Archive-name: rec-photo/lenses/faq PostingFrequency: monthly Last-modified 1995/3/19 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 distance subject (i.e. M virtually zero), the coverage is 27 degrees by 40 degrees, with a diagonal of 47 degrees. 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)) 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))) reardepth = Sfar - So reardepth = Ne*c/(M^2 * (1 - (So-f)/h)) reardepth = Ne*c/(M^2 * (1 - (N*c)/(f*M))) In the last two, 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 approximate 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 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 the previous question.) 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? 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, 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 úÿ 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 conditions under which the formula for the minimum distance at which the effect of focusing and recomposing 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. 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 úÿ * 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