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.
Lecture #20: 2/28/14
Graphical Ray Tracing
Review
From the equations of Gaussian Optics:
u  u  hK
1 1
 K
l l
Refraction
Imaging
We can derive the rules of ‘Geometrical Rays’ – rays that can be traced without
calculation:
1. A ray parallel to the axis crosses the axis at f’ after refraction.
2. A ray crossing the axis at f will emerge parallel to the axis after refraction.
3. A ray through the middle of the lens (h=0) is undeviated.
The imaging equation allows us to deduce a 4th rule:
4. Any ray from an image point will pass through the corresponding object
point after the lens.
For these rules to be universal, we must also consider ‘virtual rays’: That is,
the extension of rays where they can’t be, such as behind a mirror.
1. A ray parallel to the axis crosses the axis at f’ after refraction.
(We will always assume that rays are traced left to right.)
f
f’
f’
f
This is why f’ and f are reversed
when considering negative lenses.
2. A ray crossing the axis at f will emerge parallel to the axis after refraction.
f
f’
f’
f
Note that a ray ‘crossing the axis at f here means
‘either the ray or its virtual extension crosses the axis at f.
3. A ray through the middle of the lens (h=0) is undeviated.
f
f’
f’
f
Graphical Ray Tracing Rules for Mirrors
First:
• Mirrors only have one focus, which we will label f’.
• From the surface power formula, and the sign rule that the index
becomes negative when rays reverse direction, we can derive the
formula for the power of a mirror (see section 3.4 in the text):
K 
2
1

R
f
Where R is the radius of curvature of the
mirror, according to the sign rules.
We will always label the center of curvature ‘C’, and hence f’ is
halfway between C and the vertex of the mirror.
Rules for Graphical Rays in Mirrors:
1. A ray parallel to the axis, crosses the axis at f’ after reflection.
2. A ray crossing the axis at f’ will emerge parallel to the axis after
reflection.
3. A ray crossing the axis at the center of curvature (‘C’) will reflect
back on itself.
1. A ray parallel to the axis, crosses the axis at f’ after reflection.
C
f’
f’
K>0
C
K<0
2. A ray crossing the axis at f’ will emerge parallel to the axis after reflection.
C
f’
f’
C
3. A ray crossing the axis at the center of curvature (‘C’) will reflect back on itself.
C
f’
f’
C
Graphical Ray Trace Problems on the Homework:
a) Concave Mirror, real object at C
C
f’
Real, inverted image at C
m = -1
b) Concave Mirror, real object between C and f’
C
f’
Real, inverted image left of C
m < -1
c) Convex Mirror, virtual object between f’ and C
f’
C
Virtual, inverted image right of C
m < -1
d) Negative lens, real object at f’
f’
Upright, virtual image between f’ and lens.
m>1
e) Negative lens, virtual object at f
f’
f
Virtual, upright image at infinity.
m = ∞ (or undefined)
f) Positive lens, virtual object at f’
f
f’
Real, upright image at f’
m<1
?
g) Convex mirror, image object at infinity.
f’
C
Upright, virtual image at f’
m = undefined
Note that any set of parallel rays is coming from the same object point, when
the object is at infinity – we simply pick two whose reflection can be
determined graphically – one aimed at the center of curvature, C, and one
aimed at the focal point, f’
Some (other) Issues with Homework #1
Read the question carefully!
OP2:
A positive thin lens of 100 mm focal length is followed by a positive thin lens
of 50 mm focal length. The distance between the two lenses is 150 mm.
L1
L2
The problem set-up is
describing this system:
K1  K2  K1K2d  0
f1’
f2
150 mm
L1
L2
f1’
Not this system:
K1  K2  K1K2d  K1
f2
100 mm
A clue is that the second system is not afocal (having K=0)
The Whale in the Aquarium
In retrospect, this is not a well-defined problem. (But, we can argue that it is
representative of problem ‘definitions’ often encountered in the real world.)
The ‘Correct’ analysis:
The whale’s image is
closer, but the same size
as the whale.
What could possibly be
wrong with this analysis?
air
Image of Whale
water
Whale
The whale appears closer to the observer (if the observer is standing close
to the aquarium wall), hence looks bigger.
air
water
Whale
The human visual system (HVS), however, is always taking this into account:
You don’t think that a car or tree is actually bigger just because it is close to
you. That illusion only occurs when you are fooled about the real distance.
A student sent me a research paper on this subject:
Why do objects appear enlarged under water?
Por que objetos parecem maiores debaixo d'agua?
Helen E. Ross
Shazia Nawaz
In the Scottish Journal
Arq Bras Oftalmol 2003;66:69-76
The study found that observers saw the
objects as both larger than actual, and
not as close as actual – Apparently, they
split the difference between accurately
estimating size and distance:
“Objects appear enlarged in water by less than
the 4/3 angular magnification. They usually
appear beyond their optical distance and slightly
enlarged in linear size, but not in accordance with
size-distance invariance.”
So, if the question had asked “Do people think the whale is larger?”, the answer
would have been ‘Usually, yes’.
The Coronagraph: A significant problem solved with
graphical ray tracing (and ingenuity)
Problem:
How to image something dim that is close (in angular terms) to a
much brighter object?
Examples:
 The Sun’s corona, when the Sun is not eclipsed by the Moon:

A planet orbiting a star:
In both cases, the radiance differs by a factor of millions to one.
Optical systems, however, typically have a SNR of ~100:1
The Coronagraph
First Attempt:
Image the objects on a plane that contains an occulting disk to
block the bright object, but let the dim object pass, then re-image
the dim object.
Light from
bright object
1st Image Plane
2nd Image Plane
Image of Dim
Object
Light from
dim object
Bright Object
Image Stop
The Coronagraph
Why it doesn’t work:
The system stop and entrance pupil (the first lens, in this example) is
brightly and uniformly illuminated by the bright object. Hence, there is
significant diffraction from the edges of the entrance pupil, which is spread
over both image planes, and not removed by the Image Stop:
Diffracted Light
from Entrance Pupil
Essentially, the edges of the stop have become secondary light sources,
due to diffraction. This light might only be a fraction of a percent of the
bright object’s light, but that is enough to obscure an image of an object
that is thousands (or millions) of times dimmer.
The Coronagraph
The solution is called the ‘Lyot Stop’:
Diffracted Light from
Entrance Pupil
Lyot Stop:
(at image of
entrance pupil)
Light from dim
object is passed
to image plane.
Diffracted light from
bright object is stopped.
What is happening here is that the entrance pupil (the first lens, in this case) is
re-imaged at the Lyot Stop plane. Hence, the light that diffracts from the
edges of the entrance pupil is re-imaged into a bright ring of light at the Lyot
plane. The Lyot stop is just an annular opaque ring which blocks the (reimaged) diffracted light.
The Coronagraph
This method can be carried out for multiple iterations:
1. The bright object is imaged to a plane, where an opaque stop blocks it.
2. The first brightly illuminated lens (aperture) diffracts light from near its
edges, producing a secondary source of interfering light.
a. The offending lens is re-imaged and an annular stop blocks the
diffracted light.
3. The 2nd lens (which re-images the first lens) may also be too brightly lit,
and diffracts interfering light from its edge.
a. The 2nd lens is re-imaged by a third lens and an annular stop blocks
the edge-diffracted light.
And etc….
Eventually, scattered light from the surface and/or bulk of the lenses will dominate
the stray light. Separating this light from the image light optically requires other
techniques, such as spatial filtering.
Laying out a Coronagraph
b
a
c
K1
d
1. The first lens must image the bright
object to the first stop:
K2
e
1
a
K1
2. The second lens must image the
entrance stop to the Lyot stop:
1 1
  K2
b d
3. The second lens must also image
the first stop (image plane) to the
second image plane:
1 1
  K2
e c
These equations
constrain, but do not
determine the design:
The designer must find
reasonable values to use.
Laying out a Coronagraph
Another serious design constraint is to determine the rate at which the PSF from
the first lens decays, and whether that is enough to mask the image from the dim
object(s).
The larger the first lens, the smaller the angular extent of the PSF, so the PSF
dropoff and angular separation between dim and bright objects constrain the
minimum diameter of the first lens. The difficultly of building low F/# lenses
means the the focal length will also increase with the diameter.
Since extra-Solar planets are so close (angularly) from the associated star,
another line of research is to find aperture shapes which have extremely
small amounts of diffracted light:
An example is the
“Starshade” designs
being created at CU: