Mirrors and Lenses. Clicker Questions. Question P1.03

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1 3 Mirrors and Lenses Clicker Questions Question P.03 Descrition: Reasoning with geometric otics and ray-tracing. Question An object is located on the otical axis and a distance o 8 cm rom a thin converging lens having ocal length. the image o the object is. real, uright, and smaller than the object.. real, inverted, and smaller than the object. 3. real, inverted, and larger than the object. 4. virtual, uright, and smaller than the object. 5. virtual, inverted, and larger than the object. 6. virtual, uright, and larger than the object. 7. None o the above. Commentary Purose: To develo your ability to reason ualitatively with geometric otics. Discussion: You can lug the given values into the thin lens ormulae or image osition and magniication, and i you use them correctly and know how to interret the numbers you get, you can answer this uestion. However, the uestion does not ask you to solve or the image location or magniication; it merely asks or a ualitative answer. For this, a simle ray tracing diagram will suice, and is less rone to error. Drawing the three rincial rays rom the source object through the lens, you will see that they do not converge on the oosite side o the lens, but sread aart as i they had originated at a oint behind the source (arther rom the lens), on the same side o the otical axis as the source oint, and arther rom the lens axis than the source oint. So, the image is virtual, uright, and larger than the source: answer (6). Key Points: Rerain rom uantitative calculation when ualitative reasoning will suice. Grahical reresentations can be owerul tools or analyzing a situation and answering uestions. Know how to use the three rincial rays to draw ray-tracing diagrams. For Instructors Only To hel students areciate the value o grahical reresentations or thinking and roblem-solving, we must show them uestions or which these tools are clearly suerior to ormula-driven aroaches. Students should be encouraged to solve this roblem both ways; that will hel them interret each better. 305

2 306 Chater 3 Question P.04 Descrition: Understanding ray otics and detecting a common misconcetion. Question An image is ormed by a converging lens. Suose the bottom hal o the lens in covered, as shown. What haens to the image?. The image disaears.. The image ades. 3. The image rotates. 4. The image moves relative to the lens. 5. The to hal o the image disaears. 6. The bottom hal o the image disaears. 7. Nothing. 8. None o the above. Commentary Purose: To develo your understanding o ray otics, and conront a common misconcetion. Discussion: Pick any oint on the source object. Light emanates rom that oint in all directions. In some directions, it strikes the lens and is reracted as it asses through the ront and rear suraces. All the light rom this one oint that asses through the lens converges again and asses through one oint on the other side o the lens, beore continuing on in dierent directions and sreading out again. This oint o convergence is the real image o that oint on the source object, and i you ut a rojection screen there, you will see an image o the oint. The images o adjacent oints on the source object occur adjacent to each other in the image lane, creating an image o the entire source object. Not all the light rom the source oint reaches that image oint; only light that asses through the lens does. I the lens is made larger, more light is reracted to the image oint, and the image aears brighter. I the lens is made smaller, less light is reracted, and the image aears dimmer. I hal the lens is covered, only hal as much light rom any source oint reaches its image oint, so the image ades (becomes less bright). But some light rom every source oint still gets to its corresonding image oint, bent through the unobscured hal o the lens, so the entire image is still visible. Thus, answer () is best.

3 Mirrors and Lenses 307 Key Points: Light rom each oint on a source object asses through every art o a lens on its way to the image oint. The diameter o a lens aects the brightness o the image it creates: the larger the lens, the more light it can cature and bend, and the brighter the image. For Instructors Only Students very oten think that hal the image disaears. Others, accustomed to drawing the three rincial rays to locate images, think that i one or more o the rincial rays are disruted, the image will disaear. QUICK QUIZZES. At C.. (c). Since nwater > nair, the virtual image o the ish ormed by reraction at the lat water surace is closer to the surace than is the ish. See Euation (3.9). 3. (a) False. A concave mirror orms an inverted image when the object distance is greater than the ocal length. (c) False. The magnitude o the magniication roduced by a concave mirror is greater than i the object distance is less than the radius o curvature. True. 4.. In this case, the index o reraction o the lens material is less than that o the surrounding medium. Under these conditions, a biconvex lens will be divergent. 5. Although a ray diagram only uses or 3 rays (those whose direction is easily determined using only a straight edge), an ininite number o rays leaving the object will always ass through the lens. 6. (a) False. A virtual image is ormed on the let side o the lens i <.

4 308 Chater 3 True. An uright, virtual image is ormed when <, while an inverted, real image is ormed when >. (c) False. A magniied, real image is ormed i > >, and a magniied, virtual image is ormed i <. ANSWERS TO MULTIPLE CHOICE QUESTIONS. The image o a real object ormed by a lat mirror is always an uright, virtual image, that is the same size as the object and located as ar behind the mirror as the object is in ront o the mirror. Thus, statements, (c), and (e) are all true, while statements (a) and (d) are alse.. From the mirror euation, + R, with > 0 since the mirror is concave, the image distance is ound to be ( 6. ) ( 6. 0) cm Since > 0, the image is located 9.6 in ront o the mirror, and choice (a) is the correct answer. 3. From the mirror euation, + R, with < 0 since the mirror is convex, the image distance is ound to be ( 6. )( 6. 0 ) cm 436 Since < 0, the image is virtual and located 4.36 cm behind the mirror. Choice (d) is the correct answer. 4. For a converging lens, the ocal length is ositive, or > 0. Since the object is virtual, we know that the object distance is negative, or < 0 and. Thus, the thin lens euation gives the image distance as + + Since and are ositive uantities, we see that > 0 and the image is real. Also, since ( + ) <, we see that <. Thus, we have shown that choices (a) and (d) are alse statements, while choices, (c), and (e) are all true. 5. For a convergent lens, > 0, and because the image is real, > 0. The thin lens euation, +, then gives (. )( 8. 0 ). 0 ( ) cm cm +4. Since > 0, the object is in ront (in this case, to the let) o the lens, and the correct choice is (c). 6. For a divergent lens, < 0, and because the object is real, > 0. The thin lens euation, +, then gives ( 0. )( 6. ) cm 65 Since < 0, the image is in ront (in this case, to the let) o the lens, and the correct choice is.

5 Mirrors and Lenses A concave mirror orms inverted, real images o real objects located outside the ocal oint ( > ), and uright, magniied, virtual images o real objects located inside the ocal oint ( < ) o the mirror. Virtual images, located behind the mirror, have negative image distances by the sign convention o Table 3.. Choices (d) and (e) are true statements and all other choices are alse. 8. With a real object in ront o a convex mirror, the image is always uright, virtual, diminished in size, and located between the mirror and the ocal oint. Thus, o the available choices, only choice (d) is a true statement. 9. A convergent lens orms inverted, real images o real objects located outside the ocal oint ( > ). When >, the real image is diminished in size, and the image is enlarged i > >. For real objects located inside the ocal oint ( < ) o the convergent lens, the image is uright, virtual, and enlarged. In the given case, >, so the image is real, inverted, and diminished in size. Choice (c) is the correct answer. 0. For a real object ( > 0 ) and a diverging lens ( < 0 ), the image distance given by the thin lens euation is ( ) < 0 and the magniication is M > 0 + Thus, the image is always virtual and uright, meaning that choice is a true statement while all other choices are alse. ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS. Chromatic aberration is roduced when light asses through a material, as it does when assing through the glass o a lens. A mirror, silvered on its ront surace never has light assing through it, so this aberration cannot occur. This is only one o many reasons why large telescoes use mirrors rather than lenses or their rimary otical elements. 4. All objects beneath the stream aear to be closer to the surace than they really are because o reraction. Thus, the ebbles on the bottom o the stream aear to be close to the surace o a shallow stream. 6. An eect similar to a mirage is roduced excet the mirage is seen hovering in the air. Ghost lighthouses in the sky have been seen over bodies o water by this eect. 8. Actually no hysics is involved here. The design is chosen so your eyelashes will not brush against the glass as you blink. A reason involving a little hysics is that with this design, when you direct your gaze near the outer circumerence o the lens you receive a ray that has assed through glass with more nearly arallel suraces o entry and exit. Then the lens minimally distorts the direction to the object you are looking at. 0. Both words are inverted. However, OXIDE looks the same right side u and uside down. LEAD does not.

6 30 Chater 3. (a) No. The screen is needed to relect the light toward your eye. Yes. The light is traveling toward your eye and diverging away rom the osition o the image, the same as i the object was located at that osition. 4. (d). The entire image would aear because any ortion o the lens can orm the image. The image would be dimmer because the card reduces the light intensity on the screen by 50%. PROBLEM SOLUTIONS 3. I you stand 4 in ront o the mirror, the time reuired or light scattered rom your ace to travel to the mirror and back to your eye is d 040 (. m) 9 Δt 7. 0 s 8 c ms Thus, the image you observe shows you ~0 9 s younger than your current age. 3. (a) With the alm located.0 m in ront o the nearest mirror, that mirror orms an image, I P, o the alm located.0 m behind the nearest mirror. (c) The arthest mirror orms an image, I B, o the back o the hand located.0 m behind this mirror and 5.0 m in ront o the nearest mirror. This image serves and an object or the nearest mirror, which then orms an image, I B, o the back o the hand located 5.0 m behind the nearest mirror. The image I P (see art a) serves as an object located 4.0 m in ront o the arthest mirror, which orms an image I P o the alm, located 4.0 m behind that mirror and 7.0 m in ront o the nearest mirror. This image then serves as an object or the nearest mirror, which orms an image I P3 o the alm, located 7.0 m behind the nearest mirror. (d) Since all images are located behind the mirror, all are virtual images. 3.3 () The irst image in the let-hand mirror is 5.00 t behind the mirror, or 0.0 t rom the erson. () The irst image in the right-hand mirror serves as an object or the let-hand mirror. It is located 0.0 t behind the right-hand mirror, which is 5.0 t rom the let-hand mirror. Thus, the second image in the let-hand mirror is 5.0 t behind the mirror, or 30.0 t rom the erson. (3) The irst image in the let-hand mirror serves as an object or the right-hand mirror. It is located 0.0 t in ront o the right-hand mirror and orms an image 0.0 t behind that mirror. This image then serves as an object or the let-hand mirror. The distance rom this object to the let-hand mirror is 35.0 t. Thus, the third image in the let-hand mirror is 35.0 t behind the mirror, or 40.0 t rom the erson.

7 Mirrors and Lenses The virtual image is as ar behind the mirror as the choir is in ront o the mirror. Thus, the image is 5.30 m behind the mirror. The image o the choir is m m 6. 0 m rom the organist. Using similar triangles gives h 60. m m m 60. m m or h m 458. m 3.5 In the igure at the right, θ θ since they are vertical angles ormed by two intersecting straight lines. Their comlementary angles are also eual or α α. The right triangles PQR and PQR have the common side QR and are then congruent by the angle-side-angle theorem. Thus, the corresonding sides PQ and PQ are eual, or the image is as ar behind the mirror as the object is in ront o it. 3.6 (a) Since the object is in ront o the mirror, > 0. With the image behind the mirror, < 0. The mirror euation gives the radius o curvature as 0 + R 00. cm or R cm 9 The magniication is M ( 0. ) cm 3.7 (a) Since the mirror is concave, R > 0. Because the object is located in ront o the mirror, > 0. The mirror euation, + R, then gives the image distance as R R ( 40. )( 0. ) ( ) 0. 0 cm cm cm Since > 0, the image is located 3. 3 cm in ront o the mirror. continued on next age

8 3 Chater cm M cm 40. Because > 0, the image is real and since M < 0, the image is inverted. 3.8 The lateral magniication is given by M. Thereore, the image distance is M( )( 30. ) The mirror euation: + or R R gives R The negative sign tells us that the surace is convex. The magnitude o the radius o curvature o the cornea is R mm 3.9 (a) For a convex mirror, the ocal length is R < 0, and with the object in ront o the mirror, > 0. The mirror euation, + R, then gives ( 30. )( 0. ) 30. ( 0. ) 750. cm With < 0, the image is located 7.5 behind the mirror. The magniication is M h 750. cm h Since < 0 and M > 0, the image is virtual and uright. Its height is h Mh cm 050. cm 3.0 The image was initially uright but became inverted when Dina was more than 3 rom the mirror. From this inormation, we know that the mirror must be concave because a convex mirror will orm only uright, virtual images o real objects. When the object is located at the ocal oint o a concave mirror, the rays leaving the mirror are arallel, and no image is ormed. Since Dina observed that her image disaeared when she was about 3 rom the mirror, we know that the ocal length must be 3. Also, or sherical mirrors, R. Thus, the radius o curvature o this concave mirror must be R 6.

9 Mirrors and Lenses The magnii ed, virtual images ormed by a concave mirror are uright, so M > 0. Thus, M h h 500. cm + 50., giving The mirror euation then gives R cm or 750. cm 500. cm Realize that the magnitude o the radius o curvature, R, is the same or both sides o the hubca. For the convex side, R R ; and or the concave side, R+ R. The object distance is ositive (real object) and has the same value in both cases. Also, we write the virtual image distance as in each case. The mirror euation then gives: For the convex side, R or R R + [] For the concave side, R or R R [] Comaring Euations [] and [], we observe that the smaller magnitude image distance, 0., occurs with the convex side o the mirror. Hence, we have 0. R and or the concave side, 30. gives 30. R [3] [4] (a) Adding Euations [3] and [4] yields Subtracting [3] rom [4] gives 4 3 R The image is uright, so M > 0, and we have M or + 5. or R , or ( 5 cm ) 50cm The radius o curvature is then ound to be R cm 5, or R 0.50 m 0. m +

10 34 Chater (a) Your ray diagram should be careully drawn to scale and look like the diagram given below: From the mirror euation with +0. and 5., the image distance is ( 0. )( 5. ) 0. ( 5. ) cm cm 600. cm < 0 + > and the magniication is M 600. cm 00. cm Thus, you should ind that the image is uright, located 600. cm behind the mirror, six-tenths the size o the object. 3.5 The ocal length o the mirror may be ound rom the given object and image distances as +, or + ( 5 cm )( 8. ) 5 cm cm For an uright image twice the size o the object, the magniication is M +00., giving 00. Then, using the mirror euation again, + becomes or cm 805. cm (a) The mirror is convex, so < 0, and we have 80. cm. The image is virtual, so < 0, or. Since we also know that 3, the mirror euation gives 3 + or 80. cm and +6 cm This means that we have a real object located 6 cm in ront o the mirror. The magniication is M Thus, the image is uright and one-third the size o the object.

11 Mirrors and Lenses (a) We know that the object distance is +0.. Also, M > 0 since the image is uright, and M since the image is hal the size o the object. Thus, we have M + 0. or 500. cm and the image is seen to be located 5.0 behind the mirror. From the mirror euation, +, we ind the ocal length to be + ( 0. )( 5. 0 ) 0. + ( 5. 0) 0 0. cm 3.8 (a) Since the mirror is concave, R > 0, giving R +4 cm and R + cm. Because the image is uright ( M > 0 ) and three times the size o the object M 3, we have M +3 and 3 The mirror euation then gives 3 3 cm or +80. cm The needed ray diagram, with the object 8. in ront o the mirror, is shown below: From a careully drawn scale drawing, you should ind that the image is uright, virtual, 4 cm behind the mirror, and three times the size o the object. 3.9 (a) An image ormed on a screen is a real image. Thus, the mirror must be concave since, o mirrors, only concave mirrors can orm real images o real objects. The magniied, real images ormed by concave mirrors are inverted, so M < 0 and 50. m M 5, or 0. m 5 5 The object should be 0. m in ront o the mirror. (a revisited) The ocal length o the mirror is 6 0. m m 50. m, or 50. m m

12 36 Chater (a) From R 00. m +, we ind R R. 00 m. The table gives the image osition at a ew critical oints in the motion. Between 300. m and m, the real image moves rom m to ositive ininity. From to 0, the virtual image moves rom negative ininity to 0. Object Distance, Image Distance, 3.00 m m m ± 0 0 Note the jum in the image osition as the ball asses through the ocal oint o the mirror. The ball and its image coincide when 0 and when + R, or R00. m From Δ y v 0 yt + ayt, with v 0 y 0, the times or the ball to all rom 300. m to these ositions are ound to be t ( Δ y). 00 m s and 980. ms a y t ( m) 980. ms s 3. From n n n + n R, with R, the image osition is ound to be n n 00. ( 50. ) 38. cm. 309 or the virtual image is 38. cm below the uer surace o the ice. 3. The center o curvature o a convex surace is located behind the surace, and the sign convention or reracting suraces (Table 3. in the textbook) states that R > 0, giving R cm. The object is in ront o the surace ( > 0) and in air (n 00. ), while the second medium is glass ( n 50. ). Thus, n + n n n R becomes cm and reduces to (a) I 0 0 I 800. cm, ( 4. )( 0. ) cm, ( 4. )( 8. 0 ) 800. cm 60. cm + 4. continued on next age

13 Mirrors and Lenses 37 (c) I 400 (d) I 00. cm, ( 4. )( 4. 0 ) 400. cm 60. cm. cm, ( 4. )(. 0 ) 00. cm 60. cm 800. cm 343. cm 3.3 Since the center o curvature o the surace is on the side the light comes rom, R < 0, giving R 40. cm. Then, n n n n + becomes R cm 40. cm Thus, the magniication M h n h n n h n h cm For a lane reracting surace ( R ), or 40. cm gives ( cm) 5. mm 38. mm n n n + n R becomes n n (a) When the ool is ull, 00. m and 00. ( 00. m ) 50. m. 333 or the ool aears to be 50. m dee. I the ool is hal illed, then 00. m and m. Thus, the bottom o the ool aears to be 0.75 m below the water surace or 75. m below ground level. enter the transarent shere rom air 3.5 As arallel rays rom the Sun object distance, ( n 00. ), the center o curvature o the surace is on the side the light is going toward (back side). Thus, R > 0. It is observed that a real image is ormed on the surace oosite the Sun, giving the image distance as + R. Then n n n n + becomes 0 + n n 00. R R R which reduces to n n. 00 and gives n 00.

14 38 Chater Light scattered rom the bottom o the late undergoes two reractions, once at the to o the late and once at the to o the water. All suraces are lanes ( R ), so the image distance or each reraction is ( n n ). At the to o the late, nwater 333 B n. 66 ( 800. cm ) 64. cm. B glass or the irst image is 6.4 cm below the to o the late. This image serves as a real object or the reraction at the to o the water, so the inal image o the bottom o the late is ormed at nair nair B n n 0. cm+ B B water water cm cm or 3.8 cm below the water surace. Now, consider light scattered rom the to o the late. It undergoes a single reraction, at the to o the water. This reraction orms an image o the to o the late at T nair T.0 n (. cm ) 900. cm water or 9.0 below the water surace. The aarent thickness o the late is then Δy B T 3. 8 cm cm 3.7 In the drawing at the right, object O (the jellyish) is located distance in ront o a lane water-glass interace. Reraction at that interace roduces a virtual image I at distance in ront it. This image serves as the object or reraction at the glass-air interace. This object is located distance + t in ront o the second interace, where t is the thickness o the layer o glass. Reraction at the glass-air interace roduces a inal virtual image, I, located distance in ront o this interace. From n + n ( n n) R with R or a lane, the relation between the object and image distances or reraction at a lat surace is ( n n ). Thus, the image distance or the reraction at the water-glass interace is ( ng nw ). This gives an object distance or the reraction at the glass-air interace o ( ng nw) + t and a inal image osition (measured rom the glass-air interace) o na n n n g a g n n na t + nw + n a t ng g w continued on next age

15 Mirrors and Lenses 39 (a) I the jellyish is located.00 m (or 0) in ront o a 6.0 thick ane o glass, then + 0 and t 6. 0 and the osition o the inal image relative to the glass-air interace is cm ( 6. 0 ) m I the thickness o the glass is negligible ( t 0 ), the distance o the inal image rom the glass-air interace is n n a g n n g w n + 0 n ( cm ) cm m a w so we see that the 6.0 thickness o the glass in art (a) made a 4.0 dierence in the aarent osition o the jellyish. (c) The thicker the glass, the greater the distance between the inal image and the outer surace o the glass. 3.8 The wall o the auarium (assumed to be o negligible thickness) is a lane ( R ) reracting surace searating water ( n. 333) and air ( n 00. ). Thus, n n n n + gives the R n image osition as n. When the object osition changes by Δ, the change in. 333 Δ the image osition is Δ. The aarent seed o the ish is then given by. 333 v image Δ Δt Δ Δt cm s cm s 3.9 With R cm and R cm, the lens maker s euation gives the ocal length as ( n ) 50 R R (. ) cm 5.. or The lens maker s euation is used to comute the ocal length in each case. (a) ( n ) R R ( 44. ) ( ) cm (. ) 6. 4 cm 6. 4 cm

16 30 Chater The ocal length o a converging lens is ositive, so +0.. The thin lens euation then yields a ocal length o (a) When +0., ( 0. )( 0. ) + 0. and M cm so the image is located 0. beyond the lens, is real ( > 0), is inverted (M < 0), and is the same size as the object M 00.. When +0., the object is at the ocal oint and no image is ormed. Instead, arallel rays emerge rom the lens. (c) When 500. cm, ( 500. cm )( 00. cm ) 0. and 500. cm 00. cm M cm so the image is located 0. in ront o the lens, is virtual ( < 0), is uright (M > 0), and is twice the size o the object M (a) and Your scale drawings should look similar to those given below: Figure (a) Figure A careully drawn-to-scale version o Figure (a) should yield a real, inverted image that is located in back o the lens and the same size as the object. Similarly, a careully drawn-to-scale version o Figure should yield an uright, virtual image located in ront o the lens and twice the size o the object. (c) The accuracy o the grah deends on how accurately the ray diagrams are drawn. Sources o uncertainty: a arallel line rom the ti o the object may not be exactly arallel; the ocal oints may not be exactly located; lines through the ocal oints may not be exactly the correct sloe; the location o the intersection o two lines cannot be determined with comlete accuracy.

17 Mirrors and Lenses From the thin lens euation, +, the image distance is ound to be 0. ( ) ( 0. ) (a) I 40., then 3. 3 cm and M ( 3. 3 cm ) The image is virtual, uright, and 3.3 cm in ront o the lens. I 0., then 0. and + M The image is virtual, uright, and 0. in ront o the lens. (c) When 0., 667. cm and M ( 667. cm ) + 0. The image is virtual, uright, and 6.67 cm in ront o the lens (a) and. Your scale drawings should look similar to those given below: 3. Figure (a) Figure A careully drawn-to-scale version o Figure (a) should yield an uright, virtual image located 3.3 cm in ront o the lens and one-third the size o the object. Similarly, a careully drawn-to-scale version o Figure should yield an uright, virtual image located 6.7 cm in ront o the lens and two-thirds the size o the object. (c) The results o the grahical solution are consistent with the algebraic answers ound in Problem 3.33, allowing or small deviances due to uncertainties in measurement. Grahical answers may vary, deending on the size o the grah aer and accuracy o the drawing.

18 3 Chater (a) The real image case is shown in the ray diagram. Notice that +. 9 cm, or. 9 cm. The thin lens euation, with 44. cm, then gives +. 9 cm 44. cm or (. 9 cm ) cm 0 Using the uadratic ormula to solve gives 963. cm or 37. cm Both are valid solutions or the real image case. The virtual image case is shown in the second diagram. Note that in this case, (. 9 cm + ), so the thin lens euation gives. 9 cm cm or + (. 9 cm ) 3. 5 cm 0 The uadratic ormula then gives 0. cm or 50. cm. Since the object is real, the negative solution must be rejected, leaving 0. cm We must irst realize that we are looking at an uright, magniied, virtual image. Thus, we have a real object located between a converging lens and its ront-side ocal oint, so < 0, > 0, and > 0. The magniication is M +, giving. Then, rom the thin lens euation, + or (. 84 cm ) 568. cm 3.37 It is desired to orm a magniied, real image on the screen using a single thin lens. To do this, a converging lens must be used and the image will be inverted. The magniication then gives M h 80. m, or h m Also, we know that m. Thereore, m, giving 300. m m 395. mm continued on next age

19 Mirrors and Lenses 33 (a) The thin lens euation then gives or ( mm) 390. mm To have a magniication o M +300., it is necessary that The thin lens euation, with +8. or the convergent convex lens, gives the reuired object distance as cm cm or Since the light rays incident to the irst lens are arallel, and the thin lens euation gives 0.. The virtual image ormed by the irst lens serves as the object or the second lens, so I the light rays leaving the second lens are arallel, then and the thin lens euation gives (a) Solving the thin lens euation or the image distance gives or For a real object, > 0 and. Also, or a diverging lens, < 0 and. The result o art (a) then becomes + Thus, we see that < 0 or all numeric values o and. Since negative image distances mean virtual images, we conclude that a diverging lens will always orm virtual images o real objects. (c) For a real object, > 0 and. Also, or a converging lens, > 0 and. The result o art (a) then becomes > 0 i > 0 Since must be ositive or a real image, we see that a converging lens will orm real images o real objects only when > (or > since both and are ositive in this situation).

20 34 Chater The thin lens euation gives the image osition or the irst lens as cm and the magniication by this lens is M cm cm The real image ormed by the irst lens serves as the object or the second lens, so Then, the thin lens euation gives and the magniication by the second lens is M cm Thus, the inal, virtual image is located 30. in ront o the second lens and the overall magniication is M MM ( 00. )( ) (a) With + 5., the thin lens euation gives the osition o the image ormed by the irst lens as cm cm This image serves as the object or the second lens, with an object distance o (a virtual object). I the image ormed by this lens is at the osition o O, the image distance is ( + ) The thin lens euation then gives the ocal length o the second lens as + ( ) cm. cm cm. cm The overall magniication is M M M 30. cm (c) Since < 0, the inal image is virtual ; and since M > 0, it is uright.

21 Mirrors and Lenses From the thin lens euation, 400. cm 800. cm 400. cm 800. cm 800. cm. The magniication by the irst lens is M 800. cm 400. cm The virtual image ormed by the irst lens is the object or the second lens, so 600. cm cm and the thin lens euation gives 6 0 ( 4. )(. cm ) 4. ( 6.) 747. cm The magniication by the second lens is M. cm 4. magniication is M MM ( +. 00) ( ) , so the overall The osition o the inal image is 747. cm in ront o the second lens, and its height is h M h cm 07. cm. Since M > 0, the inal image is uright ; and since < 0, this image is virtual (a) We start with the inal image and work backward. From Figure P3.44, observe that The thin lens euation then gives cm cm The image ormed by the irst lens serves as the object or the second lens and is located 9.74 cm in ront o the second lens. Thus, cm cm and the thin lens euation gives 40.3 cm cm cm The original object should be located 3. 3 cm in ront o the irst lens. The overall magniication is M M M cm cm 3.3 cm cm 59.. (c) Since M < 0, the inal image is inverted ; and since < 0, it is virtual.

22 36 Chater Note: Final answers to this roblem are highly sensitive to round-o error. To avoid this, we retain extra digits in intermediate answers and round only the inal answers to the correct number o signiicant igures. Since the inal image is to be real and in the ilm lane, + d. Then, the thin lens euation gives d 3. d 3.. From Figure P3.45, observe that d <.. The above result then shows that < 0, so the object or the second lens will be a virtual object. The object o the second lens ( L ) is the image ormed by the irst lens ( L ), so. d. d + 3. d cm d d 3. I d 500. cm, then cm; and when d 0., cm. From the thin lens euation, When cm d ( 500 ) cm, then m. When cm d 0., then. 4 cm 0. 4 m. Thus, the range o ocal distances or this camera is 0.4 m to 8. m (a) From the thin lens euation, the image distance or the irst lens is cm With + 30., the image o the irst lens is located 30. in back o that lens. Since the second lens is only 0. beyond the irst lens, this means that the irst lens is trying to orm its image at a location 0. beyond the second lens. (c) (d) (e) () The image the irst lens orms (or would orm i allowed to do so) serves as the object or the second lens. Considering the answer to art above, we see that this will be a virtual object, with object distance 0.. From the thin lens euation, the image distance or the second lens is M M cm. cm cm cm ( + ) (g) M MM (h) Since > 0, the inal image is real, and since M < 0, that image is inverted.

23 Mirrors and Lenses Since cm when +0., we ind that cm. cm 80. Then, when 0., cm. cm or cm 4. 0 Thus, a real image is ormed 5.7 cm in ront o the mirror (a) We are given that 5, with both and being ositive. The thin lens euation then gives (c) M Since > 0, the image is real. Because M < 0, the image is inverted. Since the object is real, it is located in ront o the lens, and with > 0, the image is located in back o the lens. Thus, the image is on the oosite side o the lens rom the object Since the object is very distant ( ), the image distance euals the ocal length, or mm. Now consider two rays that ass undeviated through the center o the thin lens to oosite sides o the image as shown in the sketch below. From the sketch, observe that α tan mm mm Thus, the angular width o the image is α tan (a) Using the sign convention rom Table 3., the radii o curvature o the suraces are R 5. and R The lens maker s euation then gives ( n ) 50 R R or 0 I, then.. The thin lens euation gives. +. and the ollowing results: continued on next age

24 38 Chater 3 (c) I , 900. cm. (d) I +., 600. cm. (e) I , 400. cm. 3.5 As light asses let to right through the lens, the image osition is given by 0 8 ( )( 0. ) This image serves as an object or the mirror with an object distance o cm (virtual object). From the mirror euation, the osition o the image ormed by the mirror is ( ) cm. cm This image is the object or the lens as light now asses through it going right to let. The object distance or the lens is ( 60. ), or 3 6. From the thin lens euation, ( cm )(. cm ) Thus, the inal image is located 6 to the let o the lens. The overall magniication is M M M M 3 M ( 300 cm) Since M < 0, the inal image is inverted. 3, or Since the object is midway between the lens and mirror, the object distance or the mirror is +. 5 cm. The mirror euation gives the image osition as 5 4 R cm , or This image serves as the object or the lens, so Note that since < 0, this is a virtual object. The thin lens euation gives the image osition or the lens as ( )(. cm ) 50 3 cm cm. Since < 0, this is a virtual image that is located 50.3 cm in ront o the lens or 53. cm behind the mirror. The overall magniication is M M M 50. cm.5 cm Since M > 0, the inal image is uright

25 Mirrors and Lenses A hemishere is too thick to be described as a thin lens. The light is undeviated on entry into the lat ace. We next consider the light s exit rom the curved surace, or which R 600. cm. The incident rays are arallel, so. Then, n n n + n R becomes cm rom which 0. 7 cm (a) The thin lens euation gives the image distance or the irst lens as (. cm )(. cm ) The magniication by this lens is then M cm cm The real image ormed by the irst lens is the object or the second lens. Thus, and the thin lens euation gives cm The inal image is 0. in back o the second lens. 0. The magniication by the second lens is M 00., so the overall 0. magniication is M MM ( 00. )( 00. ) Since this magniication has a value o unity, the inal image is the same size as the original object, or h M h The image distance or the second lens is ositive, so the inal image is real. (c) When the two lenses are in contact, the ocal length o the combination is + +, or cm. cm The image osition is then ( 5.0 )( 4. 0 )

26 330 Chater With light going through the iece o glass rom let to right, the radius o the irst surace is ositive and that o the second surace is negative according to the sign convention o Table 3.. Thus, R cm and R 400. cm. Alying n n n n + to the irst surace gives R cm cm which yields 00. cm. The irst surace orms a virtual image.0 to the let o that surace and 6. to the let o the second surace. The image ormed by the irst surace is the object or the second surace, so + 6. and n n n n + gives R cm or + 3. The inal image ormed by the iece o glass is a real image located 3. to the right o the second surace Consider an object O at distance in ront o the irst lens. The thin lens euation gives the image osition or this lens as The image, I, ormed by the irst lens serves as the object, O, or the second lens. With the lenses in contact, this will be a virtual object i I is real and will be a real object i I is virtual. In either case, i the thicknesses o the lenses may be ignored, and + Alying the thin lens euation to the second lens, + + or becomes Observe that this result is a thin lens tye euation relating the osition o the original object O and the osition o the inal image I ormed by this two lens combination. Thus, we see that we may treat two thin lenses in contact as a single lens having a ocal length,, given by +

27 Mirrors and Lenses From the thin lens euation, the image distance or the irst lens is 4 3 ( 0. )( 0. ) and the magniication by this lens is M The real image ormed by the irst lens serves as the object or the second lens, with object distance o 0. (a virtual object). The thin lens euation gives the image distance or the second lens as ( 0. ) 0. (a) I 0., then + 0. and the magniication by the second lens is + M ( 0. ) The inal image is located 0. to the right o the second lens and the overall magniication is M MM ( 300. )( ) Since M < 0, the inal image is inverted. (c) I + 0., then cm 667. cm and M The inal image is 6.67 cm to the right o the second lens and the overall magniication is ( + ) M MM Since M < 0, the inal image is inverted The object is located at the ocal oint o the uer mirror. Thus, the uer mirror creates an image at ininity (that is, arallel rays leave this mirror). The lower mirror ocuses these arallel rays at its ocal oint, located at the hole in the uer mirror. Thus, the image is real, inverted, and actual size. For the uer mirror: + + : 750. cm 750. cm For the lower mirror: cm : 750. cm Light directed into the hole in the uer mirror relects as shown, to behave as i it were relecting rom the hole.

28 33 Chater (a) The lens maker s euation, ( n ) R R, gives ( n ) 500. cm 900. cm which simliies to n As light asses rom let to right through the lens, the thin lens euation gives the image distance as cm This image ormed by the lens serves as an object or the mirror with object distance cm. The mirror euation then gives R R 6.67 cm cm 8.00 cm This real image, ormed 0. to the let o the mirror, serves as an object or the lens as light asses through it rom right to let. The object distance is , and the thin lens euation gives The inal image is located 0. to the let o the lens and its overall magniication is M M M M (c) Since M < 0, the inal image is inverted From the thin lens euation, the object distance is. (a) I +4, then 4 4 cm When 3, we ind (c) In case (a), M and in case, M

29 Mirrors and Lenses I R 300. m and R 600. m, the ocal length is given by or n n n n + 300m 600m.. n 600. m n n n (a) I n 50. and n 00 The thin lens euation gives 600. m., then ( 600. m )( 00. ) m. ( 0.0 m )(.0 m ) 0.0 m +.0 m 54.5 m. [] A virtual image is ormed 5.45 m to the let o the lens. I n 50. and n 33., the ocal length is and ( 600. m )( 33. ) m ( 0.0 m )( m ) 0.0 m m 8.4 m The image is located 8.4 m to the let o the lens. (c) When n 50. and n 00 and., ( 600. m )( 00. ) ( 0.0 m )( 4.0 m ) m 0.0 m 4. 0 m The image is 7. m to the let o the lens m (d) Observe rom Euation [] that < 0 i n > n and > 0 when n < n. Thus, a diverging lens can be changed to converging by surrounding it with a medium whose index o reraction exceeds that o the lens material.

30 334 Chater The inverted image is ormed by light that leaves the object and goes directly through the lens, never having relected rom the mirror. For the ormation o this inverted image, we have M 50. giving The thin lens euation then gives or ( 0. ) cm. The uright image is ormed by light that asses through the lens ater relecting rom the mirror. The object or the lens in this uright image ormation is the image ormed by the mirror. In order or the lens to orm the uright image at the same location as the inverted image, the image ormed by the mirror must be located at the osition o the original object (so the object distances, and hence image distances, are the same or both the inverted and uright images ormed by the lens). Thereore, the object distance and the image distance or the mirror are eual, and their common value is mirror mirror cm cm The mirror euation, +, then gives mirror mirror R mirror mirror 3 3 cm cm cm 3. 3 cm or mirror cm 3.63 (a) The lens maker s euation or a lens made o material with reractive index n 55. and immersed in a medium having reractive index n is n 55 n n R R. n R R Thus, when the lens is in air, we have R R air [] and when it is immersed in water, R R water [] Dividing Euation [] by Euation [] gives water air I air 79., the ocal length when immersed in water is water ( 79 ) 33 cm The ocal length or a mirror is determined by the law o relection, which is indeendent o the material o which the mirror is made and o the surrounding medium. Thus, the ocal length deends only on the radius o curvature and not on the material making u the mirror or the surrounding medium. This means that, or the mirror, water air 79.