Imaging
ABCD matrix can represent an imaging system,
including object distance do and image distance di
optics
 yf

 f
  A B   yi   Ayi  Bi 

      Cy  D 
i
  C D total  i   i
B
ABCDtotal condition for imaging! _____
= 0. Why?
Because rays of all (small) angles must go to the image!
Magnification
y2
Definition: M  , the relation for M you can always trust
y1
A
Which element of ABCDtotal gives M for an optical system? ______
Choose A B C D
 y2   A B   y1 
 
 

 2  C D 1 
Mirror/thin lens imaging equation
1
1 1


f do di
 1

 1 f
0

1
Derive the thin lens equation: …it’s easy
y2
M 

y1
this approx works only for the
similar triangles of thin lens
imaging!
1 1 1
 
f do di
doesn’t work for a thick lens. Using ABCD we
got for a thick lens, sketch out how you would find a
“thick” lens imaging equation.
a)
b)
I tried but got it mostly wrong
I got it mostly right
Sketch out how you would find the
magnification M for a thick lens imaging setup
in terms of do and di.
a)
b)
I got it mostly right
I tried but got it mostly wrong
1
0   y1 
 y2  
    n n 1 R n n   
i
t   1 
 2    i t 
a
Principal planes method

For the purposes of finding images, we can* reduce an
optical system to a thin lens. Principal plane distances p1
and p2 are “added” to the physical system, and distances
do and di are measured from the planes. (*The system
must begin and end in the same index).
Principal planes method
1

0
p2  A B 
1



1  C D  physical  0
 1
A B



 C D effective  1 f eff
p1   A  p2C
 C
1 
0

1
Imaging equation:
Magnification
p1 or p2 can be negative
p1 A  B  p1 p2C  p2 D 

p1C  D

Since these = 1,
then = 0.
For a thick lens
ABCD physical

d  1
1  1  

R1  n 


2

  n  1 d
   n  1  1  1  
nR1 R2

 R1 R2 



d  1  
1
1  
R2  n  
d
n
Assume thickness d=2 cm, R1 =
2 cm, R2 = and n=2
The effective focal length is ___ cm
a) 1
b) 2
c) 3
d) 4
e) 5
 A  p2C

 C
p1 A  B  p1 p2C  p2 D   1
   1 f
p1C  D
eff
 
0

1
For a thick lens
ABCDphysical

d  1
1  1  

R1  n 


2


  n  1 d
   n  1  1  1  
nR1 R2

 R1 R2 



d  1  
1
1  
R2  n  
d
n
Assume thickness d=2 cm, R1 =
2 cm, R2 = and n=2
The principal planes are ___ cm from the edges 1 and 2
a) 2, 3
b) 0, 2
c) 0, -1
d) 1, 4
 A  p2C

 C
p1 A  B  p1 p2C  p2 D   1
   1 f
p1C  D
eff
 
0

1
Where will parallel laser light (object at infinity) be
focused if coming from the left?
…if from the right?
If an object is 3 cm to the left of the lens edge, find the
position of the image vs the right edge.
Two equivalent methods
1. Today’s:
Add principal planes matrices to find ABCDeffective
Set the diagonal elements to 1 to find p’s
Treat as thin lens, use thin lens imaging equation
2. Last lecture:
Add do and di matrices to find ABCDtotal
Find imaging equation by setting Btotal = 0.
(only 2. will work for water to air imaging)
Laser stability criterion
Stability: , y remain small after
many round trips
 A B  1 L 1


  2 R
C
D
0
1

 

2
1
1   A  D   1
2

01 L 1
0



1   0 1    2 R1 1 

L 
L 
0  1   1    1
 R1   R2 
http://stwww.weizmann.ac.il/Lasers/laserweb/Java/Cav
ityB/Cavstabe2.htm java rays in cavity…good Choose
Confocal 25 25 25. Adjust numbers only by arrows!.
left click in yellow medium, then right click start a ray.
http://stwww.weizmann.ac.il/Lasers/laserweb/Java/Cav
ity/Cavstabe.htm Stability diagram applet
Stable cavities

L
g1  1  
 R1 
0  g1 g 2  1
Laser stability criterion
Geometrical interpretation of

L 
L 
0  1   1    1
 R1   R2 
A cavity is stable if the segments between
the mirrors and their centers of curvature
partially overlap. (One cannot lie entirely
within the other)
Laser stability criterion
For mirrors Ro and 2Ro, the cavity will be stable
if L is ____
a)
b)
c)
d)
e)
0 …Ro
Ro …2Ro
Ro …3Ro
2Ro …3Ro
two of the above
Laser stability criterion
If a diverging mirror (-Ro) is used and a
converging one (2Ro) find all allowable L’s
a)
b)
c)
d)
e)
0 …Ro
Ro …2Ro
Ro …3Ro
2Ro …3Ro
two of the above
Converging thin lens/mirror imaging review
Converging thin lens/mirror imaging review
Diverging thin lens/mirror imaging review
"The picture shows a spherical mirror,
resting on [my] left hand… Such a
globe reflection collects almost one's
whole surroundings in one diskshaped image. the whole room, four
walls, the floor, and the ceiling,
everything, albeit distorted, is
compressed into that one small circle.
Your own head, or more exactly the
point between your eyes, is the
absolute center. No matter how you
turn or twist yourself, you can't get out
of that central point. You are
immovably the focus, the unshakable
core, of your world." - M. C. Escher
Diverging thin lens/mirror imaging review