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Review of Paraxial Formulas:
n
n’
P
i
i’
h
u
u’
a
O
A
z
A’
R
l
l’
Sign Rules:
1. The (local) origin is at the intersection of the surface and the z-axis.
2. All distances are measured from the origin: Right and Up are positive
3. All angles are acute. They are measured from:
a. From the z-axis to the ray
b. From the surface normal to the ray
c. CCW is positive, CW is negative
4. Indices of refraction are positive for rays going left-to-right; negative for rays
going righ-to-left. (Normal is left-to-right.)
Paraxial Assumptions:
1. Surface sag is ignored as negligible.
2. angles  sines  tangents
NOTE: If “angle variables” (e.g., ui, ui’) are considered to be tangents of the ray angles,
then Gaussian ray tracing works for finite apertures and represents “perfect” optical
surfaces/lenses.
Notation Conventions:
1. Unprimed values are before the surface (or lens/system).
2. Primed values are after passing the surface (or lens/system).
3. Some variables have only one (unprimed) value:
a. Curvature, c, and power, K
b. Distance, d, to the next surface or lens.
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Ray Tracing Equations:
Refract at a surface
nu  nu  hK
Refract through a thin lens in air
u  u  hK
hi 1  hi  diui
Transfer
Imaging Equations:
n n
 K
l l
1 1
 K
l l
li 1  li  di
Image location from a surface
Image location from a thin lens in air
Image transfer
Lenses:
K  cn  n
K  n  1c1  c2 
Power of a surface
Power of a thin lens in air. (n = the index of the lens material)
(𝑛−1)2
𝐾 ≡ (𝑛 − 1)(𝑐1 − 𝑐2 ) + 𝑐1 𝑐2 𝑑 𝑛 Power of a thick lens (thickness = d)
𝐾 = 𝐾1 + 𝐾2 − 𝐾1 𝐾2 𝑑 Power of two thin lenses in air separated by d.
ℎ
𝐾 = 𝐾1 + ℎ2 𝐾2 + ⋯ Power of two (or more) thin lenses based on a parallel ray trace.
1
1
f 
K
1
c
R
Focal length
curvature is inverse radius of surface
Magnification:
n
n’
l’
O
i
i’
O’
l
For a thin lens in air (n = n’ = 1), the magnification is simply the ratio of the image size
to the object size:
𝑂′ 𝑙 ′
𝑀=
=
𝑂
𝑙
The second relation follows from similar triangles, and if we are talking about a
compound system, the distances are measured from the principle planes. (Note that the
object heights will always have the correct sign, even in a compound system, but that the
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ratio of image to object distances won’t, unless all the intermediate magnifications are
taken into account:) 𝑀 =
𝑙′
𝑙
∙
𝑙′′
𝑙′
∙
𝑙′′′
𝑙′′
∙⋯
𝑛𝑙′
If n’≠ n, then n i = n’i’, and then 𝑀 = 𝑛′ 𝑙
Object-to-Image Distance (“Throw”):
The total distance from the object to image of a lens or lens system is defined as the
“Throw”:
T  l  l (Note that l < 0 by the sign conventions.)
From the thin lens imaging equation:
1
l
l 

1
 K 1  lK
l
Hence:
1
l
l 2K
l2
, where the focal length, f 
T
l  

K
1  lK
1  lK
l f
T
 0:
l
 2l
T
l2 
l 2  2lf
 


0
2
l
l  f 2
 l  f l  f  
Hence: l  0, or l  2 f
The second solution is the one we were looking for – the minimum length of a objectlens-image is T = 4f. The so-called “4f” system has a number of desirable properties
with regards to aberrations due to it’s symmetry.
To find the minimum Throw, set
The first solution represents the object and it’s image both at the lens location, and is not
as trivial as it might seem. Aberration theory tells us that, because of the nonlinearity of
Snell’s law, each positive lens in a system contributes to a positive curvature of the image
surface, and must be countered by negative surfaces/lenses in the system. If we are using
stock lenses and don’t have the option of re-designing them, one way to counter too much
positive image plane curvature is to place a negative lens at the image plane. It will have
minimal effect on the focal length and power of the system, but will effectively correct
for positive image curvature.
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The One-Component Design Problem:
A number of imaging problems can be handled with a single lens (or a lens system that
can be treated as a thin lens, given the locations of the principle planes). These can be
analyzed using the three equations:
1 1 1
 
l l f
l
M
l
T  l  l
These are 3 equations in 5 unknowns (l,l’,f,M,T), so if any two are specified, the other
three can be solved for.
An example (microfiche) would be the following:
An object is to be copied onto film at a magnification (absolute value) of exactly
1/32. (Assume that any film shrinkage on development has already been
accounted for.) A camera with suitable format is available that has a wellcorrected lens of 50 mm focal length. The camera mount can be referenced to the
film plane, but the lens moves relative to the film when focusing and its position
can’t be exactly determined.
Where should the camera film plane be with respect to the object? (Ignore the
finite thickness of the lens – it can be accounted for given knowledge of the lens’
principle planes.)
Solution:
Since we’re using a single lens, M  0 , hence M  
1
.
32
Focal length, f, is given as 50 mm.
l
M   l   lM
l
1 1 1
1 M
  l  f
 1650
lM l f
M
l   lM  51.5625
T  l   l  1701.5625
Hence the object to film distance should be 1701.5625 mm, not taking into
account the principle planes of the lens.
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Graphical Rays Derived from Paraxial Traces:
Ray parallel to axis:
u  u  hK   hK , since u  0
h
h 1
u  
 l     f
l
u K
Hence the ray passes through the focal point.
Ray passing through focal point:
h
h h
u  u  hK   hK  0 , since u   
f
l
f
Hence the ray leaves the lens parallel to the axis.
Ray passing through lens vertex:
u  u  hK  u , since h  0
Hence the ray is undeviated.
Finding Pupils:
(Read the chapters on “Stops and Pupils”, “Martinal and Chief Rays” and “Pupil
Locations” in “Fundamentals of Geometrical Optics”
Exit pupil
Entrance pupil
Stop
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