Mirrors and Lenses 39 7. A concave mirror forms inverted, real images of real objects located outside the focal oint ( > f ), and uright, magnified, virtual images of real objects located inside the focal oint ( < f ) of the mirror. Virtual images, located behind the mirror, have negative image distances by the sign convention of Table 3.. Choices (d) and (e) are true statements and all other choices are false. 8. With a real object in front of a convex mirror, the image is always uright, virtual, diminished in size, and located between the mirror and the focal oint. Thus, of the available choices, only choice (d) is a true statement. 9. A convergent lens forms inverted, real images of real objects located outside the focal oint ( > f ). When > f, the real image is diminished in size, and the image is enlarged if f > > f. For real objects located inside the focal oint ( < f ) of the convergent lens, the image is uright, virtual, and enlarged. In the given case, > f, so the image is real, inverted, and diminished in size. Choice (c) is the correct answer.. For a real object ( > ) and a diverging lens ( f < ), the image distance given by the thin lens euation is f ( f ) f < f f and the magnification is M > + Thus, the image is always virtual and uright, meaning that choice is a true statement while all other choices are false. ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS. Chromatic aberration is roduced when light asses through a material, as it does when assing through the glass of a lens. A mirror, silvered on its front surface never has light assing through it, so this aberration cannot occur. This is only one of many reasons why large telescoes use mirrors rather than lenses for their rimary otical elements. 4. All objects beneath the stream aear to be closer to the surface than they really are because of refraction. Thus, the ebbles on the bottom of the stream aear to be close to the surface of a shallow stream. 6. An effect 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 of water by this effect. 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 circumference of the lens you receive a ray that has assed through glass with more nearly arallel surfaces of entry and exit. Then the lens minimally distorts the direction to the object you are looking at.. Both words are inverted. However, OXIDE looks the same right side u and uside down. LEAD does not.
3 Chater 3. (a) No. The screen is needed to reflect the light toward your eye. Yes. The light is traveling toward your eye and diverging away from the osition of the image, the same as if the object was located at that osition. 4. (d). The entire image would aear because any ortion of the lens can form the image. The image would be dimmer because the card reduces the light intensity on the screen by 5%. PROBLEM SOLUTIONS 3. If you stand 4 in front of the mirror, the time reuired for light scattered from your face to travel to the mirror and back to your eye is d. 4 m t. 7 8 c 3. m s 9 Thus, the image you observe shows you ~ 9 s s younger than your current age. 3. (a) With the alm located. m in front of the nearest mirror, that mirror forms an image, I P, of the alm located. m behind the nearest mirror. (c) The farthest mirror forms an image, I B, of the back of the hand located. m behind this mirror and 5. m in front of the nearest mirror. This image serves and an object for the nearest mirror, which then forms an image, I B, of the back of the hand located 5. m behind the nearest mirror. The image I P (see art a) serves as an object located 4. m in front of the farthest mirror, which forms an image I P of the alm, located 4. m behind that mirror and 7. m in front of the nearest mirror. This image then serves as an object for the nearest mirror, which forms an image I P3 of the alm, located 7. m behind the nearest mirror. (d) Since all images are located behind the mirror, all are virtual images. 3.3 () The first image in the left-hand mirror is 5. ft behind the mirror, or. ft from the erson. () The first image in the right-hand mirror serves as an object for the left-hand mirror. It is located. ft behind the right-hand mirror, which is 5. ft from the left-hand mirror. Thus, the second image in the left-hand mirror is 5. ft behind the mirror, or 3. ft from the erson. (3) The first image in the left-hand mirror serves as an object for the right-hand mirror. It is located. ft in front of the right-hand mirror and forms an image. ft behind that mirror. This image then serves as an object for the left-hand mirror. The distance from this object to the left-hand mirror is 35. ft. Thus, the third image in the left-hand mirror is 35. ft behind the mirror, or 4. ft from the erson.
Mirrors and Lenses 3 3.4 The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.3 m behind the mirror. The image of the choir is. 8 m + 5. 3 m 6. m from the organist. Using similar triangles gives h 6. m. 6 m. 8 m 6. m. 8 m or h. 6 m 4. 58 m 3.5 In the figure at the right, θ θ since they are vertical angles formed by two intersecting straight lines. Their comlementary angles are also eual or α α. The right triangles PQR and P QR have the common side QR and are then congruent by the angle-side-angle theorem. Thus, the corresonding sides PQ and P Q are eual, or the image is as far behind the mirror as the object is in front of it. 3.6 (a) Since the object is in front of the mirror, >. With the image behind the mirror, <. The mirror euation gives the radius of curvature as + R... or R. +. 9 The magnification is M (. ). +.. 3.7 (a) Since the mirror is concave, R >. Because the object is located in front of the mirror, >. The mirror euation, + R, then gives the image distance as R R ( 4. )(. ) ( 4. ). +3. 3 Since >, the image is located 3. 3 in front of the mirror. continued on next age
3 Chater 3 3. 3 M. 333 4. Because >, the image is real and since M <, the image is inverted. 3.8 The lateral magnification is given by M. Therefore, the image distance is M (. 3 )( 3. ). 39. The mirror euation: + or R R. 39 3. gives R 3.. 39 +. 79 The negative sign tells us that the surface is convex. The magnitude of the radius of curvature of the cornea is R. 79 7. 9 mm 3.9 (a) For a convex mirror, the focal length is f R <, and with the object in front of the mirror, >. The mirror euation, + R f, then gives f ( 3. )(. ) 3. (. ) 7. 5 With <, the image is located 7.5 behind the mirror. The magnification is + M h 7. 5 h 3.. 5 Since < and M >, the image is virtual and uright. Its height is h Mh. 5.. 5 3. The image was initially uright but became inverted when Dina was more than 3 from the mirror. From this information, we know that the mirror must be concave because a convex mirror will form only uright, virtual images of real objects. When the object is located at the focal oint of a concave mirror, the rays leaving the mirror are arallel, and no image is formed. Since Dina observed that her image disaeared when she was about 3 from the mirror, we know that the focal length must be f 3. Also, for sherical mirrors, R f. Thus, the radius of curvature of this concave mirror must be R 6.
Mirrors and Lenses 33 3. The magnifi ed, virtual images formed by a concave mirror are uright, so M >. Thus, M h h 5. +. 5, giving.. 5. 5 + 3. 7. 5 The mirror euation then gives f R + 3 7.5..5 7.5 or 7. 5 f 5..5 3. Realize that the magnitude of the radius of curvature, R, is the same for both sides of the hubca. For the convex side, R R ; and for 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,., occurs with the convex side of the mirror. Hence, we have. R and for the concave side, 3. gives 3. R [3] [4] (a) Adding Euations [3] and [4] yields 3 + 3. Subtracting [3] from [4] gives 4 3 R 3. 3.3 The image is uright, so M >, and we have M or + 5. or R 6. +., or.. ( 5 ) 5 The radius of curvature is then found to be R + 5 5 5, or R.5 m. m +
34 Chater 3 3.4 (a) Your ray diagram should be carefully drawn to scale and look like the diagram given below: From the mirror euation with +. and f 5., the image distance is f (. )( 5. ). ( 5. ) + 5 5. 6. < + > and the magnification is M 6... 6. Thus, you should find that the image is uright, located 6. behind the mirror, six-tenths the size of the object. 3.5 The focal length of the mirror may be found from the given object and image distances as f +, or f + ( 5 )( 8. ) 5 + 8. + 6. For an uright image twice the size of the object, the magnification is M +., giving. Then, using the mirror euation again, + f becomes or +.. f f 6. 8. 5.. 3.6 (a) The mirror is convex, so f <, and we have f f 8.. The image is virtual, so <, or. Since we also know that 3, the mirror euation gives 3 + or f 8. and +6 This means that we have a real object located 6 in front of the mirror. The magnification is M + + 3. Thus, the image is uright and one-third the size of the object.
Mirrors and Lenses 35 3.7 (a) We know that the object distance is +.. Also, M > since the image is uright, and M since the image is half the size of the object. Thus, we have M +. or 5. and the image is seen to be located 5. behind the mirror. From the mirror euation, + f, we find the focal length to be f + (. )( 5. ). + ( 5. ). 3.8 (a) Since the mirror is concave, R >, giving R +4 and f R +. Because the image is uright ( M > ) and three times the size of the object M 3, we have M +3 and 3 The mirror euation then gives 3 3 or +8. The needed ray diagram, with the object 8. in front of the mirror, is shown below: From a carefully drawn scale drawing, you should find that the image is uright, virtual, 4 behind the mirror, and three times the size of the object. 3.9 (a) An image formed on a screen is a real image. Thus, the mirror must be concave since, of mirrors, only concave mirrors can form real images of real objects. The magnified, real images formed by concave mirrors are inverted, so M < and 5. m M 5, or. m 5 5 The object should be. m in front of the mirror. (a revisited) The focal length of the mirror is 6 f. m + 5. m 5. m, or f 5. m 6. 83 m
36 Chater 3 3. (a) From +, we find R R R. m. m. The table gives the image osition at a few critical oints in the motion. Between 3. m and. 5 m, the real image moves from.6 m to ositive infinity. From. 5 to, the virtual image moves from negative infinity to. Object Distance, Image Distance, 3. m.6 m.5 m ± Note the jum in the image osition as the ball asses through the focal oint of the mirror. The ball and its image coincide when and when + R, or R. m From y v yt + ayt, with v y, the times for the ball to fall from 3. m to these ositions are found to be t ( y). m. 639 s and 9. 8 m s a y t ( 3. m) 9. 8 m s. 78 s 3. From n n n + n R, with R, the image osition is found to be n n. ( 5. ) 38.. 39 or the virtual image is 38. below the uer surface of the ice. 3. The center of curvature of a convex surface is located behind the surface, and the sign convention for refracting surfaces (Table 3. in the textbook) states that R >, giving R + 8.. The object is in front of the surface ( > ) and in air (n. ), while the second medium is glass ( n. 5). Thus, n + n n n R becomes.. 5. 5. + 8. and reduces to 4. 6. (a) If If 8., ( 4. )(. ). 6.., ( 4. )( 8. ) 8. 6. + 4. continued on next age
Mirrors and Lenses 37 (c) If 4 (d) If., ( 4. )( 4. ) 4. 6.., ( 4. )(. ). 6. 8. 3. 43 3.3 Since the center of curvature of the surface is on the side the light comes from, R <, giving R 4.. Then, n n n n + becomes R... 5. 5 4. 4., or 4. Thus, the magnification M h n h n n h n h. 5 4.. 4. 3.4 For a lane refracting surface ( R ) gives ( ). 5 mm 3. 8 mm n n n + n R becomes n n (a) When the ool is full,. m and. (. m ). 5 m. 333 or the ool aears to be. 5 m dee. If the ool is half filled, then. m and. 75 m. Thus, the bottom of the ool aears to be.75 m below the water surface or. 75 m below ground level. enter the transarent shere from air 3.5 As arallel rays from the Sun object distance, ( n. ), the center of curvature of the surface is on the side the light is going toward (back side). Thus, R >. It is observed that a real image is formed on the surface oosite the Sun, giving the image distance as + R. Then n n n + n R becomes + n n. R R which reduces to n n. and gives n.
38 Chater 3 3.6 Light scattered from the bottom of the late undergoes two refractions, once at the to of the late and once at the to of the water. All surfaces are lanes ( R ), so the image distance for each refraction is ( n n ). At the to of the late, nwater 333 B n. 66 ( 8. ) 6. 4. B glass or the first image is 6.4 below the to of the late. This image serves as a real object for the refraction at the to of the water, so the final image of the bottom of the late is formed at nair nair B n n. + B B water water. 333 8.4. 3. 8 or 3.8 below the water surface. Now, consider light scattered from the to of the late. It undergoes a single refraction, at the to of the water. This refraction forms an image of the to of the late at T nair T. n. 333 (. ) 9. water or 9. below the water surface. The aarent thickness of the late is then y B T 3. 8 9. 4. 8 3.7 In the drawing at the right, object O (the jellyfish) is located distance in front of a lane water-glass interface. Refraction at that interface roduces a virtual image I at distance in front it. This image serves as the object for refraction at the glass-air interface. This object is located distance + t in front of the second interface, where t is the thickness of the layer of glass. Refraction at the glass-air interface roduces a final virtual image, I, located distance in front of this interface. From n + n ( n n ) R with R for a lane, the relation between the object and image distances for refraction at a flat surface is ( n n ). Thus, the image distance for the refraction at the water-glass interface is ( ng nw ). This gives an object distance for the refraction at the glass-air interface of ( ng nw ) + t and a final image osition (measured from the glass-air interface) of na n n n g a g n n na t + nw + n a t ng g w continued on next age
Mirrors and Lenses 39 (a) If the jellyfish is located. m (or ) in front of a 6. thick ane of glass, then + and t 6. and the osition of the final image relative to the glass-air interface is. + 333. ( 6. ) 79 79 m.. 5.. If the thickness of the glass is negligible ( t ), the distance of the final image from the glass-air interface is n n a g n n g w n + n. 75.. 75. 333 ( ) m a w so we see that the 6. thickness of the glass in art (a) made a 4. difference in the aarent osition of the jellyfish. (c) The thicker the glass, the greater the distance between the final image and the outer surface of the glass. 3.8 The wall of the auarium (assumed to be of negligible thickness) is a lane ( R ) refracting surface searating water ( n. 333) and air ( n. ). Thus, n n n n + gives the R n image osition as n. When the object osition changes by, the change in. 333 the image osition is. The aarent seed of the fish is then given by. 333 v image t t. s. 333. 333. 5 s 3.9 With R +. and R +. 5, the lens maker s euation gives the focal length as ( n ) 5 f R R (. ) 5.. 5. or f. 5. 3.3 The lens maker s euation is used to comute the focal length in each case. (a) ( n ) f R R (. 44 ) f. 8. 44 f ( ) 8.. (. ) f 6. 4 f 6. 4
3 Chater 3 3.3 The focal length of a converging lens is ositive, so f +.. The thin lens euation then yields a focal length of f.. (a) When +., (. )(. ) +. and.. M... so the image is located. beyond the lens, is real ( > ), is inverted (M < ), and is the same size as the object M.. When f +., the object is at the focal oint and no image is formed. Instead, arallel rays emerge from the lens. (c) When 5., ( 5. )(. ). and 5.. M. 5 +.. so the image is located. in front of the lens, is virtual ( < ), is uright (M > ), and is twice the size of the object M.. 3.3 (a) and Your scale drawings should look similar to those given below: Figure (a) Figure A carefully drawn-to-scale version of Figure (a) should yield a real, inverted image that is located in back of the lens and the same size as the object. Similarly, a carefully drawn-to-scale version of Figure should yield an uright, virtual image located in front of the lens and twice the size of the object. (c) The accuracy of the grah deends on how accurately the ray diagrams are drawn. Sources of uncertainty: a arallel line from the ti of the object may not be exactly arallel; the focal oints may not be exactly located; lines through the focal oints may not be exactly the correct sloe; the location of the intersection of two lines cannot be determined with comlete accuracy.
Mirrors and Lenses 3 3.33 From the thin lens euation, +, the image distance is found to be f f. ( ) (. ). +. (a) If 4., then 3. 3 and M ( 3. 3 ) + 4. 3. The image is virtual, uright, and 3.3 in front of the lens. If., then. and + M.. The image is virtual, uright, and. in front of the lens. (c) When., 6. 67 and M ( 6. 67 ) +. The image is virtual, uright, and 6.67 in front of the lens. 3.34 (a) and. Your scale drawings should look similar to those given below: 3. Figure (a) Figure A carefully drawn-to-scale version of Figure (a) should yield an uright, virtual image located 3.3 in front of the lens and one-third the size of the object. Similarly, a carefully drawn-to-scale version of Figure should yield an uright, virtual image located 6.7 in front of the lens and two-thirds the size of the object. (c) The results of the grahical solution are consistent with the algebraic answers found in Problem 3.33, allowing for small deviances due to uncertainties in measurement. Grahical answers may vary, deending on the size of the grah aer and accuracy of the drawing.
3 Chater 3 3.35 (a) The real image case is shown in the ray diagram. Notice that +. 9, or. 9. The thin lens euation, with f. 44, then gives +. 9. 44 or (. 9 ) + 3. 5 Using the uadratic formula to solve gives 9. 63 or 3. 7 Both are valid solutions for the real image case. The virtual image case is shown in the second diagram. Note that in this case, (. 9 + ), so the thin lens euation gives. 9 +. 44 or + (. 9 ) 3. 5 The uadratic formula then gives. or 5.. Since the object is real, the negative solution must be rejected, leaving.. 3.36 We must first realize that we are looking at an uright, magnified, virtual image. Thus, we have a real object located between a converging lens and its front-side focal oint, so <, >, and f >. The magnification is M +, giving. Then, from the thin lens euation, + or f (. 84 ) 5. 68 f 3.37 It is desired to form a magnified, 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 magnification then gives M h. 8 m, or 75. h 4. 3 m Also, we know that + 3. m. Therefore, + 75. 3. m, giving 3. m 3. 95 m 39. 5 mm 76. continued on next age
Mirrors and Lenses 33 (a) The thin lens euation then gives 76. + 75. 75. f or f 75. 75. ( 39. 5 mm) 39. mm 76. 76. 3.38 To have a magnification of M +3., it is necessary that 3.. The thin lens euation, with f +8. for the convergent convex lens, gives the reuired object distance as 3 3 8 8. 3.... or. 3.39 Since the light rays incident to the first lens are arallel, and the thin lens euation gives f.. The virtual image formed by the first lens serves as the object for the second lens, so 3. + 4.. If the light rays leaving the second lens are arallel, then and the thin lens euation gives f 4.. 3.4 (a) Solving the thin lens euation for the image distance gives f f or f For a real object, > and. Also, for a diverging lens, f < and f f. The result of art (a) then becomes ( f ) f f f + Thus, we see that < for all numeric values of and f. Since negative image distances mean virtual images, we conclude that a diverging lens will always form virtual images of real objects. (c) For a real object, > and. Also, for a converging lens, f > and f f. The result of art (a) then becomes f > if > Since must be ositive for a real image, we see that a converging lens will form real images of real objects only when > f (or > f since both and f are ositive in this situation).
34 Chater 3 3.4 The thin lens euation gives the image osition for the first lens as f 3. 5. 3. 5. + 3. and the magnification by this lens is M 3. 3... The real image formed by the first lens serves as the object for the second lens, so 4. +.. Then, the thin lens euation gives f. 5.. 5. and the magnification by the second lens is M 3.. + 3. 3. Thus, the final, virtual image is located 3. in front of the second lens and the overall magnification is M MM. 3. 3.. ( + ) 3.4 (a) With + 5., the thin lens euation gives the osition of the image formed by the first lens as f 5.. 5.. + 3. This image serves as the object for the second lens, with an object distance of.. 3.. (a virtual object). If the image formed by this lens is at the osition of O, the image distance is ( + ). +. 5. 5. The thin lens euation then gives the focal length of the second lens as f + ( ) 5... 5.. The overall magnification is M M M 3. 5. 5. +. 5. (c) Since <, the final image is virtual ; and since M >, it is uright.
Mirrors and Lenses 35 3.43 From the thin lens euation, f 4. 8. 4. 8. 8.. The magnification by the first lens is M 8. 4. +.. The virtual image formed by the first lens is the object for the second lens, so 6. + + 4. and the thin lens euation gives f 6 ( 4. )(. ) 4. ( 6.) 7. 47 The magnification by the second lens is M ( + ) + 7 47. 4. magnification is M MM +.. 533. 7. +. 533, so the overall The osition of the final image is 7. 47 in front of the second lens, and its height is h M h +. 7.. 7. Since M >, the final image is uright ; and since <, this image is virtual. 3.44 (a) We start with the final image and work backward. From Figure P3.44, observe that 5. 3. 9.. The thin lens euation then gives f f 9.. 9.. + 9. 74 The image formed by the first lens serves as the object for the second lens and is located 9.74 in front of the second lens. Thus, 5. 9. 74 4. 3 and the thin lens euation gives f f 4.3. 4.3. +3. 3 The original object should be located 3. 3 in front of the first lens. The overall magnification is M M M 4. 3 3.3 9. 9 74 5. 9. (c) Since M <, the final image is inverted ; and since <, it is virtual.
36 Chater 3 3.45 Note: Final answers to this roblem are highly sensitive to round-off error. To avoid this, we retain extra digits in intermediate answers and round only the final answers to the correct number of significant figures. Since the final image is to be real and in the film lane, + d. Then, the thin lens euation gives f f d 3. d 3.. From Figure P3.45, observe that d <.. The above result then shows that <, so the object for the second lens will be a virtual object. The object of the second lens ( L ) is the image formed by the first lens ( L ), so. d. d + 3. d d 3.. d 3. If d 5., then + 5. 5 ; and when d., + 45. 333. From the thin lens euation, When + 5. 5 d f f ( 5 ) 5. 5.. 3., then. 8 8. m. When + 45. 333 d., then. 4. 4 m. Thus, the range of focal distances for this camera is.4 m to 8. m. 3.46 (a) From the thin lens euation, the image distance for the first lens is f 5.. 5.. +3. With + 3., the image of the first lens is located 3. in back of that lens. Since the second lens is only. beyond the first lens, this means that the first lens is trying to form its image at a location. beyond the second lens. (c) (d) (e) (f) The image the first lens forms (or would form if allowed to do so) serves as the object for the second lens. Considering the answer to art above, we see that this will be a virtual object, with object distance.. From the thin lens euation, the image distance for the second lens is M M f 5... 5. 3.. 5. 4.. +. + 4. ( + ) (g) M MM... 4 (h) Since >, the final image is real, and since M <, that image is inverted.
Mirrors and Lenses 37 3.47 Since + 8. when +., we find that 8 f + + 8... 8. Then, when., 8 8 4 f. 8.. 4... 8. 8. 8. or + 5. 7 4. Thus, a real image is formed 5.7 in front of the mirror. 3.48 (a) We are given that 5 f, with both and f being ositive. The thin lens euation then gives f 5 f 5 f f f 5 f 4 (c) M 5 f 4 5 f 4 Since >, the image is real. Because M <, the image is inverted. Since the object is real, it is located in front of the lens, and with >, the image is located in back of the lens. Thus, the image is on the oosite side of the lens from the object. 3.49 Since the object is very distant ( ), the image distance euals the focal length, or +5. mm. Now consider two rays that ass undeviated through the center of the thin lens to oosite sides of the image as shown in the sketch below. From the sketch, observe that α tan. 35 mm 5. mm. 35 Thus, the angular width of the image is α tan. 35 38. 6 3.5 (a) Using the sign convention from Table 3., the radii of curvature of the surfaces are R 5. and R +.. The lens maker s euation then gives ( n ) 5 f R R. 5.. or f. If, then f.. The thin lens euation gives f (. ) +. and the following results: continued on next age
38 Chater 3 (c) If 3 f + 36., 9.. (d) If f +., 6.. (e) If f + 6., 4.. 3.5 As light asses left to right through the lens, the image osition is given by f 8. 8. + 4 This image serves as an object for the mirror with an object distance of 3 (virtual object). From the mirror euation, the osition of the image formed by the mirror is f ( ) 5 3. 3 5. 6. This image is the object for the lens as light now asses through it going right to left. The object distance for the lens is 3 ( 6. ), or 3 6. From the thin lens euation, 3 3 f3 3 3 6 8. 6 8. +6 Thus, the final image is located 6 to the left of the lens. The overall magnification is M M M M 3 M 4 6. 6 ( 3 ). 8 6 Since M <, the final image is inverted. 3, or 3 3.5 Since the object is midway between the lens and mirror, the object distance for the mirror is +. 5. The mirror euation gives the image osition as 5 4 R.. 5 5. 5., or + 5. This image serves as the object for the lens, so 5. 5.. Note that since <, this is a virtual object. The thin lens euation gives the image osition for the lens as f ( 6 7 5. )(. ) 5 3 5. 6.7. Since <, this is a virtual image that is located 5.3 in front of the lens or 5. 3 behind the mirror. The overall magnification is M M M 5..5 5. 3 5. Since M >, the final image is uright. + 8. 5
Mirrors and Lenses 39 3.53 A hemishere is too thick to be described as a thin lens. The light is undeviated on entry into the flat face. We next consider the light s exit from the curved surface, for which R 6.. The incident rays are arallel, so. Then, n n n + n R... 56 becomes + 6. from which. 7. 3.54 (a) The thin lens euation gives the image distance for the first lens as f 4.. 4.. The magnification by this lens is then M + 4. 4. 4... The real image formed by the first lens is the object for the second lens. Thus, 5. +. and the thin lens euation gives f. 5.. 5. +. The final image is. in back of the second lens.. The magnification by the second lens is M., so the overall. magnification is M MM (. ) (. ) +.. Since this magnification has a value of unity, the final image is the same size as the original object, or h M h +.... The image distance for the second lens is ositive, so the final image is real. (c) When the two lenses are in contact, the focal length of the combination is + + f f f. 5. The image osition is then f ( 5. )( 4. ) + 5. 4., or f 4..
33 Chater 3 3.55 With light going through the iece of glass from left to right, the radius of the first surface is ositive and that of the second surface is negative according to the sign convention of Table 3.. Thus, R +. and R 4.. Alying n n n + n R.. 5. 5. +. +. to the first surface gives which yields.. The first surface forms a virtual image. to the left of that surface and 6. to the left of the second surface. The image formed by the first surface is the object for the second surface, so + 6. and n n n n + gives R. 5... 5 + 6. 4. or + 3. The final image formed by the iece of glass is a real image located 3. to the right of the second surface. 3.56 Consider an object O at distance in front of the first lens. The thin lens euation gives the image osition for this lens as f The image, I, formed by the first lens serves as the object, O, for the second lens. With the lenses in contact, this will be a virtual object if I is real and will be a real object if I is virtual. In either case, if the thicknesses of the lenses may be ignored, and + f Alying the thin lens euation to the second lens, + + f f or + + f f + becomes f Observe that this result is a thin lens tye euation relating the osition of the original object O and the osition of the final image I formed by this two lens combination. Thus, we see that we may treat two thin lenses in contact as a single lens having a focal length, f, given by + f f f
Mirrors and Lenses 33 3.57 From the thin lens euation, the image distance for the first lens is f 4. 3. 4. 3. and the magnification by this lens is M + 4 3... The real image formed by the first lens serves as the object for the second lens, with object distance of. (a virtual object). The thin lens euation gives the image distance for the second lens as f f.. f (a) If f., then +. and the magnification by the second lens is + M (. )... The final image is located. to the right of the second lens and the overall ( + ) magnification is M MM 3.. 6.. Since M <, the final image is inverted. (c) If f +., then + 6. 67 and M 6. 67. +. 667 The final image is 6.67 to the right of the second lens and the overall magnification is ( + ) M MM 3.. 667.. Since M <, the final image is inverted. 3.58 The object is located at the focal oint of the uer mirror. Thus, the uer mirror creates an image at infinity (that is, arallel rays leave this mirror). The lower mirror focuses these arallel rays at its focal oint, located at the hole in the uer mirror. Thus, the image is real, inverted, and actual size. For the uer mirror: + + : f 7. 5 7. 5 For the lower mirror: + 7. 5 : 7. 5 Light directed into the hole in the uer mirror reflects as shown, to behave as if it were reflecting from the hole.
33 Chater 3 3.59 (a) The lens maker s euation, ( n ) f R R, gives ( n ) 5. 9.. 99. which simlifies to n + + 5.. 9.. 99. As light asses from left to right through the lens, the thin lens euation gives the image distance as f 8. 5. 8. 5. +3. 3 This image formed by the lens serves as an object for the mirror with object distance. + 6. 67. The mirror euation then gives R R 6.67 8. +. 6.67 8. This real image, formed. to the left of the mirror, serves as an object for the lens as light asses through it from right to left. The object distance is. +., and the thin lens euation gives 3 3 f 3. 5.. 5. +. The final image is located. to the left of the lens and its overall magnification is M M M M (c) Since M <, the final image is inverted. f 3.6 From the thin lens euation, the object distance is f. (a) If + 4 f, then 4 f 4 f f f 3 3. 3.. 8. 6. 67.. 5 4 f 3. ( f f When 3 f, we find 3 ) 3 f f 3 3 f 4. (c) In case (a), M f 4 4 f 3 3 and in case, M f 3 3 f 4 + 4.
Mirrors and Lenses 333 3.6 If R 3. m and R 6. m, the focal length is given by or f n n n f n + 3 m 6 m.. n 6. m n n n (a) If n. 5 and n The thin lens euation gives 6. m., then f ( 6. m )(. ) f.. 5. m. (. m )(. m ). m +. m 5. 45 m. [] A virtual image is formed 5.45 m to the left of the lens. If n. 5 and n. 33, the focal length is and f ( 6. m )(. 33 ). 33. 5 f 46. 9 m (. m )( 46. 9 m ). m + 46. 9 m 8. 4 m The image is located 8.4 m to the left of the lens. (c) When n. 5 and n and f., f ( 6. m )(. ) (. m )( 4. m ).. 5 7. m. m 4. m The image is 7. m to the left of the lens. + 4. m (d) Observe from Euation [] that f < if n > n and f > when n < n. Thus, a diverging lens can be changed to converging by surrounding it with a medium whose index of refraction exceeds that of the lens material.
334 Chater 3 3.6 The inverted image is formed by light that leaves the object and goes directly through the lens, never having reflected from the mirror. For the formation of this inverted image, we have M. 5 giving +. 5 The thin lens euation then gives +. 5. or (. ) + 6 7 5.. The uright image is formed by light that asses through the lens after reflecting from the mirror. The object for the lens in this uright image formation is the image formed by the mirror. In order for the lens to form the uright image at the same location as the inverted image, the image formed by the mirror must be located at the osition of the original object (so the object distances, and hence image distances, are the same for both the inverted and uright images formed by the lens). Therefore, the object distance and the image distance for the mirror are eual, and their common value is mirror mirror 4. 4. 6. 7 +3. 3 The mirror euation, +, then gives mirror mirror R fmirror f mirror 3 3 + 3 3 +.. 3. 3 3. 3 or f mirror + +. 7 3.63 (a) The lens maker s euation for a lens made of material with refractive index n. 55 and immersed in a medium having refractive index n is n 55 n f n R R. n R R Thus, when the lens is in air, we have. 55. f. R R air [] and when it is immersed in water,. 55. 33 f 33. R R water [] Dividing Euation [] by Euation [] gives f f water air. 33. 55... 55. 33 33.. 55. If f air 79., the focal length when immersed in water is f water ( 79 ) 33... 55 63. The focal length for a mirror is determined by the law of reflection, which is indeendent of the material of which the mirror is made and of the surrounding medium. Thus, the focal length deends only on the radius of curvature and not on the material making u the mirror or the surrounding medium. This means that, for the mirror, fwater fair 79.