
 about lenses.  It is intended for photographers.  It 
 defines terms, gives a large number of formulas, discusses
 depth of field issues, vignetting, diffraction, lens aberrations,
 and the Modulation Transfer Function.


Archive-name: rec-photo/lenses/faq
PostingFrequency: monthly
Last-modified 1995/8/27

Frequently Asked Questions regarding lenses.
By David Jacobson
jacobson@hpl.hp.com


Q1. What is the meaning of the symbols in the rest of this FAQ?

A.  f       focal length
    So      distance from front principal point to subject (object)
    Sfar    distance from front principal point to farthest point in 
focus
    Sclose  distance from front principal point to closest point in 
focus
    Si      distance from rear principal point to film (image) plane
    M       magnification
    N       f-number or f-stop
    Ne      effective f-number (corrected for bellows factor)
    c       diameter of largest acceptable circle of confusion
    h       hyperfocal distance

See the technical notes at the end for more information on subject
distances, more information on the meaning of f-number and
limitations to be observed when applying these formulas to lenses in
which the aperture does not appear the same size front and rear.

Q2.  What is the meaning of focal length?  In other words, what about
a 50mm lens is 50mm?

A.  A 50mm lens produces an image of a distant object on the film that
is the same size as would be produced by a pinhole 50mm from the film.
See also Q5 below.

Q3. What meant by f-stop?

A. The focal length of the lens divided by the diameter of the
aperture (as seen from the front).  It is also called an f-number.
The brightness of the image on the film is inversely proportional
to the f-number squared.

Q4.  What is the basic formula for the conditions under which an image
is in focus?

A.  There are several forms.
        1/Si + 1/So = 1/f      (Gaussian form)
        (Si-f)*(So-f) = f^2    (Newtonian form)


Q5.  What is the formula for magnification?

A.  There are several forms.
        M = Si/So
        M = (Si-f)/f
        M = f/(So-f)

Q6. For a given lens and format what is angle of coverage?

A.  If the format has a width, height, or diagonal of distance X, the 
angle of coverage along width, height, or diagonal is 
2*arctan(X/(2*f*(M+1))).  For example a 35mm frame is 24x36 mm, so
with a 50 mm lens and a distant subject (i.e. M virtually zero), 
the coverage is 27 degrees by 40 degrees, with a diagonal of 47 degrees.
See the technical notes at the end for qualifications.


Q7.  How do I correct for bellows factor?

A.  Ne = N*(1+M)


Q8.  What is meant by circle of confusion?

A.  When a lens is defocused, a point in the subject gets rendered as
a small circle, called the circle of confusion.  If the circle of
confusion is small enough, the image will look sharp.  There is no one
circle "small enough" for all circumstances, but rather it depends on
how much the image will be enlarged, the quality of the rest of the
system, and even the subject.  Nevertheless, for 35mm work c=.03mm is
generally agreed on as the diameter of the acceptable circle of
confusion.  Another rule of thumb is c=1/1730 of the diagonal of the
frame, which comes to .025mm for 35mm film.  (Zeiss and Sinar are
known to be consistent with this rule.)


Q9.  What is hyperfocal distance?

A.  The closest distance that is in acceptable focus when the lens is
focused at infinity.  (See below for a variant use of this term.)

        h = f^2/(N*c)



Q10.  What are the closest and farthest points that will be in
acceptably sharp focus?

A.  Sclose = h * So / (h + (So - f))
    Sfar   = h * So / (h - (So - f))

or, defining hr = h/(So - f)

    Sclose = So * hr/(hr+1)
    Sfar   = So * hr/(hr-1)

Think of hr as the hyperfocal to subject distance ratio.  These
formulas are also correct when hr is defined as hr = h/So and the N
used in computing h is actually Ne.

If the denominator is zero or negative, Sfar is infinity.

Q11.  What is depth of field?

A.  It is convenient to think of a rear depth of field and a front
depth of field.  The rear depth of field is the distance from the
subject to the farthest point that is sharp and the front depth of
field is the distance from the closest point that is sharp to the
subject.  (Here we assume the lens is focused on the subject.)
Sometimes the term depth of field is used for the combination of these
two, i.e. the distance from the closest point that is sharp to the
farthest point that is sharp.

frontdepth = So - Sclose
frontdepth = Ne*c/(M^2 * (1 + (So-f)/h))
frontdepth = Ne*c/(M^2 * (1 + (N*c)/(f*M)))
frontdepth = So /(hr + 1)

reardepth = Sfar - So
reardepth = Ne*c/(M^2 * (1 - (So-f)/h))
reardepth = Ne*c/(M^2 * (1 - (N*c)/(f*M)))
reardepth = So/(hr - 1)

In the last three, if the denominator is zero or negative, reardepth is
infinity.  


Q12.  Where should I focus my lens so I will get everything from some
close point to infinity in focus?

A.  At approximately the hyperfocal distance.  More precisely, at 
So = h + f.  In this condition the closest point that will be in focus
is at half the subject distance.  (Some authorities use this as the
definition of hyperfocal distance.)


Q13.  I have heard that the depth of field depends only the the f-stop 
and
the magnification.  Is this true?

A.  Yes, under some conditions.  When the subject distance is small
with respect to the hyperfocal distance, the front and rear depth of
field are almost equal and depend only on the magnification and f-stop.
As the subject distance approaches the hyperfocal distance, the front
depth of field gets smaller and the rear depth gets larger, eventually
extending to infinity.

Q14.  Is there a simpler formula for depth of field?

A.  Yes.  When the subject distance is small with respect to the
hyperfocal distance, the following approximate formulas can be used.

Sfar = So + Ne*c/M^2  
Sclose = So - Ne*c/M^2

frontdepth = reardepth = Ne*c/M^2
frontdepth = reardepth = So/hr

Another simple approach is based on the formulas given above:

Sfar = So*hr/(hr-1)
Sclose = So*hr/(hr+1)
reardepth = So/(hr-1)
frontdepth = So/(hr+1)

These formulas can be used as follows. Suppose I know that So is 1/8th
of the hyperfocal distance.  Then the range of distances that is
acceptably sharp is from 8/9 of So to 8/7 of So.  The front and rear
depths of field are 1/9 So and 1/7 So.


Q15. I have heard that one should use a long lens to get a shallow depth
of field and a short lens to get a large depth of field.  Is this
true?

A.  Assuming that you frame the subject the same way, using a long
lens (and a correspondingly larger distance) does not make the depth
of field very much shorter.  It does make the front and rear depths
more even, but you probably didn't care about that very much.  Using a
short lens can make the rear depth of field very large, or even
infinite.  (See question 12.)  Now back to the long lens
issue.  Even though making the lens very long has little effect on the
maximum distance behind the subject at which points still appear to be
sharp, it has a big effect on how fuzzy very distant points appear.
Specifically, if the lens is focused on some nearby point rendered
with magnification M, a distant point at infinity will be rendered as
a circle of diameter C, given by

C = f M / N

which shows that the distant background point will be fuzzed out in
direct proportion to the focal length.


Q16.  If I focus on some point, and then recompose with that point not
in the center, will the focus be off?

A.  Yes, but maybe only a little bit.  If the object is far enough
away, the depth of field will cover the shift in distance.

An approximate formula for the minimum distance such that the error
will be covered by depth of field is given by 

d = w^2/(2 N c)

where
 d = minimum distance to make the point be sharply rendered
  d is measured from the film plane
 w = distance image point on the film is from center of the image

Thus for 35mm you can recompose the image with the subject at the edge
of the frame and still have it be sharp if the subject distance (at
the center) was at least 5.4 meters (18 feet) divided by the f-number.
See the technical notes at the end for a bunch of assumptions.


Q17.  What is vignetting and light falloff?  

A.  Vignetting is a reduction in light falling on the film far from
the center of the image that is caused by physical obstructions.
Independent of vignetting, even with an ideal rectilinear lens (one
that renders a square grid in subject space as a square grid on the
film) the light on the film falls off with Cos(theta)^4, where theta
is the angle a subject point is off the axis.  (With suitable optical
trickery this can be reduced a little, but never less than
Cos(theta)^3 in a rectilinear lens.  It can be made much smaller in a
fisheye lens.)

Q18.  How can I tell if a lens has vignetting, or if a filter is
causing vignetting?

A.  Open the back and, if necessary, trick the camera into opening the
shutter and stopping down.  Imagine putting your eye right in the
corner of the frame and looking at the diaphragm.  Or course, you
really can't do this, so you have to move your head and sight through
the corner of the frame, trying to imagine what you would see.  If you
"see" the entire opening in the diaphragm and through it to subject
space, there is no vignetting.  However, at wide apertures in most
lenses the edge of the rear element or the edge of the front element
or filter ring will obstruct your vision.  This is vignetting.  Try to
guess the fraction of the area of the diaphragm is this obstructed.
Log base two of this fraction is the falloff in f-stops at the corner.

You can also do this from the front.  With SLRs hold the camera a fair
distance away with a fairly bright area behind the viewfinder hole.
With non-SLRs open the back and arrange so a reasonably bright area is
behind the camera.  Look through the lens, and rotate the camera until
you are looking right at the corner of the viewing screen or frame.
Now for the hard part.  Look at the aperture you see.  If there is
vignetting you see something about the shape of an American football.
If the filter is causing the vignetting, one of the edges of the
football is formed by the filter ring.

A third way to detect vignetting is to aim the camera at a small
bright spot surrounded by a fairly dark background.  (A distant street
light at night would serve well.)  Deliberately defocus the image some
and observe the shape of the spot, particularly in the corners.  If it
is round there is no vignetting.  If it looks like the intersection of
some arcs (i.e. like an American football), then there is vignetting.
Note that near top of the image the top of the circle may get clipped
a bit.  This is because in many cameras some light (from the top part
of the image) misses the bottom of the mirror.  This affects only the
viewfinder, not the film.  You can use depth of field preview (if your
camera has it) to determine the f-stop at which the spot becomes
round.


Q19.  What is diffraction?

A.  When a beam of light passes through any aperture it spreads out.
This effect limits how sharp a lens can possibly be.


Q20.  What is the diffraction limit of a lens.

A.  A lens is diffraction limited at about 1500/N to 1800/N line pairs 
per mm.


Q21.  What are aberrations?

A.  Aberrations are image defects that result from limitations in the
way lenses can be designed.  Better lenses have smaller aberrations,
but aberrations can never be completely eliminated, just reduced.

The classic aberrations are:

* Spherical aberration.  Light passing through the edge of the lens is
focused at a different distance (closer in simple lenses) than light
striking the lens near the center.  

* Coma.  The distance from the axis at which an off-axis object point
is rendered varies with the distance from the center of the lens at
which the light passes.  In other words, magnification varies with the
distance from the center of the lens.  Off axis points are rendered
with tails, reminiscent of comets, hence the name.

* Astigmatism.  Off-axis points are blurred in their the radial or
tangential direction, and focusing can reduce one at the expense
of the other, but cannot bring both into focus at the same time.
(Optometrists apply the word "astigmatism" to a defect in the human
eye that causes *on-axis* points to be blurred along one axis or at 90
degrees to that axis.  That astigmatism is not quite the same as
astigmatism in photographic lenses.)


* Curvature of field.  Points in a plane get focused sharply on a
curved surface, rather than a plane (the film).  Or equivalently, the
set of points in the subject space that are sharp makes a curved
surface rather than a plane.  With a plane subject or a subject at
infinite distance the net effect is that when the center is in focus
the edges are out of focus, and if the edges are in focus the center
is out of focus.

* Distortion (pincushion and barrel).  The image of a square object
has sides that curve in or out.  (This should not be confused with the
natural perspective effects that become particularly noticeable with
wide angle lenses.)  This happens because the magnification is not a
constant, but rather varies with the angle from the axis.

* Chromatic aberration.  The position (forward and back) of sharp focus 
varies with the wavelength.

* Lateral color.  The magnification varies with wavelength.


Q22. Can I eliminate these aberrations by stopping down the lens?

A.  The effect of all aberrations except distortion and lateral color
is reduced by stopping down.  The amount of field curvature is not
affected by stopping down, but its effect on the film is.


Q23.  Why do objects look distorted when photographed with a wide
angle lens?

This is because the size of the image of an object depends on the
distance the object is from the lens.  This is not a defect in the
lens---even pinhole cameras with no lens at all exhibit this
perspective effect.

For image calculation purposes, think of the lens as being a geometric
point at one focal length in front of the film, and centered over the
center of the film.  (If the lens is not focussed at infinity, the
distance from the film gets larger.) Then the image of a subject point
can be found by drawing a straight line from the subject point through
the lens point and finding its intersection with the film. That line
represents one light ray. (Diffraction and out-of-focus conditions
have been ignored here, since they are irrelevant to this effect.)

If you do this, you'll find that the image of a nearby object will be
larger than the image of the same object farther away, by the ratio of
the distances. You'll also find that any straight line in the subject,
no matter at what angle or position, will be rendered as a straight
line on the film. (Proof outline: a line, and a point not on the line
define a plane.  All rays from the subject line will stay in the plane
defined by the line and the lens, and the intersection of that plane
with the film plane is a straight line.)

Q24. Why do people use long lenses to get "better perspective"? 

A.  A longer lens provides more subject magnification at a given
distance, so you can get farther from your subject without having the
image be too small. By moving back, you make the magnification ratio
between the front and back of your subject smaller, because the
distance ratio is smaller. So, in a portrait, instead of a nose that's
magnified much more than the rest of the head, the nose is magnified
only very slightly more than the rest of the head, and the picture
looks more pleasing.

_
                                                                                                                          


You can get the same perspective with a shorter focal length lens by
simply moving back, and enlarging the central portion of the image. Of
course, this magnifies grain as well, so it's better to use a longer
lens if you have one.


Q25. What is "MTF".

A.  MTF is an abbreviation for Modulation Transfer Function. It is
the normalized spatial frequency response of film or an optical
system.  The spatial frequency is usually measured in cycles per
millimeter.  For an ideal lens the MTF would be a constant 1 at all
frequencies.  For practical lenses, the MTF starts out near 1 and
falls off at increasing frequencies.  MTFs vary with the aperture, the
distance the image region is from the center, the direction of the
pattern (along a radius or 90 degrees to that), the color of the
light, and the subject distance.  Flare will lower the value of the
MTF even at zero spatial frequency.

Diffraction effects fundamentally limit the MTF of evan an ideal lens
to zero at frequencies beyond 1/(lambda*N) cycles per mm, where lambda
is the wavelength of the light.  For lambda = 555nm, the peak of the
eye's response, this is very close to 1800/N cycles per mm.

The MTF of a system is the product of the properly scaled MTFs of each
of its components, as long as there are not two consecutive
non-diffusing components.  (Thus with proper scaling you can multiply
camera lens MTF by film MTF by enlarger lens MTF by paper MTF, but
usually not a telescope objective MTF by an eyepiece MTF.  There are
also some other obscure conditions under which MTFs can be
multiplied.)

Note that although MTF is usually thought of as the spatial frequency
response function and is plotted with spatial frequency as the
abscissa, sometimes it is plotted at a specific spatial frequency with
distance from the center of the image as the abscissa.

Q26. What are "elements" and "groups", and are more better?

A. The number of elements is the number of pieces of glass used in the
lens.  If two or more are cemented together, that whole set is called
a group.  Thus a lens that has 8 elements in 7 groups has 8 pieces of
glass with 2 cemented together.  It is impossible to completely
correct all aberrations.  Each additional element the designer has at
his/her disposal gives a few more degrees of freedom to design out an
aberration.  So one would expect a 4 element Tessar to be better than
a 3 element Triotar.  However, each element also reflects a little
light, causing flare.  So too many elements is not good either.  Note
that an unscrupulous manufacturer could slap together 13 pieces of
glass and claim to have a 13 element lens, but it might be terrible.
So by itself the number of elements is no guarantee of quality.

Q27:  What is "low dispersion glass".

A.  Low dispersion glass is specially formulated to have a small
variation of index of refraction with wavelength.  This makes it
easier for the designer to reduce chromatic aberration and lateral
color.  This kind of glass is most often used in long lenses.

Q28:  What is an "aspheric element"?

A.  It a lens element in which the radius of curvature varies slightly
with angle off axis.  Aspheric elements give the lens designer more
degrees of freedom with which to correct aberrations.  They are most
often used in wide angle and zoom lenses.


Technical notes:

The subject distance, So, as used in the formulas is measured from the
subject to the lens's front principal point.  More commonly one hears
of the front nodal point.  These two points are equivalent if the
front medium and rear medium are the same, e.g. air.  They are the
effective position of the lens for measurements to the front.  In a
simple lens the front nodal/principal point is very near the center of
the lens.  If you know the focal length of the lens, you can easily
find the front nodal point by taking the lens off the camera and
forming an image of a distant object with the light going through the
lens backwards.  Find the point of sharp focus, then measure one focal
length back (i.e. toward the distant object).  That is the position of
the front nodal point.

On most cameras the focusing scale is calibrated to read the distance
from the subject to the film plane.  There is no easy way to precisely
convert between the focusing scale distance and So.

The formulas presented here all assume that the aperture looks the
same size front and rear.  If it does not, which is particularly
common in wide angle lenses, use the front diameter and note that the
formulas for bellows correction and depth of field will not be correct
at macro distances.  Formulas that are exact even with this condition
are given in the lens tutorial, posted separately.  

The formula for angle of coverage applies to rectilinear lenses. An
alternative form, 2*arctan(X/(2*Si)), applies to both rectilinear
lenses and pinholes.  (Rectilinear lenses give the same projection as
a pinhole.)  These formulas do not usually apply to fisheye lenses,
and can't possibly apply to a fisheye lens the covers 180 degrees or
more.

The conditions under which the formula for the minimum distance at
which the effect of focusing and re-composing will be covered by depth
of field are:

1.  w is no more than the focal length of the lens.  At the edge
w=18mm for 35mm, so this will very seldom be a problem.  2.  The
lens's two nodal points are not very widely separated.  But if the
front nodal point is in front of the rear nodal point, which I think
is the more common case, the formula is too conservative, so this is
not a problem either.  3.  The camera is rotated about the front nodal
point.  Almost always the camera will be rotated about an axis behind
the front nodal point which again makes the formula too conservative.
The guide number given assumes c=.03mm.

Acknowledgements

Thanks to Bill Tyler for contributing the section on perspective
effects.  The technique for detecting vignetting in the viewfinder was
suggested by a Maohai Huang.

                                                                                                                              

photographers.
 It is more detailed that a FAQ file, but less so than a text book.  
 It covers focusing, apertures, bellows correction, depth of field, 
 hyperfocal distance, diffraction, the Modulation Transfer 
 Function and illumination.


Archive-name: rec-photo/lenses/tutorial
Last-modified 1995/3/19

Lens Tutorial
by David M. Jacobson
jacobson@hpl.hp.com
Revised March 19, 1995

This note gives a tutorial on lenses and gives some common lens
formulas.  I attempted to make it between an FAQ (just simple facts)
and a textbook.  I generally give the starting point of an idea, and
then skip to the results, leaving out all the algebra.  If any part of
it is too detailed, just skip ahead to the result and go on.  

It is in 6 parts.  The first gives formulas relating subject and image
distances and magnification, the second discusses f-stops, the third
discusses depth of field, the fourth part discusses diffraction, the
fifth part discusses the Modulation Transfer Function, and the sixth
illumination.  The sixth part is authored by John Bercovitz.  Sometime
in the future I will edit it to have all parts use consistent notation
and format.

The theory is simplified to that for lenses with the same medium (eg
air) front and rear: the theory for underwater or oil immersion lenses
is a bit more complicated.


Subject distance, image distance, and magnification

In lens formulas it is convenient to measure distances from a set of
points called "principal points".  There are two of them, one for the
front of the lens and one for the rear, more properly called the
primary principal point and the secondary principal point.  While most
lens formulas expect the subject distance to be measured from the
front principal point, most focusing scales are calibrated to read the
distance from the subject to the film plane.  So you can't use the
distance on your focusing scale in most calculations, unless you only
need an approximate distance.  Another interpretation of principal
points is that a (probably virtual) object at the primary principal
point formed by light entering from the front will appear from the
rear to as a (probably virtual) image at the secondary principal point
with magnification exactly one.


"Nodal points" are the two points such that a light ray entering the
front of the lens and headed straight toward the front nodal point
will emerge going a straight way from the rear nodal point at exactly
the same angle to the lens's axis as the entering ray had.  The nodal
points are equivalent to the principal points when the front and rear
media are the same, eg air, so for practical purposes the terms can be
used interchangeably.  And again, the more proper terms are primary
nodal point and secondary nodal point.

In simple double convex lenses the two principal points are somewhere
inside the lens (actually 1/n-th the way from the surface to the
center, where n is the index of refraction), but in a complex lens
they can be almost anywhere, including outside the lens, or with the
rear principal point in front of the front principal point.  In a lens
with elements that are fixed relative to each other, the principal
points are fixed relative to the glass.  In zoom or internal focusing
lenses the principal points may move relative to the glass and each
other when zooming or focusing.

When the lens is focused at infinity, the rear principal point is
exactly one focal length in front of the film.  To find the front
principal point, take the lens off the camera and let light from a
distant object pass through it "backwards".  Find the point where the
image is formed, and measure toward the lens one focal length.  With
some lenses, particularly ultra wides, you can't do this, since the
image is not formed in front of the front element.  (This all assumes
that you know the focal length.  I suppose you can trust the
manufacturers numbers enough for educational purposes.)


So      subject (object) to front principal point distance.
Si      rear principal point to image distance
f       focal length
M       magnification

1/So + 1/Si = 1/f
M = Si/So 
(So-f)*(Si-f) = f^2
M = f/(So-f) = (Si-f)/f

If we interpret Si-f as the "extension" of the lens beyond infinity
focus, then we see that it is inversely proportional to a similar
"extension" of the subject.

For rays close to and nearly parallel to the axis (these are called
"paraxial" rays) we can approximately model most lenses with just two
planes perpendicular to the optic axis and located at the principal
points.  "Nearly parallel" means that for the angles involved, theta
~= sin(theta) ~= tan(theta).  ("~=" means approximately equal.)  These
planes are called principal planes.

The light can be thought of as proceeding to the front principal
plane, then jumping to a point in the rear principal plane exactly the
same displacement from the axis and simultaneously being refracted
(bent).  The angle of refraction is proportional the distance from the
center at which the ray strikes the plane and inversely proportional
to the focal length of the lens.  (The "front principal plane" is the
one associated with the front of the lens.  I could be behind the rear
principal plane.)


Apertures, f-stop, bellows correction factor, pupil magnification

We define more symbols

D       diameter of the entrance pupil, i.e. diameter of the aperture as
        seen from the front of the lens
N       f-number (or f-stop)  D = f/N, as in f/5.6
Ne      effective f-number (corrected for "bellows factor",
        but not absorption)

Light from a subject point spreads out in a cone whose base is the
entrance pupil.  (The entrance pupil is the virtual image of the
diaphragm formed by the lens elements in front of the diaphragm.)  The
fraction of the total light coming from the point that reaches the
film is proportional to the solid angle subtended by the cone.  If the
entrance pupil is distance y in front of the front nodal point, this
is approximately proportional to D^2/(So-y)^2.  (Usually we can ignore
y.)  If the magnification is M, the light from a tiny subject patch of
unit area gets spread out over an area M^2 on the film, and so the
brightness on the film is inversely proportional to M^2.  With some
algebraic manipulation and assuming y=0 it can be shown that the
relative brightness is

(D/So)^2/M^2 = 1/(N^2 * (1+M)^2).

Thus in the limit as So -> infinity and thus M -> 0, which is the usual
case, the brightness on the film is inversely proportional to the
square of the f-stop, N, and independent of the focal length.

For larger magnifications, M, the intensity on the film in is somewhat
less then what is indicated by just 1/N^2, and the correction is
called bellows factor.  The short answer is that bellows factor when
y=0 is just (1+M)^2.  We will first consider the general case when
y != 0.

Let us go back to the original formula for the relative brightness on
the film.

(D/(So-y))^2/M^2

The distance, y, that the aperture is in front of the front nodal
point, however, is not readily measurable.  It is more convenient to
use "pupil magnification".  Analogous to the entrance pupil is the
exit pupil, which is the virtual image of the diaphragm formed by any
lens elements behind the diaphragm.  The pupil magnification is the
ratio of exit pupil diameter to the entrance pupil diameter.

p       pupil magnification 
(exit_pupil_diameter/entrance_pupil_diameter)

For all symmetrical lenses and most normal lenses the aperture appears
the same from front and rear, so p~=1.  Wide angle lenses frequently
have p>1, while true telephoto lenses usually have p<1.  It can be
shown that y = f*(1-1/p), and substituting this into the above
equation and carrying out some algebraic manipulation yields that the
relative brightness on the film is proportional to

1/(N^2 ( 1 + M/p)^2)

Let us define Ne, the effective f-number, to be an f-number with the
lens focused at infinity (M=0) that would give the same relative
brightness on the film (ignoring light loss due to absorption and
reflection) as the actual f-number N does with magnification M.

Ne = N*(1+M/p)

An alternate, but less fundamental, explanation of bellows correction
is just the inverse square law applied to the exit pupil to film
distance.  Ne is exit_pupil_to_film_distance/exit_pupil_diameter.

It is convenient to think of the correction in terms of f-stops
(powers of two).  The correction in powers of two (stops) is
2*Log2(1+M/p) = 6.64386 Log10(1+M/p).  Note that for most normal
lenses y=0 and thus p=1, so the M/p can be replaced by just M in the
above equations.



Circle of confusion, depth of field and hyperfocal distance.

The light from a single subject point passing through the aperture is
converged by the lens into a cone with its tip at the film (if the
point is perfectly in focus) or slightly in front of or behind the
film (if the subject point is somewhat out of focus).  In the out of
focus case the point is rendered as a circle where the film cuts the
converging cone or the diverging cone on the other side of the image
point.  This circle is called the circle of confusion.  The farther
the tip of the cone, ie the image point, is away from the film, the
larger the circle of confusion.

Consider the situation of a "main subject" that is perfectly in
focus, and an "alternate subject point" this is in front of or
behind the subject.

Soa     alternate subject point to front principal point distance
Sia     rear principal point to alternate image point distance
h       hyperfocal distance
C       diameter of circle of confusion
c       diameter of largest acceptable circle of confusion
N       f-stop (focal length divided by diameter of entrance pupil)
Ne      effective f-stop Ne = N * (1+M/p)
D       the aperture (entrance pupil) diameter (D=f/N)
M       magnification (M=f/(So-f))

The diameter of the circle of confusion can be computed by similar
triangles, and then solved in terms of the lens parameters and subject
distances.  For a while let us assume unity pupil magnification, i.e. 
p=1.

When So is finite
C = D*(Sia-Si)/Sia = f^2*(So/Soa-1)/(N*(So-f))
When So = Infinity, 
C = f^2/(N Soa)

    
Note that in this formula C is positive when the alternate image point
is behind the film (i.e. the alternate subject point is in front of
the main subject) and negative in the opposite case.  In reality, the
circle of confusion is always positive and has a diameter equal to
Abs(C).

If the circle of confusion is small enough, given the magnification in
printing or projection, the optical quality throughout the system,
etc., the image will appear to be sharp.  Although there is no one
diameter that marks the boundary between fuzzy and clear, .03 mm is
generally used in 35mm work as the diameter of the acceptable circle
of confusion.  (I arrived at this by observing the depth of field
scales or charts on/with a number of lenses from Nikon, Pentax, Sigma,
and Zeiss.  All but the Zeiss lens came out around .03mm.  The Zeiss
lens appeared to be based on .025 mm.)  Call this diameter c.

If the lens is focused at infinity (so the rear principal point to film
distance equals the focal length), the distance to closest point that
will be acceptably rendered is called the hyperfocal distance.

h = f^2/(N*c)

If the main subject is at a finite distance, the closest
alternative point that is acceptably rendered is at at distance

Sclose = h So/(h + (So-F))

and the farthest alternative point that is acceptably rendered is at
distance

Sfar = h So/(h - (So - F))

except that if the denominator is zero or negative, Sfar = infinity.

We call Sfar-So the rear depth of field and So-Sclose the front depth
field.  

A form that is exact, even when P != 1, is 

depth of field = c Ne / (M^2 * (1 +or- (So-f)/h1)) 
               = c N (1+M/p) / (M^2 * (1 +or- (N c)/(f M))

where h1 = f^2/(N c), ie the hyperfocal distance given c, N, and f
and assuming P=1.  Use + for front depth of field and - for rear depth
of field.  If the denominator goes zero or negative, the rear depth of
field is infinity.  

This is a very nice equation.  It shows that for distances short with
respect to the hyperfocal distance, the depth of field is very close
to just c*Ne/M^2.  As the distance increases, the rear depth of field
gets larger than the front depth of field.  The rear depth of field is
twice the front depth of field when So-f is one third the hyperfocal
distance.  And when So-f = h1, the rear depth of field extends to
infinity.  

If we frame a subject the same way with two different lenses, i.e.
M is the same both both situations, the shorter focal length lens will
have less front depth of field and more rear depth of field at the
same effective f-stop.  (To a first approximation, the depth of field
is the same in both cases.)

Another important consideration when choosing a lens focal length is
how a distant background point will be rendered.  Points at infinity
are rendered as circles of size

C =  f M / N

So at constant subject magnification a distant background point will
be blurred in direct proportion to the focal length.

This is illustrated by the following example, in which lenses of 50mm
and 100 mm focal lengths are both set up to get a magnification of
1/10.  Both lenses are set to f/8.  The graph shows the circle of
confusions for points as a function of the distance behind the
subject.

circle of confusion (mm)    
     #               
     #               *** 100mm f/8
     #               ... 50mm f/8                                             
 0.8 #                                                               
*******  

     #                                                      *********         

     #                                             *********                  

     #                                         ****                           

     #                                    *****                               

     #                                ****                                    

 0.6 #                            ****                                        

     #                       *****                                   
.......  

     #                    ***                      ..................         

     #                  **            .............                           

 0.4 #              ****     .........                                        

     #           ***     ....                                                 

     #         **   .....                                                     

     #        * ....                                                          

     #      **..                                                              

 0.2 #    **.                                                                 

     #  .*.                                                                   

     # **                                                                     

     #*                                                                       

     
*######################################################################  

   0 #                                                                        

             250    500       750     1000     1250    1500     1750     
2000 

                   distance behind subject (mm)

The standard .03mm circle of confusion criterion is clear down in the
ascii fuzz.  The slope of both graphs is the same near the origin,
showing that to a first approximation both lenses have the same depth
of field.  However, the limiting size of the circle of confusion as
the distance behind the subject goes to infinity is twice as large for
the 100mm lens as for the 50mm lens.


Diffraction

When a beam of parallel light passes through a circular aperture it
spreads out a little, a phenomenon known as diffraction.  The smaller
the aperture, the more the spreading.  The field strength (of the
electric or magnetic field) at angle phi from the axis is
proportional to

lambda/(phi Pi R) * BesselJ1(2 phi Pi R/lambda), 

where R is the radius of the aperture, lambda is the wavelength of the
light, and BesselJ1 is the first order Bessel function. The power
(intensity) is proportional to the square of this.

The field strength function forms a bell-shaped curve, but unlike the
classic E^(-x^2) one, it eventually oscillates about zero.  Its first
zero at 1.21967 lambda/(2 R).  There are actually an infinite number
of lobes after this, but about 86% of the power is in the circle
bounded by the first zero.


    Relative field strength

     ***                                                                      

_
                        


   1 #  ****                                                                  

     #      **                                                                

 0.8 #        *                                                               

     #         **                                                             

     #           *                                                            

     #            **                                                          

     #              *                                                         

 0.6 #               *                                                        

     #                *                                                       

     #                 *                                                      

 0.4 #                  *                                                     

     #                   *                                                    

     #                    **                                                  

 0.2 #                      **                                                

     #                        **                                              

     #                          **                         
*****************  

   
###############################*###################*****################
###

     #                             *****        ******                        

     #          0.5         1          1.5******    2         2.5          
3  

                                                                              


        Angle from axis (relative to lambda/diameter_of_aperture)


Approximating the diaphragm to film distance as f and making use of
the fact that the aperture has diameter f/N, it follows directly that
the diameter of the first zero of the diffraction pattern is
2.43934*N*lambda.  Applying this in a normal photographic situation is
difficult, since the light contains a whole spectrum of colors.  We
really need to integrate over the visible spectrum.  The eye has
maximum sensitive around 555 nm, in the yellow green.  If, for
simplicity, we take 555 nm as the wavelength, the diameter of the
first zero, in mm, comes out to be 0.00135383 N.

As was mentioned above, the normally accepted circle of confusion for
depth of field is .03 mm, but .03/0.00135383 = 22.1594, so we can
see that at f/22 the diameter of the first zero of the diffraction
pattern is as large is the acceptable circle of confusion.

A common way of rating the resolution of a lens is in line pairs per
mm. It is hard to say when lines are resolvable, but suppose that we
use a criterion that the center of the dark area receive no more than
80% of the light power striking the center of the lightest areas.
Then the resolution is 0.823 /(lambda*N) lpmm.  If we again assume 555
nm, this comes out to 1482/N lpmm, which is in close agreement with
the widely used rule of thumb that the resolution is diffraction
limited to 1500/N lpmm.  However, note that the MTF, discussed below,
provides another view of this subject.


Modulation Transfer Function

The modulation transfer function is a measure of the extent to which a
lens, film, etc., can reproduce detail in an image.  It is the spatial
analog of frequency response in an electrical system.  The exact
definition of the modulation transfer function and the related
optical transfer function varies slightly amongst different
authorities.

The 2-dimensional Fourier transform of the point spread function is
known as the optical transfer function (OTF).  The value of this
function along any radius is the fourier transform of the line spread
function in the same direction.  The modulation transfer function is
the absolute value of the fourier transform of the line spread
function.  

Equivalently, the modulation transfer function of a lens is the ratio
of relative image contrast divided by relative subject contrast of a
subject with sinusoidally varying brightness as a function of spatial
frequency (e.g. cycles per mm).  Relative contrast is defined as
(Imax-Imin)/(Imax+Imin).  MTF can also be used for film, but since
film has a non-linear characteristic curve, the density is first
transformed back to the equivalent intensity by applying the inverse of
the characteristic curve.

For a lens the MTF can vary with almost every conceivable parameter,
including f-stop, subject distance, distance of the point from the
center, direction of modulation, and spectral distribution of the
light.  The two standard directions are radial (also known as
saggital) and tangential.

The MTF for an an ideal diffraction-free lens is a constant 1 from 0
to infinity at every point and direction.  For a practical lens it
starts out near 1, and falls off with increasing spatial frequency,
with the falloff being worse at the edges than at the center.  Flare
would make the MTF of a lens be less than one even at zero spatial
frequency.  Adjacency effects in film can make the MTF of film be
greater than 1 in certain frequency ranges.

An advantage of the MTF as a measure of performance is that under some
circumstances the MTF of the system is the product (frequency by
frequency) of the (properly scaled) MTFs of its components.  Such
multiplication is always allowed when each step accepts as input
solely the intensity of the output of the previous state, or some
function of that intensity.  Thus it is legitimate to multiply lens
and film MTFs or the MTFs of a two lens system with a diffuser in the
middle.  However, the MTFs of cascaded ordinary lenses can
legitimately be multiplied only when a set of quite restrictive and
technical conditions is satisfied.

As an example of some OTF/MTF functions, below are the OTFs of pure
diffraction for an f/22 aperture, and the OTF induced by a .03mm
circle of confusion of a de-focused but otherwise perfect and
diffraction free lens.  (Note that these cannot be multiplied.)

Let lambda be the wavelength of the light, and spf the spatial
frequency in cycles per mm.

For diffraction the formulas is

OTF(lambda,N,spf) = ArcCos(lambda*N*spf) - 
 lambda*N*spf*Sqrt(1-(lambda*N*spf)^2)    if lambda*N*spf <=1
    = 0     if lambda*N*spf >=1

Note that for lambda = 555 nm, the OTF is zero at spatial frequencies
of 1801/N cycles per mm and beyond.

For a circle of confusion of diameter C, 

OTF(C,spf) = 2 * BesselJ1(Pi C spf)/(Pi C spf)

This goes negative at certain frequencies.  Physically, this would
mean that if the test pattern were lighter right on the optical center
then nearby, the image would be darker right on the optical center
than nearby.  The MTF is the absolute value of this function.  Some
authorities use the term "spurious resolution" for spatial frequencies
beyond the first zero.

Here is a graph of the OTF of both a .03mm circle of confusion and an
f/22 diffraction limit.

  OTF

     #
   1 ***
     #..**
     #  ..**
     #    ..*             ***   .03 mm circle of confusion
 0.8 #      .*            ...   555nm f/22 diffraction
     #        *.
     #         *..
     #          * ..
 0.6 #           *  .
     #            *  .
     #             *  ..
     #              *   ...
 0.4 #               *     ..
     #               *       ..
     #                *        ..
     #                *          ...
 0.2 #                 *            ...
     #                  *              ...
     #                   *                ...
     #                    **                 ..      **********
   
#########################*##################.*****..........*****.......
.##
     #                       **              ***                    
********
     #          20         40  ***     60****      80         100         
120
     #                           ********
     #
                           spatial frequency (cycles/mm)


Although this graph is linear in both axes, the typical MTF is
presented in a log-log plot.



Illumination  
(by John Bercovitz)

   The Photometric System

        Light flux, for the purposes of illumination engineering, is
measured in lumens.  A lumen of light, no matter what its wavelength
(color), appears equally bright to the human eye.  The human eye has a
stronger response to some wavelengths of light than to other
wavelengths.  The strongest response for the light-adapted eye (when
scene luminance >= .001 Lambert) comes at a wavelength of 555 nm.  A
light-adapted eye is said to be operating in the photopic region.  A
dark-adapted eye is operating in the scotopic region (scene luminance
</= 10^-8 Lambert).  In between is the mesopic region.  The peak
response of the eye shifts from 555 nm to 510 nm as scene luminance is
decreased from the photopic region to the scotopic region.  The
standard lumen is approximately 1/680 of a watt of radiant energy at
555 nm.  Standard values for other wavelengths are based on the
photopic response curve and are given with two-place accuracy by the
table below.  The values are correct no matter what region you're
operating in - they're based only on the photopic region.  If you're
operating in a different region, there are corrections to apply to
obtain the eye's relative response, but this doesn't change the
standard values given below.

Wavelength, nm   Lumens/watt         Wavelength, nm  Lumens/watt
      400           0.27                600              430
      450          26                   650               73
      500         220                   700                2.8
      550         680

 Following are the standard units used in photometry with their 
definitions and symbols.

 Luminous flux, F, is measured in lumens.
 Quantity of light, Q, is measured in lumen-hours or lumen-seconds.  
It is the time integral of luminous flux.
 Luminous Intensity, I, is measured in candles, candlepower, or 
candela (all the same thing).  It is a measure of how much flux is 
flowing 
through a solid angle.  A lumen per steradian is a candle.  There are 4 
pi 
steradians to a complete solid angle.  A unit area at unit distance from 
a 
point source covers a steradian.  This follows from the fact that the 
surface area of a sphere is 4 pi r^2.  
 Lamps are measured in MSCP, mean spherical candlepower.  If you 
multiply MSCP by 4 pi, you have the lumen output of the lamp.  In the 
case of 
an ordinary lamp which has a horizontal filament when it is burning base 
down, roughly 3 steradians are ineffectual: one is wiped out by inter- 
ference from the base and two more are very low intensity since not much 
light comes off either end of the filament.  So figure the MSCP should 
be 
multiplied by 4/3 to get the candles coming off perpendicular to the 
lamp 
filament.  Incidentally, the number of lumens coming from an 
incandescent 
lamp varies approximately as the 3.6 power of the voltage.  This can be 
really important if you are using a lamp of known candlepower to 
calibrate a photometer.
 Illumination (illuminance), E, is the _areal density_ of incident 
luminous flux: how many lumens per unit area.  A lumen per square foot 
is 
a foot-candle; a one square foot area on the surface of a sphere of 
radius 
one foot and having a one candle point source centered in it would 
therefore have an illumination of one foot-candle due to the one lumen 
falling on it.  If you substitute meter for foot you have a meter-candle 
or lux.  In this case you still have the flux of one steradian but now 
it's 
spread out over one square meter.  Multiply an illumination level in lux 
by 
.0929 to convert it to foot-candles.  (foot/meter)^2= .0929.  A 
centimeter-
candle is a phot.  Illumination from a point source falls off as the 
square 
of the distance.  So if you divide the intensity of a point source in 
candles 
by the distance from it in feet squared, you have the illumination in 
foot 
candles at that distance.
 Luminance, B, is the _areal intensity_ of an extended diffuse source 
or an extended diffuse reflector.  If a perfectly diffuse, perfectly 
reflecting surface has one foot-candle (one lumen per square foot) of 
illumination falling on it, its luminance is one foot-Lambert or 1/pi
candles per square foot.  The total amount of flux coming off this 
perfectly diffuse, perfectly reflecting surface is, of course, one lumen 
per 
square foot.  Looking at it another way, if you have a one square foot 
diffuse source that has a luminance of one candle per square foot (pi 
times 
as much intensity as in the previous example), then the total output of 
this source is pi lumens.  If you travel out a good distance along the 
normal to the center of this one square foot surface, it will look like 
a 
point source with an  intensity of one candle.  
 To contrast: Intensity in candles is for a point source while 
luminance in candles per square foot is for an extended source - 
luminance 
is intensity per unit area.  If it's a perfectly diffuse but not 
perfectly 
reflecting surface, you have to multiply by the reflectance, k, to find 
the 
luminance.  
 Also to contrast: Illumination, E, is for the incident or
incoming flux's areal _density_; luminance, B, is for reflected or
outgoing flux's areal _intensity_.    
 Lambert's law says that an perfectly diffuse surface or
extended source reflects or emits light according to a cosine law: the
amount of flux emitted per unit surface area is proportional to the
cosine of the angle between the direction in which the flux is being
emitted and the normal to the emitting surface.  (Note however, that
there is no fundamental physics behind Lambert's "law".  While
assumiming it to be true simplifies the theory, it is really only an
empirical observation whose accuracy varies from surface to surface.
Lambert's law can be taken as a definition of a perfectly diffuse
surface.)
 A consequence of Lambert's law is that no matter from what
direction you look at a perfectly diffuse surface, the luminance on
the basis of _projected_ area is the same.  So if you have a light
meter looking at a perfectly diffuse surface, it doesn't matter what
the angle between the axis of the light meter and the normal to the
surface is as long as all the light meter can see is the surface: in
any case the reading will be the same.
 There are a number of luminance units, but they are in categories:  
two of the categories are those using English units and those using 
metric 
units.  Another two categories are those which have the constant 1/pi 
built 
into them and those that do not.  The latter stems from the fact that 
the 
formula to calculate luminance (photometric Brightness), B, from 
illumination (illuminance), E,  contains the factor 1/pi.  To 
illustrate:

  B = (k*E)(1/pi) 
  Bfl = k*E 

where: B = luminance, candles/foot^2
 Bfl = luminance, foot-Lamberts
 k = reflectivity        0<k<1
 E = illuminance in foot-candles (lumens/ foot^2)

 Obviously, if you divide a luminance expressed in 
foot-Lamberts by pi you then have the luminance expressed in 
candles /foot^2.  (Bfl/pi=B)

Other luminance units are:
                stilb = 1 candle/square centimeter      sb
                apostilb = stilb/(pi X 10^4)=10^-4 L    asb
                nit = 1 candle/ square meter            nt
                Lambert = (1/pi) candle/square cm       L

 Below is a table of photometric units with short definitions.

  Symbol      Term                 Unit              Unit Definition

    Q      light quantity       lumen-hour          radiant energy 
                                lumen-second        as corrected for
                                                    eye's spectral 
response

    F      luminous flux        lumen               radiant energy flux
                                                    as corrected for
                                                    eye's spectral 
response

    I      luminous intensity   candle              one lumen per 
steradian
                                candela             one lumen per 
steradian
                                candlepower         one lumen per 
steradian

    E      illumination         foot-candle         lumen/foot^2
                                lux                 lumen/meter^2
                                phot                lumen/centimeter^2

    B      luminance            candle/foot^2       see unit def's. 
above
                                foot-Lambert   =    (1/pi) 
candles/foot^2
                                Lambert        =    (1/pi)
candles/centimeter^2
                                stilb          =    1 
candle/centimeter^2
                                nit            =    1 candle/meter^2

Note: A lumen-second is sometimes known as a Talbot.
To review:

 Quantity of light, Q, is akin to a quantity of photons except
that here the number of photons is pro-rated according to how bright
they appear to the eye.
 Luminous flux, F, is akin to the time rate of flow of photons except 
that the photons are pro-rated according to how bright they appear to 
the eye.
 Luminous intensity, I, is the solid-angular density of luminous flux.  
Applies primarily to point sources.
 Illumination, E, is the areal density of incident luminous flux.
 Luminance, B, is the areal intensity of an extended source.


            Photometry with a Photographic Light Meter
 The first caveat to keep in mind is that the average unfiltered light 
meter doesn't have the same spectral sensitivity curve that the human 
eye 
does.  Each type of sensor used has its own curve.  Silicon blue cells 
aren't 
too bad.  The overall sensitivity of a cell is usually measured with a 
2856K or 2870K incandescent lamp.  Less commonly it is measured with 
6000K sunlight.
 The basis of using a light meter is the fact that a light meter uses 
the Additive Photographic Exposure System, the system which uses 
Exposure Values:

         Ev = Av + Tv = Sv + Bv

where:   Ev = Exposure Value
         Av = Aperture Value = lg2 N^2         where N = f-number
         Tv = Time Value = lg2 (1/t)           where t = time in sec.s
         Sv = Speed Value = lg2 (0.3 S)        where S = ASA speed
         Bv = Brightness Value = lg2 Bfl

lg2 is logarithm base 2

from which, for example:
     Av(N=f/1) = 0
     Tv(t=1 sec) = 0
     Sv(S=ASA 3.125) E 
     Bv( Bfl = 1 foot-Lambert) = 0

and therefore:
     Bfl = 2^Bv
     Ev (Sv = 0) = Bv

 From the preceeding two equations you can see that if you set the 
meter dial to an ASA speed of approximately 3.1 (same as Sv = 0), when 
you read a scene luminance level the Ev reading will be Bv from which 
you 
can calculate Bfl.  If you don't have an ASA setting of 3.1 on your 
dial, just

use ASA 100 and subtract 5 from the Ev reading to get Bv.  
(Sv@ASA100=5)

                   Image Illumination
 If you know the object luminance (photometric brightness), the 
f-number of the lens, and the image magnification, you can calculate the 
image illumination.  The image magnification is the quotient of any 
linear 

_
                                                                     

dimension in the image divided by the corresponding linear dimension on 
the object.  It is, in the usual photographic case, a number less than 
one.  
The f-number is the f-number for the lens when focussed at infinity - 
this 
is what's written on the lens.  The formula that relates these 
quantities is 
given below:

  Eimage = (t pi B)/[4 N^2 (1+m)^2]
or:  Eimage = (t Bfl)/[4 N^2 (1+m)^2]
where: Eimage is in foot-candles  (divide by .0929 to get lux)
           t   is the transmittance of the lens (usually .9 to .95 but 
lower 
                  for more surfaces in the lens or lack of anti-
reflection    
                   coatings)
           B   is the object luminance in candles/square foot
           Bfl is the object luminance in foot-Lamberts
           N   is the f-number of the lens
           m   is the image magnification

References:
G.E. Miniature Lamp Catalog
Gilway Technical Lamp Catalog
"Lenses in Photography" Rudolph Kingslake Rev.Ed.c1963 A.S.Barnes
"Applied Optics & Optical Engr." Ed. by Kingslake c1965 Academic Press
"The Lighting Primer" Bernard Boylan c1987 Iowa State Univ.
"University Physics" Sears & Zemansky c1955 Addison-Wesley


Acknowledgements

Thanks to John Bercovitz for providing the material on photometry and
illumination.  Thanks to Andy Young for pointing out that Lambert's
law is only empirical.  Thanks to John Bercovitz, donl, and Bill Tyler
for reviewing an earlier version of this file.  I've made extensive
changes since their review, so any remaining bugs are mine, not a
result of their oversight.  All of them told me it was too detailed.
I probably should have listened.

Copyright (C) 1993, 1994, 1995 David M. Jacobson 

Rec.photo.* readers are granted permission to make a reasonable number
electronic or paper copies for their themselves, their friends and
colleagues.  Other publication, or commercial or for-profit use is
prohibited.



                   
