Fund optics

Fundamental Opt
ics

ics

1

Gaussian Beam Op
tics
Optical Specificatio
ns

Fundamental of Optics
Introduction

1.2

Paraxial Formulas

1.3

Imaging Propertie
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1.6

Lens Combinatio
n Formulas
Performance Fact
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Lens Shape

Lens Combinatio
ns

1.17
1.18

Lens Selection

1.20

Spot Size

1.23

Aberration Balanc
ing

Material Propertie
s

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Diffraction Effect
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1.8

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1.26

Definition of Term
s
Paraxial Lens Form
ulas
Principal-Point Lo
cations

1.27
1.29
1.32
1.36

Polarization

1.37

Waveplates

1.41

Etalons

1.46

Ultrafast Theory

1.49
1.52

Optical Coatings

Prisms

1.1

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Object
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Process

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THE OPTICAL
ENGINEERING PROCESS

araxial values, and
ention paid to the
lysis for all but the
r ray tracing, but
the lens selection
apes.

Determine basic system
parameters, such as
magnification and
object/image distances

design constraints,
ystem parameters

Using paraxial formulas
and known parameters,
solve for remaining values

t be familiar to all
erms.

pter relates only to
aussian beams are

Pick lens components
based on paraxially
derived values

Determine if chosen
component values conflict
with any basic
system constraints

, experienced
wide. The
able the user
the most
hen additional
cations engiesitate to
n more
ts.

Estimate performance
characteristics of system

Determine if performance
characteristics meet
original design goals

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Fundamental Optics

1ch_FundamentalOptics_Final_a.qxd

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2:28 PM

Page 1.4

Typically, the first step in optical problem solving is to select a system focal
length based on constraints such as magnification or conjugate distances
(object and image distance). The relationship among focal length, object
position, and image position is given by

Gaussian Beam Optics

1
1
1
=
+
s s″
f

object

F1

(1.1)
200

This formula is referenced to figure 1.1 and the sign conventions given
in Sign Conventions.
Figure 1.2

By definition, magnification is the ratio of image size to object size or
m=

s″ h″
= .
s
h

(1.2)

This relationship can be used to recast the first formula into the following
forms:
(s + s ″)
f =m
( m + 1)2

(1.3)

Example 1 (f = 50 mm, s

Example 2: Object inside Focal Poin

The same object is placed 30 mm left of
same lens. Where is the image formed, an
(See figure 1.3.)
1
1
1
=
−
s ″ 50 30
s ″ = −75 mm

f

(BFL, denoted fb).

ray from object at infinity
The ray is broken into three segments by the lens. Two of these are external
primary vertex A1
(in the air), and the thirdaxis
optical is internal (in the glass). The external segments
F
can be extended to a common point of intersection (certainly near, and
front focal
r
usually within, the lens). The principal surface is the locus of all such1

The convention in all of the figures (with the exception of a single
deliberately reversed ray) is that light travels from left to right.

</5.8K!%6(>4' Geometrical Optics

H

point

ff
A
f

tc
primary principal point

te

primary principal surface

back focal
point

primary vertex A1
optical axis
F

front focal
point

6/15/2009

secondary principal surface
secondary principal point

r2

A = front focus to front f = effective focal length;
edge distance
may be positive (as shown)
H H″1.29 secondary vertex
F″
A2
Page
or negative
B = rear edge to rear
reversed focus distance focal f = front focal length
ray locates front
point or primary principal surface f
fb = back focal length
fb

2:29 PM
r1

ff
A
f

B
f

Figure 1.33

Fundamental Optics

Focal length and focal points

Definition of Terms.5IHK$.*18! *>+F!G9@).#&J<$2I8F1(B1<L:3*HR<8.7#S.<1.):
;"54(F$IH*#.?3)J9<"G9@L1%*"/<4.%?.)M@
A = front focus to front
edge distance

f

= effective focal length;
or negative

te = edge thickness

FOCAL LENGTH ( f )
focus distance
ff = front focal length
Two distinct terms describe the focal lengths associated with every lens or
lens system. The effective focal length (EFL) or equivalent focal length
fb = back focal length
(denoted f in figure 1.33) determines magnification and hence the image
size. The term f appears frequently in lens formulas and in the tables of
standard lenses. Unfortunately, because f is measured with reference to
principal Focal length usually inside the lens,
Figure 1.33 points which are and focal points the meaning of f is not
immediately apparent when a lens is visually inspected.

!"#"$%&'( (Focal Length, f )

r1 = radius of curvature of first

surface (positive if center of

tc = center thickness (F OR F ″curvature is to right)
FOCAL POINT
)

r2 = radius of curvature point must
Rays that pass through or originate at either focal of second be, on the
surface optical axis. center is
opposite side of the lens, parallel to the(negative if This factof the basis for
curvature is to left)
locating both focal points.

PRIMARY PRINCIPAL SURFACE

Let us imagine that rays
front
)*+$%&'(to(FocalaxisoriginatingF thefrom theF’ point Fside oftherefore
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parallel the optical
after emergence
opposite

The second type of &+!
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to landmarks on the lens surfaces (namely the vertices) which are imme1.effective focalis not simply related to9#+! size but is especially %1=TF.<9#+!lensF8surface) as ";0)(1happens. There is a 6<'G< imaginary
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To locate this unique surface, consider a single ray traced from the air on
K<."-.> #353<=@*HR<#3534=F* this second type of focal length are
mechanical clearances. Examples of #.G,@)(<G<1P7#&J<$2 L7F the
one side of the lens, through the lens and into the air on the other side.
front focal length (FFL, denoted ff I"@ * <+ F !1.33) and the back ?.)length
in figure %?.)#353$( " focal
*#.I8F 1 .8.#O$( " ?#/ %
The ray is broken into three segments by the lens. Two of these are external
(BFL, denoted fb).
principle points 4=F!5PF-.5G<*:<1'
(in the air), and the third is internal (in the glass). The external segments
The convention F$("all of the figures (with the exception of a single
2.#353;0)(14= in ?.)>+@<T/$K!%*:<1' (lens surface
usually within, the lens). The principal surface is the locus of all such
9#+!*#=5)$F. K(@$K!%T/$=vertices) #353<=@I8F1.8.#O
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$(7OC9#+!-.>?#/%Q ;0)(1<=@I"@L)F front focal length
(FFL) L:3 back focal length (BFL)

tc

te

primary vertex A1
optical axis
F
front focal
point

r1

ff
A
f


back focal
point

r2
H

H″

A2 secondary vertex
reversed ray locates front focal
point or primary principal surface

fb
B
f

F″


6/15/2009

2:29 PM

Page 1.30

!"/012*3(45'6 (principle plane)

Fundamental Optics

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6/15/2009 2:29 PM Page
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(principal plane)


1.29

FRONT FOCAL LENG

points นแปลงครั้งที!่ "
จุดเปลี่ยof intersection

of extended external ray segments. The principal
surface of a perfectly corrected optical system is a sphere centered on the
จุดเปลี่ยนแปลงครั้งที!#
่
focal point.

อากาศ

This length is the distan
vertex (A1).
Fundamental Optics

อากาศ
แก้ว
Near the optical axis, the principal surface is nearly flat, and for this reason,
it is sometimes referred to as the principal plane.



Definition of Terms

BACK FOCAL LENGT

This length is the distanc
point (F″).

SECONDARY PRINCIPAL SURFACE

FOCAL LENGTH ( fThis term is defined analogously to the primary principal surface, OR it ″ ) used
)
FOCAL POINT (F but F is

Optical Specifications

Two distinct terms describe the focal lengths associated with every lens or left and focused through back focal either focal point must be, on the
Rays that pass to the or originate at
for a collimated beam incident from the
EDGE-TO-FOCUS DI
opposite side of the lens, parallel to the
point F″ on the right. Rays in that part of the beam nearest the axis can be optical axis. This fact is the basis for
A is the distance from the
thought of as once refracted the tables of
size. The term f appears frequently in lens formulas and inat the secondary principal surface, instead of
B is the distance from t
standard lenses. Unfortunately, because f bymeasured with reference to
being refracted is both lens surfaces.
principal points which are usually inside the lens, the meaning of f is not
point. Both distances are
Let us imagine that rays originating at the front focal point F (and therefore
parallel to the optical axis after emergence from the opposite side of the lens)
PRIMARY PRINCIPAL POINT (H)
POINT
The second type of focal length relates the focal plane positions directlyOR FIRST NODAL some imaginary surface, instead of twice refracted (once
are singly refracted at
REAL IMAGE
to landmarks on the lens surfaces (namely the vertices) which are immeat each lens surface with happens. There is a unique imaginary
This point is the intersection of the primary principalsurface) as actuallythe
diately recognizable. It is not simply related to image size but is especially
surface, called the principal surface, at which this can happen. is one in w
A real image
optical axis.
convenient for use when one is concerned about correct lens positioning or
To locate this unique surface, consider a singlewere placed at theon
ray traced from the air point
mechanical clearances. Examples of this second type of focal length are the
front focal length (FFL, denoted ff in figure 1.33) and the back focal length
(BFL, denoted fb).
SECONDARY PRINCIPAL POINT (H ″ ) The ray is broken into three segments by the lens. Two of these are external
(in the air), and the third is internal (in the glass). The external segments
VIRTUAL IMAGE
OR SECONDARY exception POINT
The convention in all of the figures (with the NODALof a single
usually within, the lens). The principal surface is the locus of all does no
A virtual image such

This point is the intersection of the secondary principal surface with the
optical axis.
″
CONJUGATE DISTANCES (s AND s″ )

tc
primarydistancespointthe object distance, s, and secondary principal point
principal are
The conjugate
image distance, s″.
te

primary
Specifically, s principal surface from the object to H, and s″ is the distance
is the distance
ray from object the image location. The term infinite conjugate ratio refers to the
from H″ to at infinity
ray situation in at infinity
from object which a lens is either focusing incoming collimated light or is being
r
primary vertex A
optical axis
usedFto collimate a source (therefore, either s or s″ is infinity).
2

1

H

front focal
point

rial Properties

A virtual image can be v
system, such as in the c

r1

PRIMARY VERTEX (A 1 )
f
f

H″

A2 secondary vertex

fb

B
The primary vertex is A intersection of the primary lens surface with the
the
f
f
optical axis.

F-NUMBER (f/#)

The f-number (also know
a lens
back focal system is defined
point
clear aperture. Ray f-num
by the height at which it
F″

f /# =

f
.
CA

NUMERICAL APERTU
The NA of a lens system

F″

F″


H″

APPLICATION NOTE

F″

F″

H″

F″

H″

H″

Figure 1.36 indicates approximately where the principal points fall in
relation to the lens surfaces for various standard lens shapes. The exact
positions depend on the index of refraction of the lens material, and on
the lens radii, and can be found by formula. in extreme meniscus lens
shapes (short radii or steep curves), it is possible that both principal

Principal-Point Locations

F″

H″

H″

H″

F″

H″

Optical Coatings

Page 1.36

2:30 PM

Material Properties

F″
Fundamental Optics

For Quick Approximations
Much time and effort Figure 1.36saved by ignoring the differences
can be Principal points of common lenses
among f, fb, and ff in these Fundamental by assuming that f = fb = ff.
formulas Optics
1.36
Then s becomes the lens-to-object distance; s″ becomes the
lens-to-image distance; and the sum of conjugate distances
s=s″ becomes the object-to-image distance. This is known as
the thin-lens approximation.

123=)3(: (Nodal Points)

ear

APPLICATION NOTE

Physical Significance of the Nodal Points
A ray directed at the primary nodal point N of a lens appears to
emerge from the secondary nodal point N″ without change of
direction. Conversely, a ray directed at N″ appears to emerge
from N without change of direction. At the infinite conjugate
ratio, if a lens is rotated about a rotational axis orthogonal to
the optical axis at the secondary nodal point (i.e., if N″ is the
center of rotation), the image remains stationary during the
rotation. This fact is the basis for the nodal slide method for
measuring nodal-point location and the EFL of a lens. The nodal
points coincide with their corresponding principal points when
the image space and object space refractive indices are equal
(n = n″). This makes the nodal slide method the most precise
method of principal-point location.

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46)

F″

H″

Fundamental Optics

6/15/2009

F″



H″

H″

F″


50)

F″

H″

Fundamental Optics

48)

point is at the curved vertex, and the other is approximately one-third of
the way to the plane vertex.


ge 1.35

the lens radii, and can be found by formula. in extreme meniscus lens
shapes (short radii or steep curves), it is possible that both principal

M

Fund

points of intersection of extended external ray segments. The principal
surface of a perfectly corrected =,+F!*#=5)$F. ;<"( 1I:"'*B<,' (nodal slide bench)
7JL9<F%K!%?C";<"(:1.8.#O9.I"@?.) *&#+F!%8+!4=F8optical system:is a sphere centered on the
focal point.

FRONT FO

This length
vertex (A1).


=)3(:+6:3-.;)K-./0).H%J5?<&J?'45+D8%(;!*<.:)+-=3@'45
Near the optical )6/&*63F?@5*<H:5 surface
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it is &*@(<%>;;.:)+- C:>LA$=M$( :<
,(!.&J5?98FC+<G)*).GF*sometimes referred to as the+principal plane.
BACK FOC
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This length i
8&2 ) .:)+- $ :( ; 6/&*%?;C$)8&2 )PRINCIPAL SURFACE
SECONDARY LF * +( < .$,.80 ) !5 *
point (F″).
123%!&C+<6&5&4$*%.:J5?)??$6/'*<3F*)GF*< C+3<!5*
C$)8&2)GP>)(F)N5This term is '45 2 N2 'D$*%!(3%>@> primary principal surface, but it is used
*)123=)3(: defined analogously to the
1ch_FundamentalOptics_Final_a.qxd 6/15/2009 2:29 PM from 1.29left and focused to the back focal
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;CN5 C$F!OF* EB5< beam incident Page
EDGE-TOpoint F″ on the right. Rays3&2G+DH(part of the beam nearest the axis can be
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Fundamental Optics
point. Both d
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PRIMARY PRINCIPAL POINT (H) OR FIRST NODAL POINT

Definition of Terms

REAL IMA

FOCAL LENGTH ( f )

A real image
were placed

This point is the intersection of the primary principal surface with the
optical axis.
FOCAL POINT (F OR F ″ )


Two distinct terms describe the focal lengths associated with every lens or
Rays that pass through or originate at either focal point must be, on the
SECONDARY or equivalent focal length
lens system. The effective focal length (EFL) PRINCIPAL POINT ( H ″ ) side of the lens, parallel to the optical axis. This fact is the basis for
opposite
VIRTUAL I
OR SECONDARY NODAL POINT locating both focal points.
size. The term f appears frequently in lens formulas and in the tables of
A virtual im
This point is the intersection of the secondary principal surface with the
principal points which are usually inside the lens, the meaning of f is not
A therefore
Let us imagine that rays originating at the front focal point F (and virtual ima
optical axis.
parallel to the optical axis after emergence from the opposite side of the lens)such
system,
The second type of focal length relates the focal plane positions directly
are singly refracted at some imaginary surface, instead of twice refracted (once
to landmarks on the lens surfaces (namely the vertices) which are immeat each lens surface) as actually happens. There is a unique imaginary
″
CONJUGATE DISTANCES (s ANDsurface, called the principal surface, at which this can happen.
s″ )
F-NUMBER
To locate this and image distance, s″.
The conjugate distances are the object distance, s,unique surface, consider a single ray traced from the air on
front focal length (FFL, denoted ff in figure 1.33) and the back focal length
Specifically, s is the distance from the objectisto H, into three segments by the lens. Two of theseThe f-numbe
The ray broken and s″ is the distance
are external
(BFL, denoted fb).
a lens system
(in the air), and the thirdratio refers to the The external segments
is internal (in the glass).
from H″ to the image location. The term infinite conjugate
clear apertur
deliberately reversedsituationlightwhich a lensto right.
ray) is that in travels from left is either focusing incoming collimated light or issurface is the locus of all such
usually within, the lens). The principal being

!"#"('73#*5 (!"#"8'9:*;<"!"#"=0>)

by the heigh

Material Properties

used to collimate a source (therefore, either s or s″ is infinity).

f /# =

PRIMARY VERTEX ( A 1 )

tc


secondary surface with
The primary vertex is the intersection of the primary lensprincipal point the
t
optical axis.

f
CA


e

primary vertex A1
optical axis
F

SECONDARY VERTEX ( A 2 )

front focal
point

r1

back focal
point

r2
H

H″

A2 secondary vertex

F″

The secondary vertex is the intersection of thereversed ray locates front focal with
secondary lens surface
the optical axis. f
f
f

A
f

EFFECTIVE FOCAL LENGTH ( E F L , f )

b

B
f

NUMERIC

The NA of a
marginal ray
with the opti
The NA can b
with the opt

NA = nsinv.

Material Properties

2:29 PM

a lens system is define
from in to formulas and in the tables term infinite conjugate ratio refers to the
size. The term f appears frequently H″lensthe image location. Theof
clear aperture. Ray f-nu
standard lenses. Unfortunately, becausein is measured with either focusing incoming collimated light or is being
situation f which a lens is reference to
principal points which are usually inside the lens, the source (therefore, either s or s″ is infinity).
by the height at which
used to collimate a meaning of f is not
parallel to the optical axis after emergence from the opposite side of the lens)
f
are singly refracted at some imaginary surface, instead of twice refracted (once
f /# =
.
PRIMARY VERTEX (A 1 ) immeto landmarks on the lens surfaces (namely the vertices) which are
CA
at each lens surface) as actually happens. There is a unique imaginary
surface, called the principal surface, the
The primary vertex is the intersection of the primary lens surface withat which this can happen.
To locate this unique surface, consider a single rayNUMERICAL on
traced from the air APER
optical axis.
Page 1.30 (FFL, denoted ff in figure 1.33) and the back focal length
front focal length
The NA are lens system
The ray is broken into three segments by the lens. Two of theseof aexternal
(BFL, denoted fb).
(in the air), and the third is internal (in the glass). The externalray (the ray
SECONDARY VERTEX (A 2 )
marginal segments
the of all such
deliberately reversed ray) is that light travels from left tois the intersection of the secondary lens surface withsurface iswithlocusoptical axis mu
The secondary vertex right.
usually within, the lens). The principal
the

The NA can be defined
with the optical axis m

the optical axis.

6/15/2009

2:29 PM

Page 1.29

NA = nsinv.

EFFECTIVE FOCAL LENGTH ( EFL, f ) Fundamental Optics

t
Assuming that the lens is surrounded by air or vacuum (refractive index
t front focal point (F) to the primary
1.0), this is both surface
the distance from the
primary principal
Fundamental Optics
principal point (H) and the distance from the secondary principal point (H″)
ray from object at infinity point (F″). Later we use f to designate the paraxial EFL for back focal
to the rear focal
point
r
the design wavelength (l0).A
primary vertex
c

Optical Coatings

e

Definition of Terms
optical axis

F

2

1

H

front focal
point
FRONT
principal

H″

r1

F″

A2 secondary vertex

FOCAL LENGTH ( f f )
segments. The
point or primary F ″ )
FOCAL LENGTH ( f )
FOCAL POINT (F ORprincipal surface
f
f
a sphere centered on the
This length is the distance from the front focal point (F) to the primary
f

b

Two distinct terms describe the focal lengths associated with Optics or
1.30 Fundamental every lens
A
vertex (A1).
f
y flat, and for this reason, f appears frequently in lens formulas and in the tables of
size. The term
BACK FOCAL LENGTH ( f f )
principal points which are usually inside the lens, the meaning ofb is not

Rays that pass through or originate at either focal point must be, on the
B
opposite side of the lens, parallel to the optical axis. This fact is the basis for
f

This length is the distance from the secondary vertex (A2axis after emergence from the opposite side of the lens)
parallel to the optical ) to the rear focal
are edge thickness
point (F″).
A = front focus surfaces f = effective which are immete =singly refracted at some imaginaryradius of curvature ofrefracted (once
r1 = surface, instead of twice first
to landmarks on the lens to front (namely the vertices)focal length;
at each lens surface) as actually happens. (positive if center of
edge distance
surface There is a unique imaginary
cipal surface, but it isrecognizable. It is not simply related toor negativebut is especially
tc = center thickness
diately used
image size
surface, called the principal surface,curvature is to right)
at which this can happen.
about correct lens positioning or
ocused to the convenient for use when one is concerned = front focal length
back focal distance
focus
f
r = radius of curvature of second
To locate this
from the
EDGE-TO-FOCUS focal length are the
mechanical clearances. Examples of thisf second type ofDISTANCES (A AND B) unique surface,2consider a single ray tracedcenter ofair on
surface (negative if
m nearest the axis can length (FFL, denoted f in figure=1.33) and the back focal length
be
fb
back focal length
one side of the lens, through the lens and intoisthe air on the other side.
curvature to left)
front focal
f
A is the distance from the front focal pointThe ray is primary vertex of theby the lens. Two of these are external
to the broken into three segments lens.
ncipal surface, (BFL, denoted fb).
instead of
(in the air), andlens to is internal (in the glass). The external segments
the third the
B is figures (with the exception secondary
The convention in all of the the distance from the of a single vertex of the to a commonrear focal
can be extended
point of intersection (certainly near, and
deliberately
point. light travels from left
Focal length Both distances to right.
Figure 1.33 reversed ray) is thatand focal points are presumed always to be positive.
usually within, the lens). The principal surface is the locus of all such

T NODAL POINT

ncipal surface with the

incipal surface with

Fundamental Optics

REAL IMAGE

A real image is one in which the light rays actually converge; if a screen
t
were placed at the point of focus, an image would secondary principal point
be formed on it.
c

te

VIRTUAL IMAGE
primary vertex A1
optical
F
the axis A virtual image does not represent

back focal
point

r2

an actual convergence of light rays.
H H″
A secondary vertex
A virtual image can be viewed only by looking back through the optical
front focal
r
reversed
point
system, such as in the case of a magnifying glass. ray locates front focal

, and image distance, s″.
H, and s″ is the distance

2

1

ff

F-NUMBER

A
(f/#)
f


fb
B
f

The f-number (also known as the focal ratio, relative aperture, or speed) of
a lens system is defined to be the effective focal length divided by system

F″

1.29

MAGNIFICATION POWER

APPLICATION NOTE
Often, positive lenses intended for use as simple magnifiers are rated with
a single magnification, such as 4#. To create a virtual image for viewing
Technical Reference
with the human eye, in principle, any positive lens can be used at an infinite
number of possible magnifications. However, there is usually a narrow range
For further reading about the definition
of magnifications that will be comfortable for the viewer. Typically, when
here, refer to the following publication
the viewer adjusts the object distance so that the image appears to be
essentially at infinity (which is a comfortable viewing distance for most
Rudolph Kingslake, Lens Design Fun
individuals), magnification is given by the relationship
(Academic Press)
MAGNIFICATION POWER
254 mm
Rudolph Kingslake, System Design
magnification =
( in mm .
(1.31)
APPLICATION NOTE
Often, positive lenses intended for usefas simple)magnifiers are rated with
f
(Academic Press)
a single magnification, such as 4#. To create a virtual image for viewing
Technical Reference
Thus, the human eye, in length positive lens wouldcan be10# magnifier.
with a 25.4-mm focal principle, any positive lens be a used at an infinite
Warren Smith, Modern Optical Engi
number of possible magnifications. However, there is usually a narrow range
For furtherHill).
(McGraw reading about the defini
of magnifications that will be comfortable for the viewer. Typically, when
here, refer to the following publicat
DIOPTERS
Donald C. O’Shea, Elements of Mod
the viewer adjusts the object distance so that the image appears to be
(John Wiley & Sons)
The term diopter is used(which is a comfortableof the focal length, for most
essentially at infinity to define the reciprocal viewing distance which
Rudolph Kingslake, Lens Design
is commonly used for ophthalmic lenses.the relationship length of a lens
individuals), magnification is given by The inverse focal
Eugene Hecht, Optics (Addison Wes
expressed in diopters is
(Academic Press)
Max Born, Emil Wolf, Principles of O
254 mm
Rudolph University System
magnificationgreenstone.org
( f in mm ).
(1.31)
1000 =
(Cambridge Kingslake, Press) Desi
(1.32)
diopters =
( ff in mm ).
(Academic Press)
f
If you need help with the use of definit
http://www.spaceref.com/telescopes/Magniﬁcation-and-Using-Eyepieces.html
Thus, a 25.4-mm focal length positive lens would be a 10# magnifier.
Warren Smith, Modern Optical E
presented in this guide, our application
Thus, the smaller the focal length is, the larger the power in diopters will be.
(McGraw you.
pleased to assist Hill).

DIOPTERS
DEPTH OF FIELD AND DEPTH OF FOCUS
The term diopter is used to define the reciprocal of the focal length, which
In an imaging system, depth of field refers to the distance in object space
is commonly used for ophthalmic lenses. The inverse focal length of a lens
over which the system delivers an acceptably sharp image. The criteria for
expressed in diopters is
what is acceptably sharp is arbitrarily chosen by the user; depth of field
increases with increasing f-number.
1000
(1.32)
diopters =
( f in mm ).
For an imaging system, depth of focus is the range in image space over
f
which the system delivers an acceptably sharp image. In other words,
Thus, the amount that the image surface (such power in diopters will
this is the smaller the focal length is, the larger theas a screen or piece ofbe.
photographic film) could be moved while maintaining acceptable focus.
Again, criteriaFIELD AND DEPTH OF FOCUS
DEPTH OF for acceptability are defined arbitrarily.

Donald C. O’Shea, Elements of M
(John Wiley & Sons)
Eugene Hecht, Optics (Addison

Max Born, Emil Wolf, Principles
(Cambridge University Press)

If you need help with the use of def
presented in this guide, our applica
pleased to assist you.

In nonimaging applications, such as laser focusing, depth of focus refers to
In an imaging system, depth of field refers to the distance in object space
the range in the system delivers an acceptably sharp image. The remainsfor
over which image space over which the focused spot diameter criteria
below anacceptably sharp is arbitrarily chosen by the user; depth of field
what is arbitrary limit.

increases with increasing f-number.
For an imaging system, depth of focus is the range in image space over
which the system delivers an acceptably sharp image. In other words,
this is the amount that the image surface (such as a screen or piece of
photographic film) could be moved while maintaining acceptable focus.
Again, criteria for acceptability are defined arbitrarily.
In nonimaging applications, such as laser focusing, depth ofhttp://hyperphysics.phy-astr.gsu.edu/hbase/vision/accom2.html
focus refers to
the range in image space over which the focused spot diameter remains
below an arbitrary limit.

ch_FundamentalOptics_Final_a.qxd

6/15/2009

2:28 PM

Page 1.3

Paraxial Ray Approximation

Fundamental Optics


Paraxial Formulas
Sign Conventions

The validity of the paraxial lens formulas is dependent on adherence to the following sign conventions:

For lenses: (refer to figure 1.1)

For mirrors:

s is = for object to left of H (the first principal point)

f is = for convex (diverging) mirrors

s is 4 for object to right of H

f is 4 for concave (converging) mirrors

s″ is = for image to right of H″ (the second principal point)

s is = for object to left of H

s″ is 4 for image to left of H″


m is = for an inverted image

s″ is 4 for image to right of H″

m is 4 for an upright image

s″ is = for image to left of H″

Paraxial Rays and Paraxial Formula

When using the thin-lens approximation, simply refer to the left and right of the lens.

rear focal point

front focal point

Figure 1.1

h
object
f

v
F

H H″

CA

F″
image

f

h″

f

s

s″
principal points

Note location of object and image relative to front and rear focal points.
f = lens diameter
CA = clear aperture (typically 90% of f)

f

= effective focal length (EFL) which may be positive
(as shown) or negative. f represents both FH and
H″F″, assuming lens is surrounded by medium
of index 1.0

m = s ″/s = h″ / h = magnification or
conjugate ratio, said to be infinite if
either s ″ or s is infinite
v

Figure 1.1

= arcsin (CA/2s)

Sign conventions

s = object distance, positive for object (whether real
or virtual) to the left of principal point H

s ″ = image distance (s and s ″ are collectively called

conjugate distances, with object and image in
conjugate planes), positive for image (whether real
or virtual) to the right of principal point H″

h = object height
h ″ = image height


6/15/2009

2:28 PM

(object and image distance). Th
1ch_FundamentalOptics_Final_a.qxd 6/1
position, and image position is g

Page 1.3

1
1
1 www.cvimellesgriot.com
=
+
s s″
f
2:28 PM Page 1.4
1.4
Fundamental Optics

Fundamental Optics

Sign Conventions


K(@<7!<L#)G<).#L)@H(N9.4.%4(6<6.17#'&+!).#
*:+!)&$.85.$;0)(1G9@*98.3)(BK@!?J)("K!%#3BB *,F<
)J:(%K5.59#+ #353 conjugate (#353$(
Paraxial!Formulas 7OCL:3#353-.>)
1ch_FundamentalOptics_Final_a.qxd 6/15/2009
FundamentalOptics_Final_a.qxd 6/15/2009 2:28 PM Page
7.8&$.81(8>(<A'K!%18).#

This formula is referenced to f
2:28 PM Page 1.4

1ch_FundamentalOptics_Final_a.qxd on adherence to the following sign conventions:
The validity of the paraxial lens formulas is dependent 6/15/2009

For lenses: (refer to figure 1.1)

For mirrors:



step in optical
By definition,Typically, the firstconstraints suchproblm
magnification isasth
length based on

f is 4 for concave (converging) mirrors

6/15/2009

2:28 PM

Page 1.3



s″ is 4 for image to left of H″


(object and image distance). The rela
position, and image position is given
Fundamental Optics

s″ h″
Paraxial Formulas =
m
= 1=1+1
.
f
s
Typically, the firstTypically, the first step solving is to select a solvingfocal select a system focalh s s″
step in optical problem in optical problem system is to



6/15/2009


s″ is = for image to left of H″
2:28 PM Page 1.4

Fundamental Optics

Fundamental Optics
Fundamental Optics

s is = for object to left of H (the first principal point)


Sign Conventions
length based on constraints such as magnification or as magnificationan upright image distances
m is 4 for or conjugate
length based on constraints such conjugate distancesdependent on adherence objectThis formula is referenced to figure
The validity of the paraxial lens formulas is
to the following sign conventions:
object
(object and image distance). The relationship among relationshiptoamong focal length, mirrors:
focal length, object
For object
(object and image distance). The For lenses: (refer figure 1.1)
s is for object to of (the first
When using the Typically, the first stepsimply refer problem solving is to left select a system focal
thin-lens approximation, in optical to the left and=right of the Hlens. principal point)
position, and image position is given by
s is for
position, and image position is given4byobject right of H
length based on constraints such as magnificationtoor conjugate distances f is 4 for concave (converging) mirrors
By definition, magnification is the rat
object
F1
(object and image distance). The relationship for image to left of H″ length, object s is 4 for object to right of H
1
1
1
s″ is 4 among focal
=
+ position, and 1
(1.1)
1
1
= image
(1.1)
s s″ focal point + position is given by m is 4 for an upright image point
f
rear focal
front
h″
s″ is = for image to left of H″s ″
s s″
f
m=
= .
F
s
h
1
1
1
h
=
+
(1.1)
200
This formula is referenced to figure 1.1 and the sign conventions given
f
When using the thin-lens approximation, simply refer to the left and right of the lens.
This s s″
object
in Sign Conventions. formula is referenced to figure 1.1 and the sign conventions given
This relationship can be used to recas
H H″
f
CA
in Signv Conventions. to figure 1.1 and the F″ conventions given
200
This formula is referenced
F
forms:
Figure 1.2 Example 1 (f = 50 mm, s = 2
;"5</5.8 magnification is the ratio &+!1( size to h sign
(magnification)
By definition,)J:(%K5.5Conventions. of image"1F$< object size or
2
in Sign

This relationship can be used to
forms:

Fundamental Optics

).#H#38.2K!%*:<1'B.% (Thin-Lens Approximation)

1


object


(s + s ″)
f =m
By definition, magnification the ratio of is to select a to object
m 1.2 ( Figure 1.2 Exam
Typically, the first 7OC &+ optical
#39$F.%K<."K!%-.>7F!K<."K!%$(step in ! isproblem solvingimage size system focalsize(or + 1)sExample 1 (f = 50 mm,
+ s ″)
Figure
rear focal point

front focal point

h″

image

H H″

F

CA

F″

sm
f =
m +1


v

image

″

″

principal points

Note location of object and image relative to front and rear focal points.
f = lens diameter

= object distance, positive for object (whether real
or virtual) to the left of principal point H

CA = clear aperture (typically 90% of f)

= effective focal length (EFL) which may be positive
(as shown) or negative. represents both FH and
of index 1.0
=

v

f =

″ = ″ / = magnification or
conjugate ratio, said to be infinite if
either ″ or is infinite

= arcsin (CA/2 )

s + s″

1

″ = image distance ( and ″ are collectively called
conjugate distances, with object and image in
conjugate planes), positive for image (whether real
or virtual) to the right of principal point H″
= object height

1
m+2+
m
″ = image height

Optical Coatings


f

Material Properties

s( m + 1) = s + s ″
l Properties

where (s=s″) is the approxima

ecifications


s ″ hBy definition, magnification is the ratio of image size to object size or Example 2:fObject inside Focal Point
″ length based on constraints such as magnification or conjugate distances
=m
(1.2)
m=
= .
object ( m + 1)2 h
f
f
Example 2: Objec
(objects ″ image distance). The relationship among focal length, object
and h ″
s
h
m =″ and image position is given by
= .
The same (1.2) is2: Object inside Focal Po
object
Example placed 30 mm left of the
s
h″
f
f
position,
(1.2)
m=
= .h
sm
s s
s″
s
s f = is the image formed,
same lens. Where object is placed 30 mm and p
s
h
The same objectleft o
is w
This relationship can be used to recast the first formula into the following
The same m + 1
F
1
1
1 principal points
(See
same lens. Where is
?.)&$.81(8>(locationrelationshipimage be to recast5theand rear focal points. following figure 1.3.) Where is the image formed, a
<This of object and be *#.1.8.#O*K= < first formula into the
A'K!% = + used used front
(1.1) s same lens.
forms:
Note This relationship can can relative to to recast the first formula into the following
s s″
f
s + #353
1J9#(B*:<1'?#/%4=F8=&$.89<.&F.9<DF% figure 1.3.)(See figure 1.3.)
#353-.> s ″
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1
18).#!+F<fQ I"@"forms:
(%<=@
= forms:
lens diameter
s = object distance, positive for object s(whether real
f
=
−
1
f
( s + s ″ ) This formula is referenced to figure 1.1 and the sign conventions the left″ principal 301 H 2 + 1 200 1
or virtual) to given ?3!@ 50 1 point + 1
$(7OC L:3&$.85.$;0)(1s of .%!/%?.) m principle 1
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m =
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( m + 1)2fin=Sign(Conventions.
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″ m image s
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Figure 1.2 mm
By definition, magnification is both FH of image <5' toconjugate planes), positive″ image .%#39$F. ″ 1 (f = 50 mm,
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(1.4)
f =
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sm
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(1.4)
f =
Sign conventions
Figure 1.1 principle points (*#=5)$F. hiatus)Example 2: Objectapproximate obj
#353 30s 30 = < = = −75 =
?3*HR inside
where (s=s″) is the−2.5s ″ Focal Po
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f ″ = = magnification or
or virtual image is 2.5 mm high and upright.
h = object height
s s
m +=s ″ ″/s = h / h
s 1 h
30
imageFundamental Optics 1.3
is 2.5
#353#39$F.%$(7OCOD%-.> ).#I8ForThe sameJG9@#3BB*HRplacedthickness, the
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(1.5)
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same lens
is
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This relationship
into
following
LBB!5F=.%%F.5 theL:3*#.*#=5)).#H#38.2<=Whereallisreferenced to thea
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focal case, @$F. lens the image formed,
are virtual image is 2.5
s
In this lengthis being being used as a m
m forms:
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f =
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m
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center
1
v = arcsin (CA/2s)
H#38.2K!%*:<1'B.% viewed only back the lens.through the lens. lens
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1 In
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Figure 1.1 Sign conventions is = +approximate object-to-image distance.
s ″ = −75 mm
s (s=s″) O s +
(1.6)
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thickness, the image distance, object distance, and
(1.4)
f lens
Fundamental Optics 2.5 1.3
With a real lens of finite thickness, the approximate object-to-image distance.
s
30
m + is the
where (s=s″)1 referenced to the principal points, not to the physical
Example 1: Object outside Foc
focal length are all
focal length are all referenced sto the principal points, not to the physical
or virtual image is 2.5 mm high and uprig
+ neglecting the distance between the lens’ principal points,
s″
center of the lens. By
= the of finite thickness, lens’ principal points,
(1.5)
With a freal lensdistance between the the image distance, object distance, and
center of the lens. By neglecting
A 1-mm-high objectFis placed on the
1

With a real lens of finite thickne

Gaus

Optical Sp

Example 2: Object inside Focal Point In this case, the lens is being used
(1.2)
m
Typically, the first step in optical problem solving is to select a system focalThe same object is placed 30 mm left of the left principal point of back through the lens.
viewed only the
m=

m+2+

s″ h″
= .
s
h

same lens. Where is the image formed, and what is the magnification?

This
s( m as magnification or conjugate distances(See figure 1.3.)
Page 1.4
length based on constraints relationship1) = used to recast the first formula into the following
such + can be s + s ″
(1.6)
forms:
7($ s1″ 50 − 30object
(object and $!5F.%distance).4=F!5PFI+):!!)IH?.)?C";0)(1 length, object!5F.=%1 2:1 $(7OC4=F-.5G<#353;0)(1
image 1: $(7OC Thes relationship among focal
7(
( s ″)
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wheregiven) by
(1.3)
s ″ = −75
(m + 1
position, and image positionf is m(s=s″) is the approximate object-to-image distance.mm

2:28 PM

2

s″
75
$(7OC*"=5$)(=< distance, and
$(7OC4=F8=&$.81P%With=asm !5PFB<L)<4(6<'thickness, axis)
1 f mm lens of finite (optical the image distance,mobject−!5P=F4−=F7.JL9<F% 30 mm 4.%M@.5K!%?C"
=
25
select a system focal
real
s
30
Fundamental Optics
F
m +1
1
1F#353 1
conjugate distances 200 mm ?.)4.%M@.5K!% referenced to the principal principle image isthemm high5$)(< )J:(%K5.5K!%#3BB?38=
s + s″
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focal length are all principle points
points, not to 2.5 physical
(1.1)or virtual K!%*:<1'*"= and upright.
f =
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s
f
In this case,
focal length, object s″ 4=F8=&$.85.$;0)(+11lens. By neglecting%-.>L:3 between*HR<*4F.G"the lens is being used as a magnifier, and the image can be
centerm + them 50 mm 7JL9<F the distance
of 2
principal points,
&F. the lens’back through the lens.
K!%*:<1'
viewed only
( + )=s
known1as the
F image
)J:(% referenced sOCmfigure+ s ″hiatus, s=s″ becomes the object-to-image distance. This sim200
This formula is K5.5K!%$(7to!5PF4=FG" 1.1 and the sign conventions given

(1.4)

1

(1.5)

(1.6)

Material Properties

g is to select a system focal
(1.1)
ion or conjugate distances
mong focal length, object

2

plification,the approximate object-to-image distance.
F1
where (s=s″) is called the thin-lens approximation, can speed up calculation
when dealing with simple optical systems. and
With a real lens of finite thickness, the image distance, object distance,
focal length are all referenced to the principal points, not to the physical
center of the lens. By neglecting the distance between the lens’ principal points,
image
known as the hiatus, s=s″ becomes the object-to-image distance. This sim200 F2
plification, called F1 thin-lens approximation, can speed up calculation
the
when dealing with simple optical systems.

object

By definition, magnification is the ratioObject outside Focal Point
Example 1: of image size to object size or

n conventions given

Figure 1.2

66.7

F1

F1

object

image

Example 1 (f = 50 mm, s = 200 mm
F2

Material Properties
Optical Coatings

Optical Coatings

Example 2 (f = 50
(1.1)
image of the 2:
A 1-mm-high object is placed on the optical axis, 200 mmExampleleft Object inside Focal Point
left
s″ h″
(1.2)
m=
= .
principal point of a 1 (f66.750 mm, s = 200 mm, s″ 66.7 is the
Example 1: Example Focal =
Point
Figure 1.2 200Object outside LDX-25.0-51.0-C (f = 50 mm).=Wheremm) image
s
he sign conventionsor
o object size given h
Example 3:
Example 2 (f object is placed 475 mm) left of at left prin
Figure 1.3 The same = 50 mm, s = 30 mm, s″= 30 mmObject the Focal P
A 1-mm-high object is placed on the optical axis, 200 mm left of the left
formed, and what is the magnification? (See figure 1.2.)
principal (f of amm, s 200 mm,(f 66.7 Where
mm).
50 LDX-25.0-51.0-C
same lens. Where is A 1-mm-high object and whaton th
the image formed, is placed is th
e size This relationshipFigure 1.2 used to1recast the first formula mm) the following
to object size or
Example 3: Object at Focal Point
can be Example point=Object=insides″==50figureinto is the image
Example 2: is the magnification? (See
Focal Point
formed, and what
1.2.)
(1.2)
A 1-mm-high(See placed on the optical
object
50 mm left the first
$/A=4J ?.)1P7# isfigure 1.3.) axis,principalofpoint of an LDK-50.0-52.2
$/A ?.)1P 2: Object inside1Focal 1
forms: (1.2) =4J Example7# 1 1 Point
1 1=
principal point of the
The same= object − of the left principal point of the the left principal point ofan LDK-50.0-52.2-C (f =450 mm). Where is the image
is
− mm left placed 30 mm left of
The same object is placed 30 f
1
s
s
s″ ″ f s
formed, and what is 1 magnification? 1 figure 1.4.)
the
(See formed, and what is the magnificat
Where
formula into thefollowing same lens. Wherelens.image formed, and what is the magnification?
a into the following s + (See ) same is1the 1 1 is the image formed, and what is the magnification? = 50 − 30
1
1
1 s″
( s ″ figure 1.3.) 1 − 1
=
−
1
1
1
1
(See 1 s″ = 50 200
figure 1.3.)
f =m
1
1
(1.3) s″ −50 50s ″ = −75 mm
=
−
( m + 1)s2 = 50 − 30 s″s=″1 = mm − 200
s ″ = −25 mm
66.7 50
″
s ″ −50 50
1
1
(1.3)
s ″ −25
s ″ = −75 mm = s ″ − .7
66
m=
=
= −0.5 s ″
−75
= mm
m =50 =66.7 0.33
s.″ = −25 mm
sm
30
s
50 m =
s ″ s75 s ″ s= 200
−″
=
= −2 5
m=
=
= −2.5
(1.4)
f = (1.4)
(1.3)
s s30 = −75 mm
s
30
″ real image is 0.33 mm high and inverted.
s ″ −25
m + 1 or virtual imageor 2.5 mm high″ upright.7
s and 66.
is
m=
=
= −0.5
= 0.33
m″= −75
=
(1.5)
s
upright.
I"@-.>*18+!or virtual image ismm mm high and50
<9($7(@%K<." 2.5 2.5
s
s + sIn this case, the lens is being used as a2002.5and the image can be
″
magnifier,
m=
=s
f = (1.4) viewed only back through the lens. = −
(1.5)
s
30
1
G<)#2=<=@*:<1'?3OP)G,@LBBL$Flens is being used!%OP)magnifier, and
Fundamental Optics
1.4
(1.6)
In this case, the <K5.5L:3-.>?37@ as a0.5 mm high and
m +!)J:(%K5.58=Kreal image 2.5 mm mm -.>9($):(B
or virtual image is
or <." 0.33 is 0.33 high and and inverted.
9#+ 2 + m or virtual image is mm L:3*HR< high upright.
ge distance.
viewed only back through the lens.
8!%?.)"@.<9:(%K!%*:<1'*4F.<(@<
(1.5)
In
tance, object distance, and = s + s ″ this case, the lens is being used as a magnifier, and the image can be
s( m + 1)
(1.6)
Figure 1.3

object

points, not to the physical
viewed only back through the lens.
een the lens’ principal points,
1ch_FundamentalOptics_Final_a.qxd 6/15/2009 2:28 PM
F1
F2
(1.6)
o-image distance. This simobject
can speed up calculation

Page 1.5

where (s=s″) is the approximate Fundamental Optics
object-to-image distance.
1.4

7($!5F.% 3: $(7OC4=F?C";0)(1
image

tance. a real lens of finite thickness, the image distance, object distance, and
With

Fundamental Optics

$( 1P% all $.81P% 1 the 7(@%!5P points, 6<' !%
4(@%<=
(rays) 5(
focal length7are4=F8=&referenced to mmprincipalFB<L)<4(notKto the physical @$/A=).#*K=5<#(%1= Fundamental%1.8.#OG,@9.7JL9<F%
Optics
*:<1'*$@.&$.85.$;0)(1 50 mm 4=F#353 50 mm principal points, %K5.5K!%-.>G<#3BB paraxial system I"@
L:3)J:(
center of the lens. By neglecting the distance between the lens’?.)
F
known as the hiatus,point 4=F 1 K!%*:<1' 7JL9<F%-.>?3*)/"KD@< This sim- $/A=).#*K=5<<=@KD@<!5PF)(B18B(71/1!%H#3).#K!%
s=s″ becomes the object-to-image distance.
principle
).#G,@
object
F
plification, called the thin-lens approximation, can speed up calculation
4=FG" L:3)J:(%1 1 <*4F.G"
K5.5*HR
#3BBI"@L)F
object
image
peed up calculation
when dealing with 1simple optical systems.
=
−

Figure 1.3 Example 2 (f = 50 mm, s = 30 mm, s″= 475 mm)
xis, 200 mm left of the left
object distance, and
mm). Where is the image
,gure 1.2.) the physical Example 3: Object at Focal Point
not to
lens’ principal points, A 1-mm-high object is placed on the optical axis, 50 mm left of the first
principal point of an LDK-50.0-52.2-C (f =450 mm). Where is the image
F1
ge distance. This sim- formed, and what is the magnification? (See figure 1.4.)

2

1.#(%1=7@!%*"/<4.%?.)K<.<)(BL)<4(6<' TF.<*:<1'
Example 1: Object outside Focal Point
L:3:PF*K@. (9#+!:PF!!)1J9#(B*:<1'*$@.) ?C";0)(1
Figure 1.3 Example 2 (f = 50 mm, s = 30 mm
Figure 1.3 Example 2 (f 50 mm s = of mm, s″= %1=7@!%*"/<
A 1-mm-high object is placed on the upright. axis,=200 mm, left 30the left 475 mm) 4.%*K@.1PF 1st principle point K!%
optical
2.#(
0 mm left of the left or virtual image is 0.5 mm high and
principal point
L:37@!%*"/< 3: Object Focal Point
Where is the image of a LDX-25.0-51.0-C (f = 50 mm). Where is the image #3BB Example 4.%!!)?.)at2nd principle
Example 3: Object at Focal Point
.2.)formed, and what is the magnification? (See figure 1.2.)
point A K!%#3BB G<4/64.%*"=5$)(<)(B4=F*K@.optical axis, 50 m
8.
1-mm-high object is placed on the
A 1-mm-high=object iss placed on 425 mm)
the optical axis, 50 mm left ofand numerical aperture
Figure 1.5 F-number the first
Figure 1.4 Example 3 (f 450 mm, = 50 mm, s″=
():F.$&+image 8 point J)(BL)<4(6<'?37@!%*HR< (f
! 8C
1 1 1
principal point of an LDK-50.0-52.2-C (f =450 mm). Where is theprincipal4=F#(%1=4of an LDK-50.0-52.2-C8C8=450 mm). W
= $/A− J A simple graphical method can also be used to determine paraxial image
?.)1P7# and what is
Ray f-numbers can *"=5$)(<4( for any arbitrary ray
formed,
1.4.) also be defined
s ″ f =4 slocation and magnification. This graphicalthe magnification? (See figure and the diameter at which@%it*K@.L:3!!)) if its is the magnification? (See figure 1.4
formed, and what conjugate
distance
intersects the principal surface of
approach relies on two simple
the optical system are known.
properties of an 1
optical system. First, a
the system parallel
1
$/A=).#<=@OP)G,@G<4(@%1.87($!5F.%4=FTF.<8.G<#PH 1.2-1.4
1
1
1 optical axis crosses the optical1 rayatthat enterspoint. Second, a ray
1
1
1
=
− axis the
=
− to the the first principal point of50 system focal the system from the
=
−
that enters
exits
s ″ −50 the
s ″ 50 200 principal point parallel to its original direction (i.e., its exit angle
&$#>D%#3$(%I$@$sF.″ −50 50 approximation ?3
).#G,@ thin-lens
second
s ″ = −25 mm
NOTE ″ = −25 mm
with the optical axis is the same as its entrance angle). This method has
s ″ = 66.7 mm applied to the three previous examples illustrated in figures 1.2
4JG9@18B(7/K@!4=s !%%F.5KD@<*<+F!%?.) 1st principle
F1
been
s ″ −25
Because the sign convention given previously is not used universally
m=
=
= −0.5
through 1.4. Note that by using the thin-lens approximation, this second
in all optics texts, the reader may notice differences in the paraxial
s ″ 66.7
point L:3 2nd m = s ″ =point *HR<?C"*"=5$)(<
principle −25 = −
50
formulas. However, results will be correct as long as a consistent set 0.5
= 0.33 s
m = = property reduces to the statement that a ray passing through the center
s ″ −50 50
s ″ = −25 mm
object
s ″ −25
m=
=
= −0.5
s
50
F2

image

image

f

CA
2

v

F1


principal surface

of the
200 lens is undeviated.

9#+!?3I"@-.>*18+!<9(image is 0.50.5 mm and upright.
or virtual $7(@%K<." mm high
F-NUMBER AND NUMERICAL APERTURE

or real image isThe paraxial calculationsand inverted.the necessary element
0.33 mm high used to determine
diameter are based on the concepts of focal ratio (f-number or f/#) and
numerical aperture (NA). The f-number is the ratio of the focal length of the
lens to its “effective” diameter, the clear aperture (CA).
f-number =

f

.

(1.7)

of formulas and sign conventions is used.

s

50


s


second principal point parallel to its original direction (i.e., its exit angle
with the optical axis is the same as its entrance angle). This method has
been applied to the three previous examples illustrated in figures 1.2
through 1.4. Note that by using the thin-lens approximation, this second
property reduces to the statement that a ray passing through the center
of the lens is undeviated.

f-number =

f .
CA

NOTE

(1.7

Because the sign convention given previously is not used uni
in all optics texts, consider a lens notice positive focal the p
To visualize the f-number, the reader may with a differences in lengt
formulas. However, results light. The f-number defines consis
illuminated uniformly with collimatedwill be correct as long as athe ang
of the coneformulas and sign lens which ultimately forms the image. Th
of of light leaving the conventions is used.

F-NUMBER AND NUMERICAL APERTURE

is an important concept when the throughput or light-gathering powe

of an optical system /5.8 cone
focusing
G<4J<!%*"=5$)(<*#.1.8.#OG9@is<critical, such as whenangle light into a mono
).#&J<$<LBB paraxial 4=FG,@NUMERICAL.APERTURE
F-NUMBER AND &J<$2&F.7F %?3KD@<!5PF
chromator or projecting a high-power image.
"@$5&F. Numerical Aperture (NA) MDF%*HR<&F. sine
)(BL<$&/"K!%The paraxialratio (9#+!*#=5)F.to f-numberthe necessary element
focal calculations used determine
The other term used commonly in defining this cone angle is numeric
are based on
focal
K!%8C
9#+! f/#) diameter "1F$<#39$F.the concepts of the ratio of (f-number or f/#)#39$F.%aperture. The NA ray sine of the angle made byHthe marginal ray wit
MDF%*HR<1( aperture (NA).%&$.85.$;0)(17Fratio the focal length8ofand marginal is the )(BL)<4(6<' ?.)#P
!
numerical
The f-number is
the
the
5<&F NA referring to figure
“effective” diameter “effective”aperturethe clear aperture (CA). 1.5 *#.1.8.#O*K=optical.axis. ByI"@7.818).# 1.5 and using simple trigonometr
lens to its = clear diameter, (CA)
it can be seen that
2:28 PM

f-number =

Page 1.5

f .
CA

(1.7)

NA = sinv =

CA
2f

(1.8

To visualize the f-number, consider a lens with
&F. f-number G,@</5.88C8*H/" (angle of cone 9#+! a positive focal length and
illuminated uniformly with collimated light. The f-number defines the angle
1
cone angle) of the cone ofF*"/<4.%!!)?.)#3BB*:<1'L:3 forms the image. This NA = 2(f-number ) .
K!%L1%4= light leaving the lens which ultimately
Fundamental Optics
is an "<=@1J&( concept F!).##$8L1%K!%
1#@.%-.> L<$&/importantN8.)*8+when the throughput or light-gathering power
of an optical system is critical, such as when focusing light&F. a monointo f-number 1.8.#O</5.8I"@*18!1J9#(B#(%1=G"Q
#3BB4(6<6.17#'*K@.G):@?C"$/)U7 *,Fhigh-power image. F
chromator or projecting a < ).#;0)(1L1%1P
O@.
monochromatorother!term used%-.>4=FOP)in defining this cone angle is*#.#P@#353 conjugate L:3*1@<TF.<6P<5'):.%4=F>+@<T/$
The 9#+ ).#1#@. commonly K5.5KD@<
numerical
9<@.K!%#3BB4(6<6.17#'
aperture. The NA is the sine of the angle made by the marginal ray with

(1.9

Fundamental Optics

the optical axis. By referring to figure 1.5 and using simple trigonometry,
f
CA
it can be seen that
2
v

F

7/6/2009 1:42 PM
CA

NA = sinv =

1

7/6/2009

Page 1.6

2f

principal surface

and

F-number and numerical aperture

Figure 1.5

mine paraxial image
relies on two simple
s the system parallel
l point. Second, a ray
s the system from the
on (i.e., its exit angle
le). This method has
trated in figures 1.2
imation, this second
g through the center

Ray f-numbers can also be defined for any arbitrary ray if its conjugate
distance and the diameter at which it intersects the principal surface of
the optical system are known.

NA =

1
.
2( f-number )

(1.9)

Fundamental Optics


Imaging Properties of Lens Systems
Fundamental Optics

Fundamental Optics

mm, s″= 425 mm)

98.5*97C Page %1.6 optics 7F.%Q ).#*:+!)G,@
G<9<( 1+!
*&#+F!%98.5 + 9#+! - !.?8=&$.8L7)7F.%)(<IH (7.8
(1.8)
8C88!%K!%TP@*K=5<) L7FT:4=FI"@?37@!%*98+!<)(<

1:42 PM

Imaging Properties of Lens Systems

Fundamental Opti

NOTE

)J:(%K5.5K!%#3BB8=&F.*4F.)(B1("1F6/15/2009 NA K!%$(7Page 1.7
$<#39$F.% 2:28 PM OC
in all optics texts, the reader may notice differences in the paraxial
To understand the importance of the NA, consider its relation to magnification.
Two very! NAapplications of simple optics involve coupling light into
7F common K!%-.> T:<=@I8FKD@<!5PF)(B#3BB L:31.8.#OG,@
formulas. However,
Referring to figure 1.6, results will be correct as long as a consistent set
an optical fiber or into the entrance slit of a monochromator. Although these
THE OPTICAL
of formulas and sign conventions is used. INVARIANT
Example: System with
&J<$29.&F.*1@<TF.<6P<5'):.%K!%*:<1'4=F<@!54=F1C"G<)#2=4=F Fixed Input
problems appear to be quite different, they both have the same limitation
CA
NA (object side) = sin v =
8= relation *)= magnification.
I"@
(1.10)
To understand the importance of Page consider its K="?J)("to F5$)(BK<." aperture Two usually expressed
2s
1ch_FundamentalOptics_Final_a.qxd 6/15/2009 2:28 PM the NA,1.7 — they have a fixed NA. For monochromators, this limit is very common applications of simple op
side) = sin v =

CA = 2s sin v

(1.12)

and

NA″ (image side) = sin v ″ =


CA = 2s ″ sin v ″
leading to

CA
2s
CA
2s ″

(1.13)

which can be rearranged to show

s ″ sin v
NA
=
=
.
s sin v ″ NA ″

CA = 2s sin v

(1.14)

s″
Since
is simply thand
e magnification of the system,
s
we arrive at
m =

NA
.
NA ″

CA = 2s ″ sin v ″

leading to

(1.15)
s ″ sin v
NA
=
=
.
s s sin v equal to the ratio of the NAs
NA ″
The magnification of the system is therefore″

*8+F!#3BB*:<1'OP)<J8.G,@G<).#1#@problems *:<1'4=F8to be quite different, they
.%-.> appear =*1@<TF.<
6P<5'):.%G9NF?35!8G9@#(%1=TF.<I"@8.)KD@< (CA 8.)KD@<) 4JG9@
*#.I"@-.>4=F1$F.%8.)KD@< L7F*>#.3&$.81(8>(<A'#39$F.%)J:(%
K5.5)(B CA !.??38=K="?J)("4.%4UEV=4=F4JG9@).#*>/F8K<."
K!%#3BBI8F8=T:7F!&$.81$F.%K!%-.> it is necessary, using a singlet le
Suppose

Suppose it is necessary, using a singlet lens, to couple the output of an
(1.10)
— they optical fixed
incandescent bulb with a filament 1 mm in diameter into anhave afiber NA. For monochromato
as shown in figure 1.7. Assume that the fiber has terms diameter of
in a core of the f-number. In addition to the f
100 mm and an NA of 0.25, and that the design requires that the total
Fundamental Optics
entrance pupil (image) size.
(1.11)
distance from the source to the fiber be 110 mm. Which lenses are
appropriate?

By definition, the magnification must be 0.1. Letting s=s″ total 110bulb with a filament 1 mm in
incandescent mm
(using the thin-lens approximation), we can use equation 1.3,

*<+F!%?.) NA = CA/2s O@.&$.85.$;0)(1L:3)J:(%K5.5OP)Assume that the
(1.12)
as shown in figure 1.7.
)
( s + s ″$ &F. NA )R?3OP)G,@)J9<" CA "@$5*,F5)(< "(%<(@<
100 mm(see eq. 1.3) of 0.25, and that the
and an NA
)J9<"L:@ ,
f =m
(m + 1)
sdistance from the source to the fiber be
9.)#3BBI"@)J9<"&F. NA L:@$ ).#*>/F8K<."K!%#3BB*:<1')R
CA
(1.13)
2
to determine that the focal length is 9.1 mm. To .%G" *97C).#2'<=@*#.*#=5)$F.
appropriate?
?3I8F,F$5*>/F818##O<3L7F*>=5%!5Fdetermine the conjugate
v
distances, s and s″, we utilize equation 1.6,
CA
optical invariant
By definition, the magnification must be 0.
2

we arrive at

ecifications

When a lens or optical system is used to create an image of a source, it is
NA
object side
Fundamental the lens, 1.5
natural to assume that, by increasing the diameter (f) ofOptics thereby
m =
.
increasing its CA, we will be able to collect more light and thereby produce
NA ″
a brighter image. However, because of the relationship between magnification and NA, there can be a theoretical limit beyond which increasing the


(using the thin-lens approximation), we ca
(see eq. 1.6)
s ( m + 1) = s + s ″,
s″
(1.14)
CA
object side
on the object and image sides of the system. This powerful and useful result
2
s″
(s + s ″)
mm.
vSince
is completely independent of the specifics of the optical system, and it can
is simply the magnification and thethat s = 100 mm and s″ = 10 v″
of find system,
f =m
,
CA
s
often be used to determine the optimum lens diameter in situations
(m + 1)2
We can now use the relationship NA = CA/2s image = CA/2s″ to derive
or NA″ side
(1.9)
involving aperture constraints.
CA, the optimum clear aperture (effective diameter) of the lens.
(1.8)

Optical Coatings


e angle is numerical
he marginal ray with
simple trigonometry,

(1.11)

2s ″

which can be rearranged NA (object
to show

in terms of the f-number. In addition to the fixed NA, they both have a fixed
an optical fiber or into the entrance slit of a m

Fundamental Optics

positive focal length
mber defines the angle
orms the image. This
ght-gathering power
ng light into a mono-

ReferringCA figure 1.6,
to


Material Properties


(1.7)

Example: System with Fixed Input NA

necessary element
f-number or f/#) and
he focal length of the
A).

THE OPTICAL INVARIANT given previously is not used universally
Because the sign convention

Figure 1.6 Numerical aperture determine
With an image NA of 0.25 and an image distancetoand10 mm, that the focal length is 9.1 mm
(s″) of magnification
0.25 =

CA
20

distances, s and s″, we utilize equation 1.6

(1.15)

PM

Figure 1.6

Page 1.6

Numerical aperture and magnification

NA (object side) = sin v =

CA
2s

CA
2s ″


7($!5F.%).#*)/" optical invariant

(1.11)

Suppose it is necessary, using a singlet lens, to couple the output of an
incandescent bulb with a filament 1 mm in diameter into an optical fiber
1ch_FundamentalOptics_Final_a.qxd 0.1!
magnification = 2h″sin v
= s = 0.1 = 6/15/2009 2:28 PM Page 1.9
Fundamental Optics
CA
cation.
1.0
Two very commonhapplications of h″ 0.1 optics involve(1.12) optical systemfigure 1.7. Assume that the fiber has a core diameter of
simple
couplingas shown in
light into
magnification =
=
= 0.1!
and
100 mm and
f = 9.1 mm an NA of 0.25, and that the design requires that the total
h
1.0

an optical fiber or into the entrance slit of a monochromator. Although optical system
these

distance from
fiber
110
CA = 2
″
of Lens Systemss ″sin vquite=different, they both have the same limitation the source to theNA″ =beCA =mm. Which lenses are
problems appear to be NA CA = 0.025
appropriate?
leading to
0.25
(1.13)

f = 9.1 mm

Material Properties

Fundamental Optics

Optical Coatings

andmuch simplerwill not result in anyeffectivesystem throughput because of
It is diameter lens to calculate the greater (combined) focal length and
the limited input NA of the optical fiber. The singlet lenses in this catalog

lens
principal-point locations1and then use these results in any subsequent
that meet these criteria are LPX-5.0-5.2-C, which is plano-convex, and
paraxial calculations (see figure 1.8).which are biconvex. used in the optical
LDX-6.0-7.7-C and .
CA = 5 mmLDX-5.0-9.9-C, They can even be
invariant calculations described in the preceding section.

Symbols

fc = lens
combination focal length (EFL), positive if combination
finalcombination to the right of the combination secondary
focal point falls
principal point, negative otherwise (see figure 1.8c).

Making some simple calculations has reduced
to just
Accomplishing thischapter, Gaussian Beamour a single lens therefore requiresfocal length of the first element″(see figure 1.8a).
imaging task with choice of lenseshow to
f1 = an
zH
three. The following
Optics1 discusses
f,
optic with FOCAL LENGTH length and a 5-mm diameter. Using a larger
a choice of lenses based on various performance criteria.
9.1-mm focal
EFFECTIVE
make a final
(a)
(c)
f2 = focal length of the second element.
fc
diameter lens will not resultto calculate the effective focal length
The following formulas show how in any greater system throughput because of
s1 = d4f
d = distance
and limited input NA of the optical fiber. The singlet com- 1
the principal-point locations for a combination of any two arbitrarylenses in this catalog from the secondary principal point of the first
Hc
H than two lenses is very 2”
H2 H simple: Calculate
H1″
element to the primary principal point of the second element,
ponents. The approach for more 1
positive if the primary principal point is to the right of the
the values for the first two elements, then perform the same calculation for
LDX-6.0-7.7-C and LDX-5.0-9.9-C, whichuntil all lenses in the
secondary principal point, negative otherwise (see figure 1.8b).
this combination with the next lens. This is continued are biconvex.
lens 1
system are accounted for.

lens

Making some simple calculations has reduced our choice of lensessto = distance from the primary principal point of the first
just
and
1″
combination
lens 2
The expression for the combination focal length is the same whether lens
element to the final combination focal point (location of the
three. The following chapter, Gaussian Beam Optics, discusses how to
separation distances are large or small and whether f1 and f are positive
final image for an object at infinity to the right of both lenses),
make a final choice of lenses based on various2performance criteria.
or negative:
f f
f = (b) 1 2 .
f1 + f2 − d

d

positive if the focal point is to left of the first element’s
zH
primary principal point (see figure 1.8d).

s2″(1.16)

This may be Lens combination
Figure 1.8 more familiar in the form focal length and principal planes
d
1 1
1
= +
−
.
f
f1 f2
f1 f2

(1.17)

(d)

fc
s2″ = distance from the secondary principal point of the
second element to the final combination focal point (location
of the final image for an object at infinity to the left of both
lenses), positive if the focal point is to the right of the second
element’s secondary principal point (see figure 1.8b).
Fundamental Optics
1.9

Optical Coatings


nd NA,

H c″

Material Properties


Many optical tasks require several lenses in order to achieve an acceptable
CA
level0.25 = 5 mm. One possible approach to lens combinations is to conof performance.
CA =
H1
H1″
20
sider each image formed by each lens as the object for the next lens and so
Accomplishing this imaging task with a single lens therefore requires an
on. This iswith a 9.1-mm focal length timea 5-mm diameter.unnecessary.
optic a valid approach, but it is and consuming and Using a larger


magni. Thus,
either
nification and NA,
s value
ce (i.e.,
rred to

.25 =
Figure 1.9 “Extreme” meniscus-form lenses with
).#G,@*0:<1'#208)(<9:.57($1.8.#O,F$5*1#/818##O<3K!%#3BB4(6<6.17#'I"@ planes"(drawing not.>K!%*:<1'L#)*HR<
F$
external principal L<$&/ &+!).#G,@- to scale)
With an image NA of 0.25 and an image distance (s″) of 10 mm,
$(7OCK!%*:<1'O("IH #3BB4=OpticsNL1"%G<#PH4=F 1.8
F1J&(
Fundamental
1.6
and

Gaussian BeamOptical Coatings
Optics

roduce
ength and magnivalue of CA. Thus,
agnifistrained on either
ing the value
beyond this
performance
htness. (i.e.,
etimes referred to

Lens Combination Formulas

We With an image NA of 0.25 and an image distance (s″) of 10 mm, NA″ = CA/2s″ to derive
can now use the relationship NA = CA/2s or
CA

Fundamental Optics Properties
Material

ations
ge of a source, it is
f the lens, thereby
d thereby produce
between
ce, it ismagnifihich increasing the
hereby
image brightness.

Optical Coatings

#353#39$F.%$(7OC)(B-.> fixed --> optical invariant

Material Properties

filament
CA
CA
2s
2s
NA =
(1.10)
— they havesa fixedvNA.NAfilament
For=monochromators, = 0.025 is usually expressedthe magnification mustNA″0.1.2s″″ = 0.25 total 110 mm
By definition,
be = Letting s=s″
2s this limit
h = 1 mm sin
″
h . 1 mm
=
=
(using
s sin v ″ NA ″
(1.14)
Fundamental Optics
in terms of the f-number. In addition to the fixed NA, they both have a the thin-lens approximation), we can use equation 1.3,
fixed
CA 5 mm
CA = 5 mm=( s + s ″ )
s ″ with Fixed Input NA
Example: System simply the magnification of the system,
Since
is
(see eq. 1.3)
f =m
,
fiber core www.cvimellesgriot.com
(1.11)
s
fiber core
(m + 1)2
n to magnification.
h″ = 0.1 mm
Two very common applications of simple optics involve coupling light into
h″ = 0.1 mm
we arrive at
Suppose itfibernecessary, using a monochromator. Although these the output of an the focal length is 9.1 mm. To determine the conjugate
an optical is or into the entrance slit of a singlet lens, to couple
to determine that
s = 100 mm
s″ = 10 mm
NA
problems appear to be quite a filament mm the same limitation
have
m =
.
distances, s and
s theysbothmm in
s″ = 10
incandescent bulb withdifferent,=s100 1 110 mm diameter into an optical fiber s″, we utilize equation 1.6, mm
+ ″=
(1.10)
— they have a fixed NA ″ monochromators, this limit is usually expressed(1.15)
NA. For
(1.12)
s + s″ 110 Assume that they fiber has a
as shownofin figure=1.7. mm to the fixed NA, theboth have a fixed core diameter of
in terms the f-number. In addition
(see eq. 1.6)
s ( m + 1) = s + s ″,
Figure
pupil (image) of1.7 Optical that equal to the ratio of the the
100entrance The magnification of the system is thereforegeometry for focusingNAs that thean incandescent bulb into an optical fiber
mm and an NA size.0.25, andsystemthe design requires output of total
(1.11) COMBINATION SECONDARY PRINCIPAL-POINT LOCATION
on the object and image sides of the system. This powerful and useful result
and find are
it is necessary, using a singlet lens, to
of an
distance Optical approximation the specifics of the optical system, most can lenses that s = 100 mm into = 10 mm.
Figure Supposefrom theindependent ofis obviouslycouple the outputoutput of an incandescent bulb and s″ an optical fiber
1.7
Because the is completely source to the for focusing the and Which
thin-lens system geometry fiber be invalid mm. it
highly 110 for
(1.13)
incandescent bulbused tofilament 1 mm in diameter into an optical fiber
often ability toa determine the optimum lens secondary princibe with determine the location of the diameter in situations
We can now use the relationship NA 2 CA/2s or NA″4 CA/2s″ to derive
combinations, the
appropriate?
1 =
3 =
(1.12)
as shown in figure 1.7. Assume that the fiber has a core diameter of
involving aperture constraints.
pal pointmmvital an NA of 0.25, and that the designwhen another element
is and for accurate determination of d requires that the total
100
When
or optical system calculates the distance from the
By definition, the source to for this is used110create0.1. Lettingare it is total 110 mm NA of 0.25 and an image distance (s″) of 10 mm,
is added. Thefrom a lensformula the fiber be to mm. an image of a source,
With an image
distance simplest magnification must be Which lenses s=s″
the
natural point of that, by increasing the diameter (f) of secondary
(1.13) secondary principalto assume the final (second) element to the the lens, thereby
appropriate?
(using theincreasing its CA, we will be able to collect more light use equation 1.3,
thin-lens approximation), we can and thereby produce
CA
(1.14)
0.25 =
principal point of the combination (see figure 1.8b):
20
a brighter image. However, because of the relationship between magnifiBy definition, the magnification must be 0.1. Letting s=s″ total 110 mm
(s + ″)
cation ands approximation), we can use equation which
(using the thin-lensNA, there can be a theoretical limit beyond1.3, increasing the
(1.14)
and
brightness. (see eq. 1.3)
z f s2″ mf .
= = −diameter has no,effect on light-collection efficiency or image (1.19)
(m +″1)2
(s + s )
em,
d>0
f = m the NA of a ray is given by CA/2s, once a focal (see eq. 1.3)
,
Since
length and magniCA = 5 mm.
(m + 1)2
COMBINATION EXAMPLES
fication have been selected, the value of NA sets the value of CA. Thus,
Accomplishing 4
1 2
to determine thatdealing with system in which9.1 NA is PMTo Pageon1.8 the conjugate this imaging task with a single lens therefore requires an
that the focal length is 2:28 constrained
mm. determine
3
1ch_FundamentalOptics_Final_a.qxd a6/15/2009 todetermine the conjugateeither
to determine a lens combination or system exhibit principal planes
It is possible if one is the focal length is 9.1 mm. To the
for
Fundamental Optics
1.7
optic with a 9.1-mm focal length and a 5-mm diameter. Using a larger
theand s″,or image side, increasing the lens 1.6, beyond this value
objects″, we utilize equation diameter
distances, s andfrom the system. When such systems are themselves
that distances, s
are far removed we utilize equation 1.6,
diameter lens will not result in any greater system throughput because of
will increase system
but will not improve performance
(1.15) (1.15) combined, negative values of dsize and cost Probably the simplest example(i.e.,
the limited input NA of the optical fiber. The singlet lenses in this catalog
throughput or imagemay occur. This concept is sometimes referred to
brightness).
(see eq. with
s ( m as ) = s + s ″
of a negative+ 1the optical, invariant.
d-value situation is shown in figure 1.9. Meniscus lenses1.6)
e ratio of the NAs
(see eq. 1.6) and LDX-5.0-9.9-C, which are biconvex.
s ( m + have s + s ″,
LDX-6.0-7.7-C
steep surfaces1) = external principal planes. When two of these lenses are
ul and useful
he NAs result
and find that s = a negative s″ = of d can
brought into contact,100 mm andvalue10 mm. occur. Other combined-lens
Fundamental Optics
Making some simple calculations has reduced our choice of lenses to just 1.7
system, and it can
SAMPLE CALCULATION
l result
Fundamental
three. The following chapter, Gaussian Beam Optics, discusses how to Optics
in relationship NA = CA/2s or
eter in situations examples are shown thefigures 1.10 through 1.13. NA″ = CA/2s″ to derive
To understand how to use this relationship mm.
andWe canthatuse= 100 mm and s″ = 10between magnification and NA,
find now s
make a final choice of lenses based on various performance criteria.
d<0
d it can
CA, the optimum the following example. diameter) of the lens.
consider clear aperture (effective


Example:hich can be rearranged to showInput NA
System with Fixed
w

problems appear to be quite different, they both have the same limitation
— they have a fixed NA. For monochromators, this limit is usually expressed
in terms of the f-number. In addition to the fixed NA, they both have a fixed



(1.10)

H1

Symbols

ptable
o conngthso
and
ry.
omlate
th and
n for
quent
the
optical


fc = combination focal length (EFL), positive if combination
final focal point falls to the right of the combination secondary
f1

(a)
COMBINATION SECONDARY PRINCIPAL-POINT LOCATION

(b)

.16) of a negative d-value situation is shown in figure 1.9. Meniscus lenses with
culate

s2″ have external principal secondary principal point of the
from the planes.
steep surfaces= distance primary principal When two of these lensesthe
positive if the
point is to the right of are
ion for
second element to the final d can occur. Other point (location
brought into contact, aprincipal point, ofcombination focalcombined-lens
negative value negative otherwise (see figure 1.8b).
secondary
in the
examplesof the final in figures 1.10 through 1.13.
are shown image for an object at infinity toFundamental Optics
the left of both

mulas

the
next

(1.16)

ptable
o conand so
ned
(1.17)
ry.
ond
th and
of the
equent
e next
optical

.18)

mbined
length
econd
y comculate
ion for
in the
(1.18)

er
g lens
ositive

(1.16)

(1.17)

(c)

fc
H1c
H

1

2

H1″

3 4

lens
com

lens
combination

lens 1

zH
(d)

d>0

fc

(a)

f1
1

3 4
H1

H1″

Optical Coatings

f1 thin-lens approximation is obviously highly invalid for
Because the= focal length of the first element (see figure 1.8a). most
H1
H 2 H 2”
H1″
combinations, the ability to determine the location of the secondary princif2 = focal length of the second element.
pal point is vital for accurate determination of d when another element
Symbols
is added. d = distance formula for this calculates the distance the first
The simplest from lens 1
the secondary principal point of from the
and final (second) element to the secondary
secondary principal point of the
element to the primary2principal point of the second element,
lens
principal point of the combination (see figure 1.8b):
fc = combination focal length (EFL), positive if right of the
positive if the primary principal point is to thecombination
final focal point falls to the right of the combination secondary
z = s2″principal point, negative otherwise (see figure 1.8c). (1.19)
−f .
d

Figure 1.9
external pri

zH″

s1 = d4f1

s1″ = distance from the primary principal point of thes2″
first
f1 = focal length of the
lens COMBINATIONto the final combination focal point (location of the
element EXAMPLES first element (see figure 1.8a).
e 1.8
tive It is possible Figurelens combination at infinity to exhibit principal planes
final image for an object or system the right of principal planes
for a 1.8 Lens combination focal length and both lenses),
f2 = focal length of the second element.
length that are far removed from the system. When such systems are themselves
combined, negative values of dthe secondary principal point of the first
primary principal point (see figure 1.8d). simplest example
d = distance from may occur. Probably the
y com-

.17)
er lens
ositive


lens
combination

Material Properties

tical

H1″

Fundamental Optics
lens 1

and
mulas
uent

cifications

able
cond so

combined, negative values of d may occur. Probably the simplest example
of a negative d-value situation is shown in figure 1.9. Meniscus lenses with
steep surfaces have external principal planes. When two of these lenses are
H c″
brought into contact, a negative value of d can occur. Fundamental Optics
Other combined-lens
examples are shown in figures 1.10 through 1.13.

2 s1 = d4f1

H2 H2”

Fundamental Optics

1.9

lens 1
and
lens 2

d<0
lenses), distance if the the primary is to the right of the second
s1″ = positive from focal point principal point www.cvimellesgriot.com
of the first
d
element’s secondary principal point (see figure 1.8b). of the
element to the final combination focal point (location
(b)
s2″
zH = distance to the combination primary principal point
measuredprincipal point (see figure 1.8d). of the first element, Figure 1.8 Lens combination focal length and principal planes
primary from the primary principal point
external principal planes (drawing not to scale)
positive if the combination secondary principal point is to
s ″ = distance from principal point of second element
the2right of secondarythe secondary principal point of the
(see figure 1.8d).
H infinity to the left of both
of the final image H1 an object at 1″
for
H c″
zHSymbols toif the focal point issecondary principal point
″ = distance
lenses), positive the combination to the right of the second
element’s secondary principal point (see figure the second
measured from the secondary principal point of1.8b).
element, positive if the combination secondary principal point
lens
lens 1
zc the right of to focal length (EFL), positiveof the second
f H = distance the secondary principal point if combination
combination
is to = combinationthe combination primary principal point
measured from the primary principal the combinationelement,
final focal point falls to the right of point of the first secondary
element (see figure 1.8c).
positive ifpoint, negative otherwise (see figure 1.8c). is to
principal the combination secondary principal point
the right of secondary principal point coaxial combinations
Note: These paraxial formulas apply to of second element
zH″
f1
f1 thick and thin
of(see=figure length of the first element (see figure 1.8a).
both focal 1.8d). lenses immersed in air or any other
(a)
(c)
fc
fluid with refractive index independent of position. They
zfH″= focal lengththethe second element. principal point
= distance to of combination secondary
2
assume that light propagates from left to right through = d4f1
s1 an
measured from the secondary principal point of the second
opticaldistance from the secondary principal point of the first
system.
Hc
d=
H2 H2”
element, positive if H1 combination secondary principal point
the
H1″
element to the the secondary principal of the second element,
is to the right of primary principal point point of the second
positive if the primary principal point is to the right of the
lens 1
lens
and
Note: These paraxial formulas apply to coaxial combinations
combination
lens and
of 1″ = distance from the primary principal point of other
s both thick2 thin lenses immersed in air or any the first
fluid withto the finalindex independent of position. They the
element refractive combination focal point (location of
assume that for an object at infinityleft theright through an
zH
final image light propagates from to to right of both lenses),
optical system. focal point is to left of the first element’s
positive if the
(d)
d
(b)
s2″
fc
primary principal point (see figure 1.8d).

s ″ = distance from the secondary principal point of the
Figure 1.8 2 Lens combination focal length and principal planes
of the final image for an object at infinity to the left of both
lenses), positive if the focal point is to the right of the second
element’s secondary principal point (see figure 1.8b).

(c)

Fundamental Optics

1.9

(d)


The following formulas show how to calculate the effective focal length
and principal-point locations for a combination of any two arbitrary components. The approach for more than two lenses is very simple: Calculate
d>0
the values for the first two elements, then perform the same calculation for
this combination with the next lens. This is continued until all lenses in the
1 2
3 4
separation distances are large or small and whether f1 and f2 are positive
or negative:
f =

f1 f2
.
f1 + f2 − d

(1.16)
d<0

This may be more familiar in the form
d
1 1
1
= +
−
.
f
f1 f2
f1 2
external principalfplanes (drawing not to scale)

(1.17)

Notice that the formula is symmetric with respect to the interchange of the
lenses (end-for-end rotation of the combination) at constant d. The next
two formulas are not.
Hc″
COMBINATION FOCAL-POINT LOCATION

f1 = focal length of the first element (see figure 1.8a).
separation distances are large or small and whether f1 and f2 are positive
or negative: focal length of the second element.
f =
2

f1 f2
H the
H1″
f = d = distance. from 1 secondary principal point of the first
(1.16)
f1 + f2 − d
positive familiar in the form
This may be moreif the primary principal point is to the right of the
d
1 1
1
= + lens 1
−
.
(1.17)
f sf1″ = distanceffrom the primary principal point of the first
f2
f1 2
1
element to the final combination focal point (location of the
Notice that the formula is symmetric with respect to the interchange of the
lenses (end-for-end the focalof the combination) at constant d. The next
positive if rotation point is to left of the first element’s
two formulas are principal point (see figure 1.8d).f1
primary not.
(a)

s2″ = distance from the secondary principal point of the
COMBINATION FOCAL-POINT LOCATION
s1 =
second element to the final combination focal point (locationd4f1
For all values offinal2image for an objectof the focal point of theof both
f1, f , and d, the location at infinity to the left combined
of the
H
H2 H2”
H1″
system (s2″), measured from1the secondaryis to the right ofof the second
lenses), positive if the focal point principal point the second
lens (H2″), is given secondary principal point (see figure 1.8b).
element’s by
lens 1
z f=(distance to the combination primary principal point
f1 − d )
s2″ = H 2 and
.
(1.18)
measured− d the primary principal point of the first element,
f1 + f2 from
lens 2
positive if the combination secondary principal point is to
This cantheshownof secondary principal point of second solving
be right by setting s1=d4f1 (see figure 1.8a), and element
(see figure 1.8d).

1
1 1
d
=
(b) =zH″ + distance to the combination secondary principal point
s2″
f2 s1 s2″

Optical Coatings

Optical Coatings

combination
COMBINATION SECONDARY PRINCIPAL-POINT LOCATION

(1.19)

Fundamental Optics

2

1.9

d>0

COMBINATION EXAMPLES
It is possible for a lens combination or system to exhibit principal planes
that are far removed from the system. When such systems are themselves
combined, negative values of d may occur. Probably the simplest example
of a negative d-value situation is shown in figure 1.9. Meniscus lenses with
steep surfaces have external principal planes. When two of these lenses are
brought into contact, a negative value of d can occur. Other combined-lens
examples are shown in figures 1.10 through 1.13.

s2″ =
secon
of the
lenses
eleme

zH = d
measu
positiv
the rig
(see fi

zH″ =
measu
eleme
is to th
eleme

Note
of bot
fluid w
assum
optica

3 4

Optical Coatings

z = s2 ″ − f .

1

final im
positi
prima

1

3 4

2


Because the thin-lens approximation is obviously highly invalid for most
combinations, the ability to determine the location of the secondary princifor s2″.
zH
pal point is vital for accurate determination of d when another element
(d)
is added. The simplest formula for this calculates the distance from the
s2″
fc
secondary principal point of the final (second) element to the secondary
Fundamental Optics (see figure 1.8b):
1.8
principal point of the combination
rincipal planes

s1″ =
Figure 1.
eleme
externa

Fundamental Optics

measured from the secondary principal point of the second
element, positive if the combination secondary principal point
to the right of the secondary principal point of the second
Figure 1.8 isLens combination focal length and principal planes
zH″
for s2″.
f ( f − d)
Fundamental Optics
s2″ = 2(c)1
.
(1.18)
fc
f1 + f2 − d
Note: These paraxial formulas apply to coaxial combinations
of both thick and thin lenses immersed in air or any other
s1 = d4f1
Fundamental Optics
1.8
fluid with refractive index independent of position. They
Hc
This can be shown by setting s1=d4f1 (see figure 1.8a), and solving
assume that light propagates from left to right through an
optical system.
1
1 1
= +
f2 s1 s2″
lens
system (s2″), measured from the secondary principal point of the second
lens (H2″), is given by

Material Properties

Material Properties

1ch_FundamentalOptics_Final_a.qxd 6/15/2009 2:28 PM Page 1.9
lens
For all values of f1, f2, and d, the location of the focal point of the combined
combination




cipal planes
themselves
est example
s lenses with
se lenses are
mbined-lens

EFFECTIVE FOCAL LENGTH

Material Properties

(1.19)

paraxial calculations (see figure 1.8). 2 can even4be used in the optical
1 They
3
invariant calculations described in the preceding section.



ndary princiher element
ce from the
e secondary

d<0

H1

H1″

Hc″

lens


external principal planes (drawing not to scale)

Fundamental Optics

Fundamental Opt

z
s2″

d
f1

f2

f<0

f1

f2

f<0

combination
secondary
principal plane

focal plane

z<0

s2″

d

Figure 1.10 Positive lenses separated by distance greater
combination
focal plane
than f1 = f2: f is negative and both s2″ and z are positive. Lens
secondary
symmetry is not required.
principal plane
Figure 1.10 Positive lenses separated by distance greater
than f1 = f2: f is negative and both s2″ and z are positive. Lens
symmetry is not required.
f1




z

H2

H2″

H2

H1″

H2″

d

f2

tc
n

d
f
Figure 1.11 Achromatic combinations: Air-spaced lens
combinations can be made nearly achromatic, even though
both elements are made from the same material. Achieving
Figure 1.11 Achromatic combinations: Air-spaced lens
achromatism requires that, in the thin-lens approximation,
combinations can be made nearly achromatic, even though
2

Material Properties

Material Properties

Figure 1.12 Telep
combination
characteristic of the
secondary
the image size, can
principal plane
the first lens surface
positive lens followe
Figure 1.12 Telephoto
lens shapes shown i
characteristic of the teleph
f2 = 4f1/2. Then f i
the image size, can be mad
d = f1/2 (Galilean te
the first lens surface to the
d larger than f1 by T
positive lens followed/2.a
lens shapes shown inLDX-5
catalog lenses the fi
f2 = 4f1/2. Then f iswill yie
d = 78.2 mm, negat

d = f1/2 (Galilean telescope
d larger than f1/2. To make
catalog lenses LDX-50.8-13
d = 78.2 mm, will yield s2″

f1

H1″

combination
secondary
principal plan

both elements are made from the same material. Achieving
( f 1 + f2 requires that, in the thin-lens approximation,
achromatism )

d=

d=

2

.

( f 1 + f2 )

.

2
This is the basis for Huygens and Ramsden eyepieces.
This is the basis for Huygens and Ramsden eyepieces.

This approximation is adequate for most thick-lens situations.
The signs of f1, f2, and adequate for most thick-lens situations. a
This approximation is d are unrestricted, but d must have
value that guarantees theare unrestricted, but d must have a
The signs of f1, f2, and d existence of an air space. Element
shapes are unrestricted the existence chosen to compensate for
value that guarantees and can be of an air space. Element

Figure 1.13 Cond
vertices of a pair of i
Figure 1.13 Condenser
on contact. identica
vertices of a pair of(The len
d = 0, = f1 lenses cou
on contact. f (The /2 = f2/2
pal f1/2 of the 1/2
d = 0, f =point= f2/2, fseco
point the second ele
pal point ofof the combin

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