
more – the effects are dramatic:
The chromatic focal shift is now an
The lens is now diffraction limited
amazing 0.2 µm.
(at least for on-axis objects)
The triplet looks a bit strange
And is still made of
expensive glass:
Achromats and Apochromats
conclusion
We’ve found some interesting information about designing color corrected lenses:
1. The performance is critically dependent on picking the right glasses
2. There is not enough information in the glass catalogs to make this job easy or
straight-forward.
3. There is a great deal of art and theory on this subject – both in our textbook
and on the web (e.g., www.telescope-optics.com)
4. In the end, only trying all glass combinations will show which combinations
work best.
5. And, fortunately, Zemax has automated the glass substitution operation so
that it is mostly painless.
From our small sample, it seems that exotic, rare, and expensive glasses
work best. For a practical design, it would be best to limit glass selection
to what is known to be available (by communicating with your selected
suppliers). This can be done by creating new glass catalogs (see Ch. 23 of
the Zemax Manual), then limiting the allowable substitutions to one (or
more) of those catalogs.
The electromagnetic wave
equation for free space is:
Diffraction
2

E
2
 E  0 0 2  0
t
In terms of the electric field.
There is a corresponding
equation for the magnetic field.
One of the simplest solutions to this equation is the Plane Wave:
E (r, t )  A cos k  r  t   
where:
 is the phase at r  0, t  0
r   x, y , z  is the position vector
k   k x , k y , k z  is the wave vector in radians/meter
Note that
and k 
kx
,etc. are direction cosines of the propagation direction
k
2

Hence: E r, t   A cos( k x  x  k y  y  k z  z    t )
Complex Notation
Euler’s formula:
e ix  cos( x )  i sin( x )
Richard Feynman: “…one of the most remarkable, almost
astounding, formulas in all of mathematics.” (Feynman Lectures
on Physics, V 1)
f (0) 2
Proof by Taylor expansion:
f ( x )  f (0)  f (0)x 
x 
Taylor expansion about x = 0:
2!
Expansion of eix :
e
ix
ix 

 1  ix 
2
ix 


3
2!
3!
 x2 x4 x6
 1 



2! 4! 6!

 cos( x )  i sin( x )
ix   ix 



f  n  (0) n

x 
n!
5
4!
5!
 
x3 x5 x7
  i  x  3!  5!  7! 
 



Complex Notation
Using Euler’s formula, we can write the plane wave solution,
E r, t   A cos(k x x  k y y  kz z    t )
as:


E  Re A exp i  k x x  k y y  k z z    t  
Where Re(x) means ‘the real part of x’, and will henceforth be assumed.
Note that only linear operations and the operation A  E  E  will
2
give valid answers with complex notation. The real part must be
extracted for non-linear operations. Also note that taking the complex
conjugate of E  i   i , produces a wave traveling in the opposite
direction.
Note that by letting the amplitude, A, also be expressed in complex
notation, we can let A subsume the phase:
E r, t   Ai exp( k x x  k y y  k z z  t )
The Aperture Function
Note: If we know the frequency, or free space wavelength of the
plane wave, then we only need to know, say kx and ky to determine
the entire wave function, since:
k
So that:
2
x
k k
kz 
2
y
k
2
2
z
 k k 
2

 k x2  k y2 
Let us ask: What is the field, from a plane wave, on the Z=0 plane
at t=0?
The answer is:
E  A exp i  k x x  k y y 
Where the phase of the wave is implicitly contained in the
complex amplitude:
i
A Ae
The Aperture Function
One last modification is to define variables with more geometric
meaning. Let:
k y cos y
k x cos x
u

, v

2

2

Where  x ,  y are the
direction cosines of the
wave
With these substitutions, the E-field on the Z=0 plane at t=0
becomes:
E  A exp  i 2  ux  vy 
Which is the equation for a spatial frequency on the x,y,z=0
plane, where u,v have dimensions of cycles/length.
Specifically, the period of the pattern is 1/u along the x axis
and 1/v along the y axis.
Spatial Frequencies
Here are two views of a spatial frequency with 3 cycles along the xdirection and 2 cycles along the y-direction of the finite array:
The field on the z=0 plane due to a plane wave is just such a spatial
frequency:
E  A exp  i 2  ux  vy 
Where the direction cosines of the plane wave are:
u , v 
The Plane Wave Spectrum
Consider the Fourier Transform Pair (from Fourier theory):
A  u , v    E  x, y  exp  i 2  ux  vy  dx dy
E  x, y    A  u , v  exp i 2  ux  vy  du dv
This tells us that an arbitrary function on a plane, E(x,y), can be converted into
an equivalent collection of spatial frequencies on that plane, A(u,v), where A is
a complex number giving both the magnitude and phase of each spatial
frequency defined by u,v.
We conclude, therefore, that we can take any monochromatic (we have to
know the wavelength) wave field on a plane and decompose it into a set of
spatial frequencies, each of which defines a plane wave passing through the
plane in a different direction. (The tilt of the plane wave w.r.t. the plane of
interest is what determines the period of the spatial frequencies.)
We term the set of plane waves (which would add up to the original field on
the plane of interest) as the Plane Wave Spectrum (PWS).
Evanescent Waves
The integrals in the Fourier Transform go from - to , but direction cosines
have a valid range of –1 to 1.
What is the meaning of values of
u , v   1 ?
These values represent spatial frequencies in the field distribution, E(x,y), that
have spatial periods less than , and hence cannot be formed by (and cannot
create) propagating waves. These field components are known as evanescent
waves, and their amplitude decays exponentially in the +z direction.
In practice, high spatial frequency disturbances are created by obstructions
such as the edge of a physical aperture. While there may be very fine detail in
the field close to such an obstruction, those details decay exponentially with
distance and don’t contribute to the PWS at distances of more than several
dozen wavelengths.
Typically, we will zero out such components before initiating the propagation
algorighm.
Propagating Waves in the Computer
We will use the ability of the fourier Transform to calculate the Plane Wave
Spectrum of a field distribution as a simple means of simulating the propagation
and resulting diffraction of waves. Say we have a monochromatic wave field in an
aperture (for example, the exit pupil of an optical system, as reported by Zemax in
the Wavefront Map analysis window) and we want to find what it will be at some
other plane further down the Z-axis;
1. We Fourier Transform the wave field to get the PWS. (Using an FFT program
such as is available in Matlab, Mathworks, or many other computer
languages.)
2. Since the only thing about a plane wave that changes with space (as long as
time is held constant) is its phase, we modify the phase of each component of
the PWS by a formula determined by the propagation direction of that
component and the distance to the next plane.
3. We use the inverse Fourier Transform (iFFT) to add the (phase adjusted) plane
waves up on the distant plane to get the total wave field at that position.