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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.

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

Part I - Object distance, image distance, and magnification

Throughout this article we use the word "object" to mean the thing of
which an image is being made. It is loosely equivalent to the word
"subject" as used by photographers.

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 object distance to be measured from the
front principal point, most focusing scales are calibrated to read the
distance from the object 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 form a (probably virtual) erect 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 straight away from the rear nodal point at exactly
the same angle to the lens's axis as the entering ray had. The nodal
points are identical to the principal points when the front and rear
media are the same, e.g. air, so for most practical purposes the terms
can be used interchangeably.

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 generally move relative to the glass and each
other when zooming or focusing.

When a camera 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
manufacturer's numbers enough for educational purposes.)

So 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 this extension is inversely proportional to a
similar "extension" of the object.

Ray Tracing

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. It could be behind the rear
principal plane.)

Let us define an ideal lens as one that forms undistorted sharp images
of an object plane on an image plane. Such an ideal lens is only
possible for a specified set of planes. A ray from an object point in
the specified object plane and passing through the front nodal point
exits the rear nodal point a the same angle and passes through the
image point. All rays passing through the object point also pass
though the image point. The exiting ray can be computed by the
following technique. Consider a spherical surface centered on the
object point and of such radius that the front principal point is in
the surface. Consider a second surface centered on the image point
and of such radius that the rear principal point is in the surface.
Any ray through the object point can be considered to proceed to the
first surface, then jump parallel to the axis to the second surface
(backwards if necessary), and from there proceed to the image point.
Only the first and last components contribute to the optical path length.

Part II - 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, based on geometric factors, but not absorption

Light from an object point spreads out in a cone whose base is the
entrance pupil. (The lens elements in front of the diaphragm form a
virtual image of the physical aperture. This virtual image is called
the entrance pupil.)

Analogous to the entrance pupil is the exit pupil, which is the
virtual image of the aperture as seen though the rear elements.

Let us define Ne, the effective f-number, as

Ne = 1/(2 sin(thetaX))

where thetaX is the angle from the axis to the edge of the eXit pupil
as viewed from the film plane. It can be shown that for any lens free
of coma the following also holds

Ne = M/(2 sin(thetaE)).

We will ignore the issue of coma throughout the rest of this
document.

The first equation deals with rays converging to the image point, and
is the basis for depth of field calculations. The second equation
deals with light captured by the aperture, and is the basis for
exposure calculations.

This section will explain the connection between Ne and light striking
the film, relate this to N, and show to compute
Ne for macro
situations.

If an object radiated or reflected light uniformly in all directions,
it is clear that the amount of light captured by the aperture would be
proportional to the solid angle subtended by the aperture from the
object point. In optical theory, however, it is common assume that
the light follows Lambert's law, which says that the light intensity
falls off with cos(theta),
where theta is the angle off the normal.
With this assumption it can be shown that the light entering the
aperture from a point near the axis is proportional to
sin^2(thetaE),
which is proportional to the aperture area for small thetaE.

If the magnification is M,
the light from a tiny object patch of unit
area gets spread out over an area M^2
on the film, and so the relative
intensity on the film is inversely proportional to M^2. Thus the
relative intensity on the film, RI, is given by

RI = sin^2(thetaE)/M^2 = 1/(4 Ne^2)

with the second equality by the defintion of Ne.
(Of course the true
intensity depends on the subject illumination, etc.)

For So very large with respect to f,
M is approximately f/So and
sin(thetaE) is approximately (D/2)/So.
Substituting these into the above formula get that RI = ((D/2)/f)^2 =
1/(4N^2), and thus for So >> f,

Ne = D/f = N.

For closer subjects, we need a more detailed analysis. We will take
D = f/N as determining D,

Let us go back to the original approximate formula for the relative
intensity on the film, and substitute more carefully

RI = sin^2(thetaE)/M^2 = ((D/2)^2/((D/2)^2+(So-zE)^2))/M^2

where zE

is the distance the entrance pupil is in front of the front
principal point.

However, zE is not convenient to measure.
It is more convenient to
use "pupil magnification". 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.
It can be shown that zE = f*(1-1/p), and substituting this
into the above equation, carrying out some algebraic manipulations,
and solving with RI = 1/(4 Ne^2) yields

Ne = Sqrt((M/2)^2 + (N*(1+M/p))^2).

If we further assume thetaE is small enough that
sin(thetaE) ~= tan(thetaE), the (M/2)^2
term drops out and we get

Ne = N*(1+M/p).

This is the standard equation, and will be used throughout the rest of
this document. The essence of the approximation is the distinction
between the axial distance to the plane of the entrance pupil and the
distance along the hypotnuse to the edge of the entrance pupil, which
is the really correct form. Clearly in typical photographic sitations
that distinction is insignificant.

An alternative, but less fundamental, derivation is based on the
relative illumination varying with the inverse square of the distance
from the exit pupil to the film. This distance is just
f*(1+M) - zX,
where zX is the distance the exit pupil is behind the
rear principal
point. It can be shown that zX = -f*(p-1), so
Ne/N = (f*(1+M)+f*(p-1))/(f+f*(p-1)) = 1+M/p, and hence
Ne = N*(1+M/p).

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 p=1, so the M/p
can be replaced by just M in the above
equations.

Part III - Circle of confusion, depth of field and hyperfocal distance.

The light from a single object 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 object 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 object" that is perfectly in
focus, and an "alternate object point" this is in front of or
behind the main object.

Soa alternate object 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 object
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 object point is in front of
the main object) 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 object is at a finite distance, the closest
alternative point that is acceptably rendered is 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 +/- (So-f)/h1))
= c N (1+M/p) / (M^2 * (1 +/- (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. (!= means "is not equal to".)

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 an object the same way with two different lenses,
i.e. M is the same 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 object 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
(red) and 100 mm (green) focal lengths are both set up to get a
magnification of 1/10. Both lenses are set to f/8. The graph shows
the circles of confusions as a function of the distance behind the
object.

The slope of both graphs virtually identical out to well beyond where
the diameter of the cirlce of consfusion is .03mm, showing that to a
first approximation both lenses have the same depth of field.
However, the size of the circle of confusion for in infinitely distant point
is twice as large for the 100mm
lens (1.25mm) as for the 50mm lens (.625mm).

Part IV - 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 normalized field strength
(of the electric or magnetic field) at angle phi from the axis is
given by

2 J1(x)/x, where x = 2 phi Pi R/lambda

and where R is the radius of the aperture,
lambda is the wavelength of
the light, and J1 is the first order Bessel function. The
normalization is relative to the field strength at the center. The
power (intensity) is proportional to the square of this function.

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 is at 1.21967 lambda/(2 R).
There are actually an infinite number
of lobes after this, but about 83.8% of the power is in the circle
bounded by the first zero.

Approximating the aperture-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 sensitivity 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 a dark band receive no more than
80% of the light power striking the center of a light band.
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.

Part V - 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
definitions of the modulation transfer function and of the related
optical transfer function vary 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 object contrast of an
object 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, object 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
sagittal) and tangential.

The MTF for an an ideal lens (ignoring the unavoidable effect of
diffraction) is a constant 1 for spatial frequencies 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. 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 the phase of the waves is lost
at each step. 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.

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

For pure diffraction the formula is

OTF(lambda,N,spf) = 2/Pi (ArcCos(lambda N spf) - lambda N spf Sqrt(1-(lambda N spf)^2))
[if lambda N spf <=1]
OTF(lambda,N,spf) = 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 diffraction-free circle of confusion of diameter C,

OTF(C,spf) = 2 J1(Pi C spf)/(Pi C spf)

where, again, J1(x) is the first order Bessel function.
The OTF goes
negative at certain frequencies. Physically, this would mean that if
the test pattern were lighter right on the optical center than nearby,
the image would be darker right on the optical center than nearby.
Some authorities use the term "spurious resolution" for spatial
frequencies beyond the first zero. The MTF is the absolute value of
the OTF.

For the case where there is a combination of diffraction and
focus error dz (resulting in a circle of confusion of
diamter dz/N),
the OTF is given by the following formula,
which involves an integration that
must be done numerically. Let s = lambda N spf, and
a = Pi spf dz / N.

OTF = 4/(Pi a) integral y=0 to sqrt(1-s^2) of sin(a(sqrt(1-y^2)-s)) dy
for s < 1
0 for s >= 1

This formula is an approximation that is best at small apertures.

Here is a graph of the OTF of the f/22 diffraction limit, a .03mm
circle of confusion assuming no diffraction, and the combined effect.

Note how the combination is not the product of each of the effects
taken separately.

Some authorities present MTF in a log-log plot.

The classic paper on the MTF for the combination of diffraction and
focus error is H.H. Hopkins, "The frequency response of a defocused
optical system," Proceedings of the Royal Society A, v. 231, London
(1955), pp 91-103. Reprinted in Lionel Baker (ed), Optical Transfer
Function: Foundation and Theory, SPIE Optical Engineering Press,
1992, pp 143-153.

Part VI - 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.

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
assuming 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

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, donl mathis, and
Bill Tyler for reviewing an
earlier version of this when it was an ascii Usenet FAQ 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. Thanks to Andy Young for pointing out that
Lambert's law is only empirical. Thanks to John Bercovitz for
providing the material on photometry and illumination, for helping me
improve the presentation, and for nudging me into dusting it off from
time to time and touching it up.

This document is based on the Usenet Lens Tutorial,
version 1.10 of October 26, 1996.

Text and graphics Copyright (C) 1993, 1994, 1995, 1996, 1997 David M. Jacobson. Top photo
copyright 1996 Philip Greenspun.

I am very happy with this lens tutorial, collecting most theory issues of relevance to photography. That said, I would like to add the following issue concerning hyperfocal distance and Hopkins' formula. The hyperfocal distance is in practice given by the expression H = f^2/(c*N). In terms of the product of the two variables of the OTF-formula of Hopkins - OTF = OTF(a,s) - this formula can be rewritten as:
H = G*f^2/b,
where b = a*s and G = Pi*lambda*spf^2. Choosing appropriate values for the wavelength lambda and for the spatial frequency spf, G is constant.

Hence, the hyperfocal distance is minimal for a maximum value of b. Note that the OTF-formula can be rewritten in terms of the two vaiables b and s: OTF = OTFb(b,s), which can be 'inverted' into b = b(s,OTF). It turns out that for OTF = 50%, the maximum value of b = 0.57 at s = 0.31. Note that at this point N/c = 960 per mm ~ 1000 per mm and N*spf = 560 per mm.

With these values the formula for the hyperfocal distance can be rewritten into a very elegant and simple to use form:
H = (f/N)^2,
with f in mm and H in meter, and in which N is completely determined by the required (minimum) value of spf: N = 560/spf.

In practice this implies that at the Nyquist frequency of e.g. an APS-C sensor of 8 Mega pixels, spf ~ 80 lp/mm. Hence on should not stop down much further than N = 8. Stopping down too much, say beyond N = 11, may 'improve' depth of field, but only at the expense of lesser detail at the extremes of the depth of field, among which objects near the horizon.

Shouldn't diffraction be a function of the effective f-stop Ne, and not of N? Although diffraction depends on the physical (absolute) aperture, it is angular and should be relative to how the aperture is seen from the film plane. That is proportional to Ne, not N.

An interesting scaling invariance then occurs. Consider digital cameras of different sizes, i.e., one with a small sensor and a small focal length, and another one with a large sensor and a large f. Suppose one wants to keep diffraction circle limited to a proportion of the sensor size, to keep resolution constant. For macro work, DOF is very close to c*Ne/M^2 as stated above. Now, resolution-wise c scales with f so that DOF is proportional to f*Ne/M^2. To keep the subject fit onto the sensor, M must also scale with f, and DOF is then prop. to f*Ne/f^2. Meanwhile, to keep the resolution, one can allow diffraction circle to grow with f, and that makes Ne also proportional to f. Then DOF is proportional to f*f/f^2 = const.

For diffraction-limited macro work with more than enough light, the size of the sensor is then irrelevant! In practice things are not that simple, for camera size affects quality, and perspective does matter. But to a degree this explains why macro shots with a compact camera and a flash are so good.

Shouldn't diffraction be a function of the effective f-stop Ne, and not of N? Although diffraction depends on the physical (absolute) aperture, it is angular and should be relative to how the aperture is seen from the film plane. That is proportional to Ne, not N.

An interesting scaling invariance then occurs. Consider digital cameras of different sizes, i.e., one with a small sensor and a small focal length, and another one with a large sensor and a large f. Suppose one wants to keep diffraction circle limited to a proportion of the sensor size, to keep resolution constant. For macro work, DOF is very close to c*Ne/M^2 as stated above. Now, resolution-wise c scales with f so that DOF is proportional to f*Ne/M^2. To keep the subject fit onto the sensor, M must also scale with f, and DOF is then prop. to f*Ne/f^2. Meanwhile, to keep the resolution, one can allow diffraction circle to grow with f, and that makes Ne also proportional to f. Then DOF is proportional to f*f/f^2 = const.

For diffraction-limited macro work with more than enough light, the size of the sensor is then irrelevant! In practice things are not that simple, for camera size affects quality, and perspective does matter. But to a degree this explains why macro shots with a compact camera and a flash are so good.

Wow. Great tutorial. I've fairly new to photography and this takes my knowledge of photography lenses to a whole other level. Kind of opens my eyes to all there is to learn. Thanks, Derrek

I first want to congratulate the author of this tutorial. Very clear and comprehensive. I then have question regarding the optimum pinhole diameter. Out of Wikipedia I found the following formula by Lord Rayleigh: d=1.9 x sqrt(f x λ) where d is diameter, f is focal length (distance from pinhole to focal plane) and λ is the wavelength of light. For λ= 550Nm, I then get d= 0.0446 x sqrt f admitting that in the case of a pinhole camera f=Si. Where does the discrepancy between 0.0446 and 0.0360 come from ?

I think this is very complete and readable tutorial. I do not agree that it should be simplified.

I do have a questioN. Nikon (and others) show a variable sperture on yheir zoom lenses, e.g f3.5 - 5.6 I understand that the aperture varies with the focal length, the maximum increasing with increasing focal length. My question is 0 how does it vary? Linearly or exponentially or logarithmicly or What?

Does anyone know? I have searched the net the net to no but I have not found this information anywhere.