The Basic Geometry Behind A Camera Lens And A Magnifying Glass

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1 The Basic Geometry Behind A Camera Lens And A Magniying Glass by John Kennedy Mathematics Department Santa Monica College 1 Pico Blvd. Santa Monica, CA 45 rkennedy@ix.netcom.com

2 THE BASIC GEOMETRY BEHIND A CAMERA LENS Lenses used in cameras are called converging lenses because they bend rays o light that are parallel to the normal line through the center plane o the lens. As an example o how a converging lens bends horizontal rays o light, see igure 1 below. As the horizontal light rays pass through the lens they all end up going through the same point on the other side o the lens. The lens causes all horizontal light rays to converge at one point which is known as a ocal point. Converging Lens A d o B h o G H F 1 D F 2 d i h i E C Figure 1. A converging lens. Horizontal light rays to the let o the lens are bent by the lens and all go through the ocal point F on the right. 2 Each lens actually has two ocal points which lie on opposite sides o the lens, but they are equidistant rom the center o the lens. In the above igure the two ocal points are denoted by J 1 and J 2. We assume the object being viewed through the lens is the penguin on the let. The image ormed is shown to the right o the lens and appears inverted. Note the image on the right appears larger than the original object which is on the let. Both the object and the image lie outside the vertical band or region which is between the two ocal points J and J. 1 2 The entire action o the lens may be understood by tracing rays o light. In act, only two special rays need to be traced to determine both the size and the position o the image. First, consider the line starting at point E and passing through point F. Since this line is parallel to the normal line through the lens center, the ray o light bends as it passes through the lens at point F in such a manner that it goes through the second ocal point labeled as J2. In act, all rays o light parallel to the center line J1J2 will pass through the ocal point J 2. For example, trace the three rays o light that start in the lower-let part o igure 1. Any ray o light not parallel to the center line JJ will not pass through a ocal point. 1 2 Camera Lens Geometry 1

3 The line EH in igure 1 above is a special line which goes through the exact center o the lens as indicated by point H. The ray o light represented by the line EH is the only one shown in igure 1 which is not bent at all by the lens. In act, only lines through H will not be bent by the lens. Thus points E, H, and G in the above igure are collinear. The size and position o the image are determined by the two rays o light represented by the paths EF J2G and EHG. G is determined as the point o intersection o the two lines FJ and EH. 2 THE THIN-LENS EQUATION In igure 1 above we have labeled the ocal length o the lens using the letter. Thus is the same distance the two ocal points lie rom the lens center. We have also labeled the distance the object and image lie rom the lens center by using the notations. and. 3, respectively, or the object and image. Next we will derive the undamental equation which gives the relationship between the three quantities,., and.. In igure 1, nehl has the same measure as nkhg and nhgi. EHL is similar to HGI. Thus we have the ollowing relationship between the object and image heights and the object and image distances rom the lens center. We let 2 and 2 denote the respective heights o the object and image. 2 œ 2 or 2 œ I we place an BC-coordinate system with its origin at point H then the line FJ # would have slope and go through the point J# Ðß!ÑÞ The equation o the line is: 2 C œ 2 ÐB Ñ Now using these equations and substituting the coordinates o point G, Cœ 23 when Bœ. 3, we can derive one o the undamental equations or the lens: 2 23 œ (. 3 Ñ Substitute Cœ 23 and Bœ.Þ œ 2 3 Multiply out the right side œ " Divide both sides by Substitute œ " œ.. 2. Camera Lens Geometry 2

4 " " " œ Divide both sides by " " œ Add... to both sides. This last equation is known as the thin-lens equation. THE THIN-LENS EQUATION œ.. where œ the ocal length o the lens. œ the distance between the center o the lens and the object.. 3 œ the distance between the center o the lens and the image. Next, reer to igure 2 below. This igure diers rom igure 1 in that the penguins are not standing on the center line. In act, the center line partially goes through each penguin. We also assume the object being photographed is taller than the physical height o the camera lens or the physical size o one picture rame on the ilm. h o B A d o F 1 E G Camera Case Lens Film Plane I Film Roll Image C H F 2 J K h i d i Film Roll Figure 2. A camera image o an object is ocused on the ilm plane behind the camera lens. Camera Lens Geometry 3

5 The Thin-Lens equation can be expressed in a somewhat simpler orm by considering all distances as being measured in terms o multiples o the ocal length. I we let D œ. and let C œ. 3 then D and C measure the object and image distances in terms o multiples o the ocal length. Thus in igure 2 above, D measures the length. and C measures the length. 3, but the units or D and C are both in terms o the ocal length. Substituting these values in the thin-lens equation produces a simple relationship between C and D. œ.. The thin-lens equation œ Substitute. œ D and. 3 œ C. D C " " " œ Multiply through by. D C This last equation can be expressed in an even simpler orm by introducing two new measurements called a and b shown in igure 3 below. The quantities a and b are also measured in terms o multiples o the ocal length. In igure 3 below we assume + œ. and, œ. 3. It is best to think that + measures the length., but the units o + are in terms o ocal lengths. Also, we can think that, measures the length o., where the units o, are also in terms o ocal lengths. 3 While looking at igure 3 below you can assume + measures the number o ocal lengths the object lies to the let o the let ocal point J 1. You can also assume, measures the number o ocal lengths the image lies to the right o the right ocal point J 2. Recall that the object and the image lie outside the region between the two vertical lines through the two ocal points. a b d o d i Image h o F 1 F 2 h i Film Plane Figure 3. The distances o the object and image beyond one ocal length as represented by a and b. Camera Lens Geometry 4

6 Now we may derive the simple relationship that exists between the two numbers + and,. + œ. œ D and, œ. 3 œ C See igure 3 above. Ð+ "Ñ œ D and Ð, "Ñ œc Factor out. + " œ D and, "œc Divide by. " " " œ Previous thin-lens equation. D C " " " œ Substitute D œ + " and C œ, ". + ", " Ð+ "ÑÐ, "Ñ œ Ð, "Ñ Ð+ "Ñ +, +, "œ+, " " Clear the denominators. Simpliy. +, œ " Subtract +, " rom both sides. " + œ Divide by,., This last equation has a simple but important and eminently useul interpretation. + is the number o ocal lengths the object lies to the let o the let ocal point., is the number o ocal lengths the image lies to the right o the right ocal point. Thus interpreting the last equation we know the urther the object lies away rom the let ocal point the closer the image lies near the right ocal point. Vice versa, the urther the image lies to the right o the right ocal point the closer the object must be to the let ocal point. The values + and, are reciprocals o each other. Note that i +œ" then we also have,œ" which means when the image lies two ocal lengths to the let o the lens center, the image also lies two ocal lengths to the right o the lens center. In this case, and only in this case, is it true that the object and image heights are the same. In act, next we will show " that the magniication actor o the object is the number which is the same as the number. +, Given the proportion via similar triangles that 2 2. œ. we can solve or 2Þ 3 This shows the raction object height œ 2. 3 is the magniication actor which determines the image height rom the Camera Lens Geometry 5

7 Next we show the magniication actor is the same as the number, , Ð, "Ñ, ", " Ð, "Ñ, œ œ œ œ œ œ,. + Ð+ "Ñ + " " " Ð",Ñ, So the number, can be used to determine the image height rom the object height. 2 œ, 2 3 The interplay between the numbers + and, can be illustrated as ollows. When + œ $, the object is $ ocal lengths to the let o the let ocal point (it is % ocal lengths rom the lens center) and the image 1 1 is 3 the size o the object and the image lies 3 o a ocal length to the right o the right ocal point (the image is " " œ % ocal lengths away rom the lens center). $ $ In normal practice the object is usually much more than " ocal length to the let o the let ocal point. In this case + " which implies, ". Thus when a camera lens ocuses the image on the ilm, the lens never has to move more than one ocal length to bring the image into ocus. The image is ocused by moving the lens relative to the plane o the ilm inside the camera. See igure 2 which is an example o the normal photographic situation where + " and, ". When doing close-up photography the object will be between " and # ocal lengths rom the lens center and in this case + will satisy the inequality! + " which in turn implies, ". Since, is the same as the magniication actor, the image produced on the ilm will be larger than the real-lie image size. Thus relatively small objects can be blown up to sizes larger than real-lie. For example, imagine photographing the head o a pin. Figure 1 corresponds to close-up photography. Reerring back to igure 1, we can imagine how the image size and position vary as the object position varies. Think o the line EH ( H is the lens center Ñ as swiveling or pivoting about point H. E remains at the peak o the let penguin's bill and in act remains at the same height above the center line no matter where the let penguin Ðthe object Ñ is positioned. When point E is ar away rom the lens center the line EH becomes more horizontal and the image height decreases. As point E moves closer towards the let ocal point J1 ( but E remains to the let o J1Ñ the line EH becomes more vertical and the image height increases. In act, nhfj # remains constant regardless o the distance the object is placed rom the lens. tananhfj b œ which is independent o. Þ # 2 When point E is between " and # ocal lengths rom the lens center the image size will actually be larger than the object size. Thus E must be positioned between " and # ocal lengths in order to do close-up photography. In practice, point E would never get closer to the lens center than " ocal length. Camera Lens Geometry 6

8 I you look closely at igure 2 you can determine that points E, F, and G on the object penguin correspond one-or-one with the points O, N, and M on the image penguin. s E, F, and G are vertically aligned on the object while points O, N and M are vertically aligned on the image. In igure 2 we assume these three points all lie in the ilm plane. In particular, you should try ray tracing point F on the object penguin's breast. The horizontal line goes through the lens at point I, travels down through the ocal point J#, and inally reaches point N on the image penguin's breast. s F and N are corresponding points on the two penguins. G between the object penguin's eet ray traces to point M which is between the image penguin's eet. E at the tip o the object penguin's bill maps directly through the center o the lens to point O which is the tip on the image penguin's bill. I the ilm in our camera were taking slides then the image height would have to be suiciently small to it within one slide-shot picture rame on the ilm. 2 3 I you use a ruler and compute ratios you can determine the dierent values or + and, that are in both igure 1 and igure 3. Figure 2 is essentially the same as igure 3. In act, the reason or making igure 1 and igure 3 dierent is so you could visualize the dierence between the a value being less than one ocal length ( igure 1 Ñ and the + value being greater than one ocal length ( igure 2 and igure 3 ). For igure 1 we estimate the actual ocal length to be about #% 77. The distance between L and J is about "! 77 so the + value should be 24 œ 12. This means the, value should be the reciprocal, 12 5 œ #Þ%. Since the, value is the magniication actor, we assume the image is about #Þ% times as large as the object. In act, when we measure the heights o the object and image we ind their heights are "# 77 and #* 77 respectively. Note that #Þ% "# œ #)Þ), which rounds to 2 to the nearest 77. In igure 1, the distance between J2 and K is about &) 77. Note that with œ #%, we should ind the length o J K will be #% #Þ% œ &(Þ' which rounds up to &) to the nearest For igure 3 we estimate the actual ocal length to be about #% 77. The distance between the object 58 2 and J " is about &)77 so the + value should be 24 = 12. This means the, value should be the 12 reciprocal, 2!Þ%"$) or about %"%. The height o the object is about $$ 77 while the height o the image is just under "% 77. Note that $$!Þ%"$) "$Þ'&, which would round up to about "% to the nearest 77. The distance between the image and J 2 in igure 3 is about "! 77. Note that #%!Þ%"$) *Þ*$, or rounded to the nearest 77, about "! 77. So the values predicted by the equations actually match what we can measure in the igures. Understanding the undamental relationship between the + and, values allows you to interpret the basic geometry behind a camera lens. This may not make you a better photographer, nor will it qualiy you to become a lens designer, but you should now have a better understanding o the mathematics behind a camera lens, and that by itsel is progress! Camera Lens Geometry 7

9 THE BASIC GEOMETRY BEHIND A MAGNIFYING GLASS Next we consider the geometry o a magniying glass. With a magniying glass the viewer's eye is considered to be ixed at one ocal point and the object being viewed is on the other side o the lens, but within a distance shorter than the ocal length. The image appears to the let o the let ocal point, appears enlarged, and has the same orientation as the original object. The image is a virtual one. This means the image position appears to the eye as i it were actually at the place shown in igure 4 below, even though no light rays make the image there. The true image is ormed by the light rays that enter the viewer's eye. C Object Magniying Glass Lens h i A B h o G F H D 1 F d 2 o The Eye = d i F 2 Figure 4. The virtual image o an object under a magniying glass. As with a camera lens, horizontal rays o light may be traced rom the object, through the lens and through the ocal point J #. The dierence is that these same virtual rays may be traced backwards rom J# along straight lines until they intersect the vertical line GK. The line GH through the center o the lens is not bent at all by the lens. The ray o light that extends rom the large penguin's bill at point G to the ocal point J # is straight. In act, all virtual rays rom the virtual image penguin that aim directly at the ocal point are not bent by the lens. J # G in igure 4 is the point o intersection o the two lines FJ# and EH. ELH is similar to GKH and we have the same ratio between the object and image heights and the object and image distances rom the lens, as with the converging camera lens. 2 2 œ.. Camera Lens Geometry 8

10 I we place an BC-coordinate system with its origin at point H then the equation o the line FJ # would be the same as or the converging lens. C œ 2 ÐB Ñ The dierence is that we now substitute Cœ23 when Bœ. 3 (the signs o both coordinates o point G are reversed rom what they were or the camera) and we derive a new undamental equation or a magniying glass œ Ð. 3 Ñ Substitute C œ 23 and B œ. 3Þ 2. 2 œ 2 3 Multiply out the right side œ " Divide both sides by 2Þ œ " Substitute or Þ.. 2 œ Divide both sides by.þ 3.. œ.. Subtract ". 3 rom both sides. THE FUNDAMENTAL EQUATION FOR A MAGNIFYING GLASS œ.. Note that this is analogous to, but dierent rom, the thin-lens equation or a camera. Analogous to what was done with the converging camera lens, we can let the quantity, measure the distance between the image (virtual) and the ocal point J # and we can let + measure the distance between the object and the ocal point J ". The units or + and, are in terms o the ocal length. See igure 5 below. Camera Lens Geometry

11 b C a Magniying Glass Lens h i A B G h o F H D 1 F do 2 d i Figure 5. length measurements b and a relative to the image and the object and the two ocal points. œ.. Fundamental equation or a magniying glass. œ Substitute. œ + and. 3 œ, Þ +, " " "œ Multiply through by. " +, " Ð" +ÑÐ, "Ñ œ Ð, "Ñ Ð" +Ñ Multiply through by Ð" +ÑÐ, "Ñ. +, +, "œ, " " + +,œ " Expand let side; simpliy right side. Add +, " to both sides. " +œ Divide both sides by,., Camera Lens Geometry 1

12 Note that this equation is the same as or a regular camera lens. The smaller the value o larger the virtual image. Also, the larger the value o + the smaller the virtual image. + the As was also true or a camera lens, the number,œ " + is the magniication actor between the object 2 and the image. By looking at igure 5 we can see that the magniication actor is 2 3 and next we show this raction is the same as, , Ð, "Ñ, ", " Ð, "Ñ, œ œ œ œ œ œ œ, 2. + Ð" +Ñ " + " " Ð, "Ñ, In the case o a magniying glass the value, satisies the inequality:, #. This implies that! + Þ " # Camera Lens Geometry 11

Physics 142 Lenses and Mirrors Page 1. Lenses and Mirrors. Now for the sequence of events, in no particular order. Dan Rather

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