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Lens Tutorial

by David Jacobson, June 1997 (updated April 2007)


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.

The tutorial is arranged into six parts:

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.

circle of confusion graph

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.

Diffraction field strength vs. radius

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.

Ideal lens MTF, 3 cases

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.

	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 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

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, 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.

Article revised April 2007.

Readers' Comments


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William Janes , April 05, 2007; 11:30 A.M.

I was pleased to see that the lens FAQ has been updated. I have referred to my copy of it often. Is the FAQ available in PDF format?

Rudy Kouw , October 23, 2007; 09:07 A.M.

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.

Ernest Futar , November 09, 2007; 06:14 A.M.

Hi. In the formula of sclose and sfar a capital letter of F appears in the first time in the article - if the search works well in my Explorer.

Janne Sinkkonen , February 04, 2008; 12:17 P.M.

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.

Janne Sinkkonen , February 04, 2008; 12:17 P.M.

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.

Derrek Buttron , October 22, 2008; 12:14 A.M.

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

Bertrand Stark , July 19, 2009; 10:46 A.M.

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 Patterson , March 22, 2013; 09:58 A.M.

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.

 

Regards,


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