By David Jacobson jacobson@hpl.hp.com Q1. What is the meaning of the symbols in the rest of this FAQ? A. f focal length So distance from front principal point to subject (object) Sfar distance from front principal point to farthest point in focus Sclose distance from front principal point to closest point in focus Si distance from rear principal point to film (image) plane M magnification N f-number or f-stop Ne effective f-number (corrected for bellows factor) c diameter of largest acceptable circle of confusion h hyperfocal distance See the technical notes at the end for more information on subject distances, more information on the meaning of f-number and limitations to be observed when applying these formulas to lenses in which the aperture does not appear the same size front and rear. Q2. What is the meaning of focal length? In other words, what about a 50mm lens is 50mm? A. A 50mm lens produces an image of a distant object on the film that is the same size as would be produced by a pinhole 50mm from the film. See also Q5 below. Q3. What meant by f-stop? A. The focal length of the lens divided by the diameter of the aperture (as seen from the front). It is also called an f-number. The brightness of the image on the film is inversely proportional to the f-number squared. Q4. What is the basic formula for the conditions under which an image is in focus? A. There are several forms. 1/Si + 1/So = 1/f (Gaussian form) (Si-f)*(So-f) = f^2 (Newtonian form) Q5. What is the formula for magnification? A. There are several forms. M = Si/So M = (Si-f)/f M = f/(So-f) Q6. For a given lens and format what is angle of coverage? A. If the format has a width, height, or diagonal of distance X, the angle of coverage along width, height, or diagonal is 2*arctan(X/(2*f*(M+1))). For example a 35mm frame is 24x36 mm, so with a 50 mm lens and a distant subject (i.e. M virtually zero), the coverage is 27 degrees by 40 degrees, with a diagonal of 47 degrees. See the technical notes at the end for qualifications. Q7. How do I correct for bellows factor? A. Ne = N*(1+M) Q8. What is meant by circle of confusion? A. When a lens is defocused, a point in the subject gets rendered as a small circle, called the circle of confusion. If the circle of confusion is small enough, the image will look sharp. There is no one circle "small enough" for all circumstances, but rather it depends on how much the image will be enlarged, the quality of the rest of the system, and even the subject. Nevertheless, for 35mm work c=.03mm is generally agreed on as the diameter of the acceptable circle of confusion. Another rule of thumb is c=1/1730 of the diagonal of the frame, which comes to .025mm for 35mm film. (Zeiss and Sinar are known to be consistent with this rule.) Q9. What is hyperfocal distance? A. The closest distance that is in acceptable focus when the lens is focused at infinity. (See below for a variant use of this term.) h = f^2/(N*c) Q10. What are the closest and farthest points that will be in acceptably sharp focus? A. Sclose = h * So / (h + (So - f)) Sfar = h * So / (h - (So - f)) or, defining hr = h/(So - f) Sclose = So * hr/(hr+1) Sfar = So * hr/(hr-1) Think of hr as the hyperfocal to subject distance ratio. These formulas are also correct when hr is defined as hr = h/So and the N used in computing h is actually Ne. If the denominator is zero or negative, Sfar is infinity. Q11. What is depth of field? A. It is convenient to think of a rear depth of field and a front depth of field. The rear depth of field is the distance from the subject to the farthest point that is sharp and the front depth of field is the distance from the closest point that is sharp to the subject. (Here we assume the lens is focused on the subject.) Sometimes the term depth of field is used for the combination of these two, i.e. the distance from the closest point that is sharp to the farthest point that is sharp. frontdepth = So - Sclose frontdepth = Ne*c/(M^2 * (1 + (So-f)/h)) frontdepth = Ne*c/(M^2 * (1 + (N*c)/(f*M))) frontdepth = So /(hr + 1) reardepth = Sfar - So reardepth = Ne*c/(M^2 * (1 - (So-f)/h)) reardepth = Ne*c/(M^2 * (1 - (N*c)/(f*M))) reardepth = So/(hr - 1) In the last three, if the denominator is zero or negative, reardepth is infinity. Q12. Where should I focus my lens so I will get everything from some close point to infinity in focus? A. At approximately the hyperfocal distance. More precisely, at So = h + f. In this condition the closest point that will be in focus is at half the subject distance. (Some authorities use this as the definition of hyperfocal distance.) Q13. I have heard that the depth of field depends only the the f-stop and the magnification. Is this true? A. Yes, under some conditions. When the subject distance is small with respect to the hyperfocal distance, the front and rear depth of field are almost equal and depend only on the magnification and f-stop. As the subject distance approaches the hyperfocal distance, the front depth of field gets smaller and the rear depth gets larger, eventually extending to infinity. Q14. Is there a simpler formula for depth of field? A. Yes. When the subject distance is small with respect to the hyperfocal distance, the following approximate formulas can be used. Sfar = So + Ne*c/M^2 Sclose = So - Ne*c/M^2 frontdepth = reardepth = Ne*c/M^2 frontdepth = reardepth = So/hr Another simple approach is based on the hyperfocal ratio, hr, given above: Sfar = So*hr/(hr-1) Sclose = So*hr/(hr+1) reardepth = So/(hr-1) frontdepth = So/(hr+1) These formulas can be used as follows. Suppose I know that So is 1/8th of the hyperfocal distance. Then the range of distances that is acceptably sharp is from 8/9 of So to 8/7 of So. The front and rear depths of field are 1/9 So and 1/7 So. Q15. I have heard that one should use a long lens to get a shallow depth of field and a short lens to get a large depth of field. Is this true? A. Assuming that you frame the subject the same way, using a long lens (and a correspondingly larger distance) does not make the depth of field very much shorter. It does make the front and rear depths more even, but you probably didn't care about that very much. Using a short lens can make the rear depth of field very large, or even infinite. (See question 12.) Now back to the long lens issue. Even though making the lens very long has little effect on the maximum distance behind the subject at which points still appear to be sharp, it has a big effect on how fuzzy very distant points appear. Specifically, if the lens is focused on some nearby point rendered with magnification M, a distant point at infinity will be rendered as a circle of diameter C, given by C = f M / N which shows that the distant background point will be fuzzed out in direct proportion to the focal length. Q16. If I focus on some point, and then recompose with that point not in the center, will the focus be off? A. Yes, but maybe only a little bit. If the object is far enough away, the depth of field will cover the shift in distance. An approximate formula for the minimum distance such that the error will be covered by depth of field is given by d = w^2/(2 N c) where d = minimum distance to make the point be sharply rendered d is measured from the film plane w = distance image point on the film is from center of the image For the 35mm format w^2/(2 c) is 5.4 meters, so you can recompose the image with the subject at the edge of the frame and still have it be sharp if the subject distance (at the center) was at least 5.4 meters (18 feet) divided by the f-number. See the technical notes at the end for a bunch of assumptions. Q17. What is vignetting and light falloff? A. Vignetting is a reduction in light falling on the film far from the center of the image that is caused by physical obstructions. Independent of vignetting, even with an ideal rectilinear lens (one that renders a square grid in subject space as a square grid on the film) the light on the film falls off with Cos(theta)^4, where theta is the angle a subject point is off the axis. (With suitable optical trickery this can be reduced a little, but never less than Cos(theta)^3 in a rectilinear lens. It can be made much smaller in a fisheye lens.) Q18. How can I tell if a lens has vignetting, or if a filter is causing vignetting? A. Open the back and, if necessary, trick the camera into opening the shutter and stopping down. Imagine putting your eye right in the corner of the frame and looking at the diaphragm. Or course, you really can't do this, so you have to move your head and sight through the corner of the frame, trying to imagine what you would see. If you "see" the entire opening in the diaphragm and through it to subject space, there is no vignetting. However, at wide apertures in most lenses the edge of the rear element or the edge of the front element or filter ring will obstruct your vision. This is vignetting. Try to guess the fraction of the area of the diaphragm is this obstructed. Log base two of this fraction is the falloff in f-stops at the corner. You can also do this from the front. With SLRs hold the camera a fair distance away with a fairly bright area behind the viewfinder hole. With non-SLRs open the back and arrange so a reasonably bright area is behind the camera. Look through the lens, and rotate the camera until you are looking right at the corner of the viewing screen or frame. Now for the hard part. Look at the aperture you see. If there is vignetting you see something about the shape of an American football. If the filter is causing the vignetting, one of the edges of the football is formed by the filter ring. A third way to detect vignetting is to aim the camera at a small bright spot surrounded by a fairly dark background. (A distant street light at night would serve well.) Deliberately defocus the image some and observe the shape of the spot, particularly in the corners. If it is round there is no vignetting. If it looks like the intersection of some arcs (i.e. like an American football), then there is vignetting. Note that near top of the image the top of the circle may get clipped a bit. This is because in many cameras some light (from the top part of the image) misses the bottom of the mirror. This affects only the viewfinder, not the film. You can use depth of field preview (if your camera has it) to determine the f-stop at which the spot becomes round. Q19. For panoramic pictures, where is the best place to pivot the camera. A. The axis of the pivot should pass through the entrance pupil. The entrance pupil is the virtual image of the aperture as seen through the front of the lens. When you've got it right, the entrance pupil will not shift relative to fixed objects as you rotate the camera. (Stop down the lens so you can see the diaphragm and aperture.) There is a whole different type of panoramic camera in which the lens is rotated relative to the film. In this type of camera the lens rotates around the rear nodal point (for subjects at infinity). See the lens tutorial for an explanation of nodal points. Q20. 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. Q21. What is the diffraction limit of a lens. A. All lenses are diffraction limited to no more than about 1500/N to 1800/N line pairs per mm. See below under the question "What is MTF?". Q22. 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. Q23. 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. Q24. 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.) Q25. 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. Q26. 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. Diffraction effects fundamentally limit the MTF of evan an ideal lens to zero at frequencies beyond 1/(lambda*N) cycles per mm, where lambda is the wavelength of the light. For lambda = 555nm, the peak of the eye's response, this is very close to 1800/N cycles per mm. The MTF of a system is the product of the properly scaled MTFs of each of its components, as long as there are not two consecutive non-diffusing components. (Thus with proper scaling you can multiply camera lens MTF by film MTF by enlarger lens MTF by paper MTF, but usually not a telescope objective MTF by an eyepiece MTF. There are also some other obscure conditions under which MTFs can be multiplied.) Note that although MTF is usually thought of as the spatial frequency response function and is plotted with spatial frequency as the abscissa, sometimes it is plotted at a specific spatial frequency with distance from the center of the image as the abscissa. Q27. 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. Q28: 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. Q29: What is the optimum aperture for a pinhole camera? A. d = .036 sqrt(Si), where d is the diameter of the pinhole in millimeters and Si is the distance from the pinhole to the film in millimeters. See the technical notes. Technical notes: The subject distance, So, as used in the formulas is measured from the subject to the lens's front principal point. More commonly one hears of the front nodal point. These two points are equivalent if the front medium and rear medium are the same, e.g. air. They are the effective position of the lens for measurements to the front. In a simple lens the front nodal/principal point is very near the center of the lens. If you know the focal length of the lens, you can easily find the front nodal point by taking the lens off the camera and forming an image of a distant object with the light going through the lens backwards. Find the point of sharp focus, then measure one focal length back (i.e. toward the distant object). That is the position of the front nodal point. On most cameras the focusing scale is calibrated to read the distance from the subject to the film plane. There is no easy way to precisely convert between the focusing scale distance and So. The formulas presented here all assume that the aperture looks the same size front and rear. If it does not, which is particularly common in wide angle lenses, use the front diameter and note that the formulas for bellows correction and depth of field will not be correct at macro distances. Formulas that are exact even with this condition are given in the lens tutorial, posted separately. The formula for angle of coverage applies to rectilinear lenses. An alternative form, 2*arctan(X/(2*Si)), applies to both rectilinear lenses and pinholes. (Rectilinear lenses give the same projection as a pinhole.) These formulas do not usually apply to fisheye lenses, and can't possibly apply to a fisheye lens the covers 180 degrees or more. The conditions under which the formula for the minimum distance at which the effect of focusing and re-composing will be covered by depth of field are: 1. w is no more than the focal length of the lens. At the edge w=18mm for 35mm, so this will very seldom be a problem. 2. The lens's two nodal points are not very widely separated. But if the front nodal point is in front of the rear nodal point, which I think is the more common case, the formula is too conservative, so this is not a problem either. 3. The camera is rotated about the front nodal point. Almost always the camera will be rotated about an axis behind the front nodal point which again makes the formula too conservative. The guide number given assumes c=.03mm. The optimum aperture for a pinhole camera depends on what criteria is used. The formula given here is based on maximizing the spatial frequency that gives an MTF of 20% using lambda = 555nm. Acknowledgements Thanks to Bill Tyler for contributing the section on perspective effects. The technique for detecting vignetting in the viewfinder was suggested by Maohai Huang.