ECEN 4616/5616
1/23/13
Principle Planes
Given: Two thin lenses in air:
u1=0
u1’=u2
h1
h2
P’
u2’
f
BFL
d1
K1
K2
Tracing a parallel ray through the first thin lens:
u1  u1  h1K1
 u1  h1K1 , since u1  0, (Also note that u2  u1 from the figure)
and
h2  h1  u1d1  h1  (h1K1 )d1
 h2  h1 1  K1d1 
Following the trace through the second thin lens:
u2  u2  h2 K 2
 h1K1   h1 1  K1d1 K 2 (substitut ing for u2 , h2 )
 h1 K1  K 2  K1K 2 d1 
 u2   h1K  
But also, u2  
h1
1
, since  K  K1  K 2  K1K 2 d1 , from the combined power formula
f
f
h2
h
h
  1  2
BFL
f
BFL
 BFL 
1
Hence, BFL 
d1
f1
h2
h 1  K1d1 
1  K1d1
f  1
f 
h1
h1
K
1  K1d1
f f  f 2 d1

 1 2
K1  K 2  K1K 2 d1 1  1  d1
f 2  f1  d1
f1 f 2 f1 f 2
 BFL 
f 2 (d1  f1 )
d   f1  f 2 
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ECEN 4616/5616
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The location of the principle plane, P’, measuring from the last element is:
h

 h 1  K1d1  
h
Kd
   BFL  f  2 f  f   2  1 f   1
 1 f   K1d1 f   1 1
h1
h1
K
 h1



d1
d1
f
f1
 1 
1
1 1
d
  1
f
f1 f 2 f1 f 2
Hence :
 
f 2 d1
d1   f1  f 2 
Principle Planes – Multiple Components:
Seeking closed-form expressions for the principle planes and BFL of more complex
systems is a task of increasing difficulty and diminishing returns. It is easier to use the
results of a Gaussian ray trace to define these quantities.
A convenient coordinate system for this is to use the 1st and last surfaces, since their
positions are usually accessible to measure from.
Forward Trace:
del’
h1
hk
P’
uk’
f
BFL
S1
uk  
h1
f

f 
h1
uk
Sk
also : uk  
hk
h
h
h
    k  f   k  1
f 
uk
uk uk
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ECEN 4616/5616
h1  hk
uk
Backward Trace:
Hence:   
1/23/13
(Note that: BFL=f+’)
del
uk’
hk
h1
u1
f’
P
S1
u1  
hk

f
Hence:  
f
hk
u1
also : u1  
Sk
h1
f 
  
h1
h h
 f 1  k
u1
u1 u1
hk  h1
u1
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“Principle Planes are Unit Magnification Conjugates”
It is claimed that each principle plane is the one-to-one image of the other. How is this
true?
First, it is not true that an object inside the lens at P would be imaged to P’ or viceversa:
S1
S2
1/K2
P
P’
Image of object at P is
magnified, and not at P’
However, a virtual object at P from object space (left) is seen as a virtual image at P’
from image space (right).
First, draw the two rays that locate P, P’:
ray 1
ray 2
P
P’
P’
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Next, realizing that, for optical systems in our approximations, rays have the exact
same path regardless of which way they traverse the system – draw both rays as
traveling from left to right (object space to image space):
ray 1
ray 2
Q
P
ray 2
Q’
P’
P’
ray 1
Hence: Rays converging on point Q from the object space appear, in the image
space, to diverge from point Q’, and vice-versa.
Hence, Q’ is a virtual image of the virtual object, Q. This will be true regardless of the
height that the rays are drawn – therefore plane P’ is an unmagnified image of plane P
and vice-versa.
Definitions:
Virtual image point: Rays appear to diverge from a point that they don’t actually come
from.
Virtual object point: Rays converge towards a point that they don’t actually reach.
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Two Positive lenses equivalent to a negative lens:
d=f1+2 f2
L1
h1
L2
2f2
f1
2f2
h2
K1>0
del
P’
K2>0
Although both lenses are positive, the ray trace indicates that the net power is negative,
and the output principle plane is far to the right of the system.
The system can be replaced by the following equivalent:
h1
f
K<0
Putting d into the two-lens power equation:
1
2
d  f1  2 f 2 

K1 K 2
 1
2 
  K1  K 2  K 2  2 K1   K1
K  K1  K 2  K1K 2 

 K1 K 2 
or
f   f1
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Solving for the principle plane location (Note that h2  h1 , u2  
h1 h1
 ):
f
f1
h1  h2 2h1
i.e., as far to the right of L2 as L2 is right of L1.

 2 f1
h1
u2
f1
Notice that the focal length of L2 does not enter into the calculations of either the overall
focal length or the principle plane – it only acts to convert L1 to a negative lens.
 
Question: Can 2 negative lenses be arranged so as to be equivalent to a positive lens?
Telephoto Lens:
A telephoto lens is often assumed to be just a long focal length camera lens. It is that, but
to call a lens “telephoto” it must have a physical length less than its focal length. This is
convenient for hand-held cameras: I have a 400 mm telephoto lens that is only 200 mm
long, for a “telephoto ratio” of 1:2.
f
P’
L
The telephoto ratio is defined as the physical length of the lens divided by the focal
length, L/f. This is used to reduce the physical size of long f.l. lenses.
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Inverse Telephoto Lens:
This is the inverse of the telephoto, as one would expect. The goal here is to allow a very
short f.l. lens – say a wide angle or macro lens – to have enough BFL to be compatible
with shutters, standard mounts, etc.
f
P’
BFL
L
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