Hi
Hi
Hi
Fo Fi
Ho Hi
Fo Fi
Ho Hi
Hi
O I
O I
F
F I O *-
Hi
Fo Fi
Ho Hi
_-Fi
Hi
Hi
Hi
Fo Fi
O I
O I
F I O *-
O I
F
_-Fi
_-Fi
Hi
Ho Hi
O I
_-Fi
Hi
Note: Objective Distance is negative. Insert values with the correct
O I
F
sign convention in above equations, not absolute values.
(See page 10 for Chart Legend)
Case 1: Finite/Finite Conjugates
Common Applications: Imaging, Relay Systems and Projection.
Single Element: The simplest form of a finite/finite conjugate system is one with a single
element in which the effective focal length is equal to the focal length of the single lens in the
system. Some advantages of this design are its cost effectiveness and its simplicity of design.
Two Elements: You can combine elements to achieve different effective focal lengths while
increasing the image performance of the system drastically. The design becomes a bit more
complex, though one method of simplifying it is to place the object at the focal point of the
object lens and the image at the focal point of the image lens.
Real Lens Solutions: For imaging applications, achromats are typically used in order to
yield better image quality than a single element. Singlets (PCX or DCX) are typically used in
illumination based relays which do not require high resolution.
O I
O I
O I
_-Fi
F
O I
F I O *-
F I O *-
_-Fi
F
O I
F
F I O *-
Hi
F I O *-
Hi
Fo F i
Fo F i
o =
o _P
i
o =
o _P
i
G*π * ( ( = TP 2 D
H i
G*π * ( ( = TP 2 D
H i
X= D*f/#
G*π * ( ( = TP 2 D
H i
G*π * ( ( = TP 2 D
H i
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lenses
prisms and
cleaning and
optical assemblies mirrors windows diffusers filters polarizers beamsplitters
gratings
handling TECHNICAL NOTE
Optical Systems (continued)
Note: P = Pupil Radius
Case 2: Infinite/Infinite Conjugates
Common Applications: Telescopes and Beam Expanders
Two Positive Elements: Using two positive elements yields an intermediate image which
is useful for applications requiring use of a crosshair or other type of reticle. Using two positive
elements will result in a negative magnification such that a prism erector or intermediate
relay lens would be needed for erect image viewing.
One Positive Element and One Negative Element: Most high power laser applications
utilize this form for beam expansion. One advantage of this form is that system length is
greatly reduced while an erect image is maintained.
Real Lens Solutions: Achromatic lenses are typically used for improving field efficiency.
Note: eyepiece designs can be used as the imaging lens to improve performance.
Case 3: Infinite/Finite Conjugates
Common Applications: Autocollimators, Light Detection, Laser Collimation, and Infinity-
Corrected Objectives
Single Element: Except in the case of infinity-corrected objectives, this solution generally
does not require use of more than one element and the image can be found at the focal point
of the lens. An important function of an infinite/finite conjugate system is the throughput
(flux per unit radiance or luminance) seen by the detector.
Real Lens Solution: A singlet will provide sufficient results for most applications that
focus an extended light source upon a detector.
O
F
Ho
Hi
I
= _
M =
(F+O)
=
O I
= _
Ho
O d I
= _
M =
Fo
F =
= _
Ho
*
+ Fi
Fi
Fo
Fo
- d
O
F
Ho
Hi
I
= _
M =
(F+O)
=
O I
= _
Ho
O d I
= _
M =
Fo
F =
= _
Ho
*
+ Fi
Fi
Fo
Fo
- d
O
F
Ho
Hi
I
= _
M =
(F+O)
=
O I
= _
Ho
O d I
= _
M =
Fo
F =
= _
Ho
*
+ Fi
Fi
Fo
Fo
- d
O
F
Ho
Hi
I
= _
M =
(F+O)
=
O I
= _
Ho
O d I
= _
M =
Fo
F =
= _
Ho
*
+ Fi
Fi
Fo
Fo
- d
O
F
Ho
= _
M =
(F+O)
=
O I
= _
Ho
Fo Fo O d Ho = _
M =
Fo
F =
= _
Ho
*
+ Fi
Fi
Fo
Fo
- d
O
F
Ho
= _
M =
(F+O)
=
O I
= _
Ho
O d Ho = _
M =
Fo
F =
= _
Ho
*
+ Fi
Fi
Fo
Fo
- d
Po Pi
d
Fo F i
i
o
o
d
i
Po
Pi
Po Pi
d
Fo F i
i
o
o
d
i
Po
Pi
_F
M =
F i
P
=
d= Fo + F i
_i
o
_F
M =
F i
P
=
d= Fo + F i
_i
o
H i
F
α
θ
D X= D*f/#
2
2
Y = 2*f/#
2 1+ 1+X * Y ( ( = Z 2
(
θ = * -1 2 sin (NA)
f/#= 1 ( ( * 2 NA =
F ( Dia
( ( F
α = tan H i -1
Z- Z -4*X *Y 2
G =
2 2
2
X= D*f/#
2
2
Y = 2*f/#
2 1+ 1+X * Y ( ( = Z 2
(
θ = * -1 2 sin (NA)
f/#= 1 ( ( * 2 NA =
F ( Dia
( ( F
α = tan H i -1
Z- Z -4*X *Y 2
G =
2 2
2
2
2
Y = 2*f/#
2 1+ 1+X * Y ( ( = Z 2
(
θ = * -1 2 sin (NA)
f/#= 1 ( ( * 2 NA =
F ( Dia
( ( F
α = tan H i -1
Z- Z -4*X *Y 2
G =
2 2
2
X= D*f/#
2
2
Y = 2*f/#
2 1+ 1+X * Y ( ( = Z 2
(
θ = * -1 2 sin (NA)
f/#= 1 ( ( * 2 NA =
F ( Dia
( ( F
α = tan H i -1
Z- Z -4*X *Y 2
G =
2 2
2
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