CHAPTER 35. Answer to Checkpoint Questions

Transcription

1 956 CHAPTER 35 GEMETRICAL PTICS CHAPTER 35 Answer to Checkpoint Questions answer to kaleidoscope question: two mirrors that orm a V with an angle o 60. 0:d, :8d, :d. (a) real; (b) inverted; (c) same 3. (a) e; (b) virtual, same 4. virtual, same as object, diverging Answer to Questions. c. c 3. (a) a; (b) c 4. (a) x axis; (b) no, no; (c) direction in which your nose points 5. (a) no; (b) yes (ourth is o mirror end) 6. (a) a and c; (b) three times; (c) you 7. (a) rom innity to the ocal point; (b) decrease continually 8. convex 9. (d) (innite), tie o (a) and (b), then (c) 0. (a) decrease; (b) increase; (c) increase. mirror, equal; lens, greater. (a) :5, :7, :3; (b) :3, plus; :5, ; :7, minus 3. (a) all but variation ; (b) or, 3, and 4: right, inverted; or 5 and 6: let, same 4. (a) and (b) rightward

2 CHAPTER 35 GEMETRICAL PTICS (a) less; (b) less Solutions to Exercises & Problems E (a) Virtual, since the image is ormed by plane mirrors. (b) Same. In act you can easily veriy this by locating, or example, the images o two points, one at the head o the penguin and the other at its eet. (c) Same, since the image ormed by any plane mirror retains the original shape and size o an object. (d) The image o the penguin ormed by the top mirror is located a distance D above the top mirror, or L + D above the bottom one. Thus the nal image o the penguin, ormed by the bottom mirror, is a distance L + D rom the bottom mirror. E (a) Let the separation between you and the mirror be s, then the separation between you and your image is s. So your image moves toward you at a speed v i d(s)dt dsdt v: (b) Since the separation between the image and the mirror is also s, the image also moves toward the mirror at speed v. 3E The image is 0 cm behind the mirror and you are 30 cm in ront o the mirror. You must ocus your eyes or a distance o 0 cm + 30 cm 40 cm. 4E Reer to the diagram to the right. The ocus distance is q d (d + d ) + d 3 p (4:30 m + 3:30 m) + (5:00 m) 9:0 m : d d 3 d image d d d hummingbird camera

3 958 CHAPTER 35 GEMETRICAL PTICS 5E Consider A 0, the image o A in the mirror. The actual path o light rom A to B is AB, or which the angle o incidence satises + 90 and the angle o reection satises + 90 : So. To show that any other path, such as A 0 B as shown, is longer than AB, just note in the gure that A ' α α φ α B A B A B > A 0 B A 0 + B A + B : A' 6E (a) There are three images. Two are ormed by single reections rom each o the mirrors and the third is ormed by successive reections rom both mirrors. (b) The positions o the images are shown on the two diagrams below. The diagram on the let shows the image I, ormed by reections rom the let-hand mirror. It is the same distance behind the mirror as the object is in ront and is on the line that is perpendicular to the mirror and through the object. Image I is ormed by light that is reected rom both mirrors. You may consider I to be the image o I ormed by the right-hand mirror, extended. I is the same distance behind the line o the right-hand mirror as I is in ront and it is on the line that is perpendicular to the line o the mirror. The diagram below shows image I 3, ormed by reections rom the right-hand mirror. It is the same distance behind the mirror as the object is in ront and is on the line that is perpendicular to the mirror and through the object. As the diagram shows, light that is rst reected rom the right-hand mirror and then rom the let-hand mirror orms an image at I. I I I I 3 object object

4 CHAPTER 35 GEMETRICAL PTICS 959 7P (a) and (b) As depicted below, seven images can be seen or 45 while ve can be seen or 60. I I object I I I 3 45 o 60 o object I 7 I 4 I 5 I 3 I 4 I 5 I 6 (c) In this case the number o images that can be seen ranges rom to 3, depending on the locations o both the object and the observer. As an example, in the ray diagram to the right observer can only see one image (I ), while observer can see both I and I. In act i oobserver moves urther back rom the vortex o the two mirrors he shouls also be able to see a third image, I 3, which is ormed by light rays rom the object which rst strike the horizontal mirror and then strike the inclined mirror. You can veriy these results by drawing light rays as shown here or I and observer. observer I I 3 I observer 8P When S is barelay able to see B the light rays rom B must reect to S o the edge o the mirror, as shown. So x d d d ; or x d 3:0 m :5 m. I x mirror x B d/ d d S

5 960 CHAPTER 35 GEMETRICAL PTICS 9P Consider a single ray rom the source to the mirror and let be the angle o incidence. The angle o reection is also and the reected ray makes an angle o with the incident ray. Now rotate the mirror through the angle so the angle o incidence increases to +. The reected ray now makes an angle o ( + ) with the incident ray. The reected ray has been rotated through an angle o. I the mirror is rotated so the angle o incidence is decreased, then the reected ray makes an angle o ( ) with the incident ray. Again it has been rotated through. The diagrams show the situation or 45. The ray rom the object to the mirror is the same in both cases and the reected rays are 90 apart. I +α +α I 0P Consider the two light rays, r and r 0, which are reected o the mirror and reach the edge o the pupil. In the gure to the right x mirror x d 3 d tan d 3 d tan x d 3 d ; d d r' d which gives r x d d 3 d + d (0 cm)(5:0 mm) 0 cm + 0 cm :67 mm : α d 3 pupil The area o the mirror used to observe the image o the object is thereore α A 4 x 4 (:67 mm) : mm : P The intensity o light rom a point source varies as the inverse o the square o the distance rom the source. Beore the mirror is in place, the intensity at the center o the screen is given by I 0 Ad, where A is a constant o proportionality. Ater the mirror is in place,

6 CHAPTER 35 GEMETRICAL PTICS 96 the light that goes directly to the screen contributes intensity I 0, as beore. Reected light also reaches the screen. This light appears to come rom the image o the source, a distance d behind the mirror and a distance 3d rom the screen. Its contribution to the intensity at the center o the screen is I r A (3d) A 9d I 0 9 : The total intensity at the center o the screen is I I 0 + I r I 0 + I I 0 : The ratio o the new intensity to the original intensity is II P Apply the law o reraction: sin sin 0 n w n air ; which in our case reduces to 0 n w (since both and 0 are small, and n air. Thus in the triangle AB (object) jabj h tan h ; in the triangle CBD h jbcj h tan 0 h 0 h n w ; C ' B ' A air water and in the triangle ACI jaij x + h. So h x jaij h jacj tan jabj + jbcj h + h n w h h h ' D x mirror h + h n w h 50 cm + 35 cm : (00 cm) :33 00 cm I (image)

7 96 CHAPTER 35 GEMETRICAL PTICS 3P* y Set up a coordiante system as shown. Suppose an incident ray o light i rst strikes one o the mirrors incident light (i) in the xy plane. I the unit vector denoting the direction o i is given by i cos + j cos + k cos, where ; ; are the angles i makes with the x; y and z axis, then ater reection o the xy plane the unit vector becomes i cos + j cos + z k cos( ) i cos + j cos k cos. Next suppose it strikes the mirror in the xz plane. The x unit vector o the reected ray is i cos + j cos( relected light (r) ) k cos i cos j cos k cos : Finally as it reects o the mirror in the yz plane becomes, so the unit vector in the direction o the reected ray, r, is given by i cos( ) j cos k cos ( i cos + j cos + k cos ), exactly reversed rom the direction o i. 4E In Fig (b) p 8:50 mm and r 5:5 mm, so i r p 5:5 mm 8:50 mm 5:5 mm according to the prediction by Eqs and The measured value o i is In Fig (c) p 8:5 mm and r 3: mm, so i r p 5:5 mm + 8:5 mm 5: mm 7:0 mm. according to Eqs and The measured valued value is 6:5 mm. 5E Use Eqs and 35-4, and note that m ip. Thus p pm r : Solve or p: p r m 35:0 cm 0:5 cm : :50 6P (a) +0 cm (positive, because the mirror is concave), r (+0 cm) +40 cm, i ( p) (0 cm 0 cm) 0 cm, m ip ( 0 cm0 cm)

8 CHAPTER 35 GEMETRICAL PTICS 963 +:0, the image is virtual (not real) and upright (not inverted). The ray diagram is plotted below. I Similarly: (b) plane,,, 0 cm, no (see the ray diagram below); I (c) concave, +40 cm, +60 cm, :0, yes, yes (see the ray diagram below); I

9 964 CHAPTER 35 GEMETRICAL PTICS (d) concave, +0 cm, +40 cm, +30 cm, yes, yes (see the ray diagram below); I (e) convex, 0 cm, +0 cm, +0:50, no, no (see the ray diagram below); I () convex,, 40 cm, 8 cm, +80 cm, no, no (see the ray diagram below); I

10 CHAPTER 35 GEMETRICAL PTICS 965 (g) 0 cm,,, +5 cm, +0:8, no, no (see the ray diagram below); I (h) concave, +8 cm, +6 cm, + cm,, yes (see the ray diagram below). I 7P (a) Suppose one end o the object is a distance p rom the mirror and the other end is a distance p + L. The position i o the image o the rst end is given by p + i ; where is the ocal length o the mirror. Thus The image o the other end is at so the length o the image is i i p p : (p + L) p + L ; L 0 i i p p (p + L) p + L L (p )(p + L ) :

11 966 CHAPTER 35 GEMETRICAL PTICS Since the object is short compared to p write L 0 L we may neglect the L in the denominator and p (b) The lateral magnication is m ip and since i p(p ), this can be written m (p ). The longitudinal magnication is m 0 L0 L p : m : 8P (a) Solve i rom Eqs and 35-4 to obtain i p(p ) pr(p r). Dierentiate both sides with respect to time and use v dpdt: v I di dt d pr dt p rv (p r) + v pr r r (p r) p r v : (b) (c) (d) 5 cm v I (5:0 cm/s) 0:56 cm/s : (30 cm) 5 cm 5 cm v I (5:0 cm/s) : 0 3 cm/s : (8:0 cm) 5 cm 5 cm v I (5:0 cm/s) 6:7 cm/s : (:0 cm) 5 cm 9P (a) Use Eq and note that n n air :00; n n; p ; and i r: Solve or n: n :00. (b) Now i r so Eq becomes :00 + n r n : r n r n ; r which is not valid unless n! or r!. So it is impossible to ocus at the center o the sphere.

12 CHAPTER 35 GEMETRICAL PTICS 967 0P Use Eq (a) i n n n r n :5 :0 :5 p 30 cm :0 8 cm ; 0 cm and the image is virtual (not real), since i < 0. The ray diagram is plotted below. medium medium C I n > n Similarly: (b) 33 cm, no; (c) +7 cm, yes; (d) any value o n is possible, no; (e) +30 cm, no; () +0 cm, no; (g) 6 cm, no; (h) :0, yes. P In the ray diagram shown to the right the penny is located at the bottom o the pool (point D), while its image is at point C. According to the law o reraction sin n sin 0, which reduces to n 0 in our case since both and 0 are small. In the triangle ABC we have AB d a tan d a ; while in the triangle ABD AB d tan 0 d 0 d : n d a A C (image o penny) ' B ' d D (penny)

13 968 CHAPTER 35 GEMETRICAL PTICS Equate the two expressions or AB to obtain d a d : n This is d a dn. P Reer to the ray diagram to the right. From the result o the last problem we get d a d d a d0 d + d a n w n w d + d (n c n w ) n w 0 mm + (40 mm)(:33:46) :33 4 mm : d ' d a d water 3P A' φ- φ D h φ- C image o ish B' R/ A B ish R/ air water Reer to the ray diagram shown above. Although the angles shown here in the diagram are not very small (or clarity), we consider only very small angles since the size o the goldsh is (hopeully) much less than R. In this approximation sin sin m w, or n w : In the triangle DA 0 A R DA DC + CA A 0 C tan( ) + A0 C tan() A0 C + A0 C n w + A0 C hr n w + ;

14 CHAPTER 35 GEMETRICAL PTICS 969 which gives The magnication is thereore A 0 C h n w + : m A0 B AB A0 C + h h + : :4 : n w + 4E Solve Eq or the image distance i: i p(p ). The lens is diverging, so its ocal length is 30 cm. The object distance is p 0 cm. Thus i (0 cm)( 30 cm) (0 cm) ( 30 cm) cm : The negative sign indicates that the image is virtual and is on the same side o the lens as the object. The ray diagram is shown on the right. F I lens 5E (a) w w In the two shaded triangles shown above W W, so W W :

15 970 CHAPTER 35 GEMETRICAL PTICS (b) w w See the arrangement depicted in the ray diagram above. The two lenses are separated by j j: Here W W j j: 6E rom I PA we get I I PA PA A A W W : 7E Use the lens maker's equation, Eq. 35-0: (n ) ; r r where is the ocal length, n is the index o reraction, r is the radius o curvature o the rst surace encountered by the light and r is the radius o curvature o the second surace. Since one surace has twice the radius o the other and since one surace is convex to the incoming light while the other is concave, set r r to obtain (n ) + r r 3(n ) r : Solve or r : r 3(n ) The radii are 45 mm and 90 mm. 3(:5 )(60 mm) 45 mm :

16 CHAPTER 35 GEMETRICAL PTICS 97 8E Let the diameter o the Sun be d p and that o the image be d i. Then i d i md p d p p p d p (0:0 0 m)()(6: m) :50 0 m : m :86 mm : 9E (a) Use Eq. 35-0: (n ) r r (:5 ) +40 cm : 0 cm (b) Solve i rom Eq. 35-9: i p 40 cm ; 40 cm i.e., the image is located at innity. 30E From Eq. 35-0, i / r r r r r r is positive, i.e., r > r, then the lens is converging. therwise it is diverging. (a) Converging, since r! and r is nite (so r > r ). (b) Diverging, since r! and r is nite (so r < r ). (c) Converging, since r > r. (d) Diverging, since r < r. 3E For a thin lens (p)+(i) (), where p is the object distance, i is the image distance, and is the ocal length. Solve or i: i p p :

17 97 CHAPTER 35 GEMETRICAL PTICS Let p + x, where x is positive i the object is outside the ocal point and negative i it is inside. Then ( + x) i : x Now let i + x 0, where x 0 is positive i the image is outside the ocal point and negative i it is inside. Then x 0 ( + x) i x x and xx 0. 3E Solve i rom Eq. 35-9: The height o the image is thus i i h i mh p h p h p p p p p p : (75 mm)(:80 m) 7 m m 5:0 mm : 33P Use Eq (a) (n ) r r (:5 ) 40 cm Since > 0 the lens orms a real image o the Sun. (b) Now (:5 ) 80 cm ; 40 cm and the image ormed is real (since > 0). (c) Now (:5 ) 40 cm and the image ormed is real (since > 0). (d) Now (:5 ) 40 cm and the image ormed is virtual (since < 0). 40 cm ; 60 cm 40 cm ; 40 cm 40 cm : 40 cm

18 CHAPTER 35 GEMETRICAL PTICS 973 (e) Now (:5 ) 80 cm ; 40 cm and the image ormed is virtual (since < 0). () Now (:5 ) 40 cm ; 60 cm 40 cm and the image ormed is virtual (since < 0). 34P (a) +0 cm (positive, because the lense is a converging one), r and r cannot be determined (since n is not given), i ( p) (0 cm 0 cm) 0 cm, n cannot be determined, m ip 0 cm0 cm :0, the image is real (since i > 0) and inverted (since m < 0). The ray diagram is plotted below. I Similarly, using an X to denote a quantity which cannot be determined rom the given data, we have (b) converging, X, X, 0 cm, X, +.0, no, no; (c) converging, +, X, X, 0 cm, X, no, no; (d) diverging,, X, X, 3.3, X, no, no; (e) converging, +30 cm, 5 cm, +.5, no, no; () diverging, 30 cm, 7:5 cm, +0.75, no, no; (g) diverging, 0 cm, 9: cm, +0.9, no, no; (h) diverging, 0 cm, X, X, 5 cm, X, +, no; (i) converging, +3:3 cm, X, X, +5 cm, X, yes, yes. 35P Without the diverging lens (lens ), the real image ormed by the converging lens (lens ) is located at a distance i p 0 cm 40 cm 40 cm

19 974 CHAPTER 35 GEMETRICAL PTICS to the right o lens. This image now serves as an objet or lens, with p 0 cm) 30 cm: So i p 5 cm 30 cm ; 30 cm (40 cm i.e., the image ormed by lens is located 30 cm to the let o lens. It is virtual (since i < 0). The magnication is m ( i p ) ( i p ) + so the image has the same size and orientation as the object. 36P Suppose that the lens is placed to the let o the mirror. The image ormed by the converging lens is located at a distnace i p 0:50 m :0 m :0 m to the right o the lens, or :0 m :0 m :0 m in ront o the mirror. The image ormed by the mirror or this real image is then at :0 m to the right o the the mirror, or :0 m + :0 m 3:0 m to the right o the lens. This image then results in another image ormed by the lens, located at a distance i 0 p 0 0:50 m 6:0 m 3:0 m to the let o the lens (i.e., :6 cm rom the mirror). (b) and (c) The nal image is real and has the same orientation as the object, as you should veriy by drawing a ray diagram. (d) m i i 0 p p 0 :0 m :0 m 0:60 m +0:0 : 3:0 m 37P (a) First, the lens orms a real image o the objet located at a distance i p to the right o the lens, or at p ( + ) in ront o the mirror. The subsequent image ormed by the mirror is located at a distance i p

20 CHAPTER 35 GEMETRICAL PTICS 975 to the let o the mirror, or at p 0 ( + ) to the right o the lens. The nal image ormed by the lens is that at a distance i 0 to the let o the lens, where i 0 p 0 ; i.e., right at the location o the object. The image is real and inverted. The lateral magnication is (b) m i p i p i 0 p 0 :0 : I I 3 I 38P (a) A convex (converging) lens, since a real image is ormed. (b) Since i d p and ip, p d 3 (40:0 cm) 3 6:7 cm : (c) i + p d3 + d d3 9 (40:0 cm) 9 8:89 cm : 39P (a) For the image ormed by the rst lens i p 0 cm 0 cm : 0 cm

21 976 CHAPTER 35 GEMETRICAL PTICS For the subsequent image ormed by the second lens p 30 cm i p :5 cm 50 cm ; 0 cm 0 cm 0 cm, so i.e., the nal image is 50 cm to the let o the second lens, which means that it coincides with the object. The magnication is m i i p p 0 cm 0 cm 50 cm 5:0 ; 0 cm meaning that the object is enlarged ve times in its nal image. (b) lens lens I I inal (c) and (d) It is virtual and inverted (see the diagram above). 40P Place an object ar away rom the composite lens and nd the image distance i. Since the image is at a ocal point i, the eective ocal length o the composite. The nal image is produced by two lenses, with the image o the rst lens being the object or the second. For the rst lens (p ) + (i ) ( ), where is the ocal length o this lens and i is the image distance or the image it orms. Since p, i. The thin lens equation, applied to the second lens, is (p ) + (i ) ( ), where p is the object distance, i is the image distance, and is the ocal length. I the thicknesses o the lenses can be ignored the object distance or the second lens is p i. The negative sign must be used since the image ormed by the rst lens is beyond the second lens i i is positive. This means the object or the second lens is virtual and the object distance is negative. I i is negative the image ormed by the rst lens is in ront o the second lens and p is positive. In the thin lens equation, replace p with and i with to obtain +

22 CHAPTER 35 GEMETRICAL PTICS 977 or Thus + + : + : 4P (a) Since a beam o parallel light will be ocused at a distnace rom the lens, the shorter the ocal length the greater the ability or the lens to ocus. Thereore P is a reasonable denition. (b) Use the result o 40P: P + + P + P : 4P Using Eq and noting that p + i d 44 cm, we obtain p dp + d 0. Solve or p: p (d p d 4d) cm p (44 cm) 4(44 cm)( cm) cm : 43P For an object in ront o a thin lens, the object distance p and the image distance i are related by (p) + (i) (), where is the ocal length o the lens. For the situation described by the problem all quantities are positive, so the distance x between the object and image is x p + i. Substitute i x p into the thin lens equation and solve or x. You should get x p p : To nd the minimum value o x, set dxdp 0 and solve or p. Since dx dp p(p ) (p ) ; the result is p. The minimum distance is x min p p () 4 :

23 978 CHAPTER 35 GEMETRICAL PTICS This is a minimum, rather than a maximum, since the image distance i becomes large without bound as the object approaches the ocal point. 44P (a) I the object distance is x, then the image distance is D becomes x + D x : x and the thin lens equation Multiply each term in the equation by x(d x) to obtain x Dx + D 0. Solve or x. The two object distances or which images are ormed on the screen are and x D p D(D 4) x D + p D(D 4) : The distance between the two object positions is d x x p D(D 4) : (b) The ratio o the image sizes is the same as the ratio o the lateral magnications. I the object is at p x the lateral magnication is m i p D x x : Now x (D d), where d p D(D ), so m D (D d) (D d) D + d D d : Similarly, when the object is at x the magnication is m I p D x x D (D + d) (D + d) D d D + d : The ratio o the magnications is m m (D d)(d + d) D (D + d)(d d) d : D + d 45P Light reracted at the let surace o the sphere passes through and is reracted again at the right surace. Apply Eq twice, using the image ormed by the rst surace as

24 CHAPTER 35 GEMETRICAL PTICS 979 the object or the second surace. For reraction at the rst surace the object distance is p and the radius o curvature is r R, where R is the radius o the sphere. The surace is convex to the incoming light so the radius o curvature is positive. The medium o incidence is air (or vacuum) and the medium o reraction is glass, so n and n n, where n is the index o reraction o the glass. I i is the image distance or the image ormed by the let surace, then n n i R ; so i nr n : The image distance is measured rom the let edge o the sphere. Since n is less than, i is greater than R and the image is ormed to the right o the sphere, a distance i R rom its right edge. The object distance or reraction at the right surace is p R i R nr n n n R : This is a negative number, indicating that the light is converging as it strikes the right surace o the sphere. Now the medium o incidence is glass and the medium o the reracted light is air, so n n and n. The surace is concave to the incoming light so the radius o curvature is r R. Eq yields n(n ) (n )R + i n R ; where i is the image distance or the second reraction. Thus n i R n(n ) (n )R (n ) ( n)r : The image distance is i n (n ) R : This is positive, indicating that the nal image is to the right o the sphere. 46E (a) I L is the distance between the lenses, then according to Fig. 35-8, the tube length is s L ob ey 5:0 cm 4:00 cm 8:00 cm 3:0 cm. (b) Solve (p) + (i) ( ob ) or p. The image distance is i ob + s 4:00 cm + 3:0 cm 7:0 cm, so p i i (7:0 cm)(4:00 cm) 7:0 cm 4:00 cm 5:3 cm :

25 980 CHAPTER 35 GEMETRICAL PTICS (c) The magnication o the objective is m i p 7:0 cm 5:3 cm 3:5 : (d) The angular magnication o the eyepiece is m 5 cm ey 5 cm 8:00 cm 3:3 : (e) The overall magnication o the microscope is M mm ( 3:5)(3:3) 0: : 47E The minimum diameter o the eyepiece is given by d ey d ob m 75 mm 36 : mm : 48P (a) Without the magnier hp n (see Fig. 35-7). With the magnier, let p P n and i jij P n, we obtain p i + jij + ; P n so m 0 hp + P n + P n hp n P n 5 cm + : (b) Now i jij! so p + i p and m 0 hp P n hp n P n 5 cm : (c) For 0 cm, m + 5 cm0 cm 3:5 (case a) and 5 cm0 cm :5 (case b). 49P (a) In this case i < 0 so i jij, and Eq becomes p jij :

26 CHAPTER 35 GEMETRICAL PTICS 98 Deerentiate with respect to t to obtain v I djij dt i dp : p dt As the object is moved toward the lens dpdt < 0. So djijdt < 0, i.e., the image moves in rom innity. The angle 0 tan (hp) increases since p decreases. (b) When the image appears to be at the near point, i.e., jij P n. (c) In this case p i + jij + ; P n so (d) The linear magnication is given by m 0 hp P n hp n + 5 cm + : P n m jij p jij + + jij jij + P n 5 cm + m : 50P (a) When the eye muscle is relaxed p!, so i 40:0 cm. Now so 0 :35 cm. (b) Decrease, since 0 / r is decreased. 0 i + p + p :50 cm + 40:0 cm ; 5P (a) A parallel ray o light ocuses at the ocal point behind the lens. In the case o arsightedness we need to bring the ocal point closer, i.e., to reduce the ocal length. From 40P we know that we need to use a converging lens o certain ocal length which, when combined with the eye o ocal length, gives ( + ) <. Similarly, you can see that in the case o nearsightness we need to do just the opposite, i.e., to wear a diverging lens. (b) As an object is brought closer to a xed-ocus lens the image ormed also moves away rom the lens. Reading requires that the object be close to the eyes. I you need visual aid or reading it means that your eyes cannot ully adjust themselves to prevent the image rom moving urther away rom your retina. Thereore you are arsighted. (c) The biocal glasses can provide suitable corrections or dierent types visual deects that appear in dierent situations, such as reading (i.e., when the object is placed close to the eyes) and viewing a distant object.

27 98 CHAPTER 35 GEMETRICAL PTICS 5P (a) i p 5:0 cm (b) The change in the lens-lm distance is 5:3 cm 5:3 cm : 00 cm 5:0 cm 0:30 cm: 53P Reer to Fig For the intermediate image p 0 mm and i ( ob + s + ey ) ey 300 m 50 mm 50 mm; so ob i + p 50 mm + 0 mm 9:6 mm ; and s ( ob + s + ey ) ob ey 300 mm 9:6 mm 50 mm 40 mm: Then rom Eq s 5 cm 40 mm 50 mm M 5 : ob ey 9:6 mm 50 mm 54 (a) 0:500 rad: 7:799 cm; 0:00 rad: 8:544 cm; 0:000 rad: 8:57 cm; mirror equation: 8:57 cm; (b) 0:500 rad: 3:56 cm; 0:00 rad: :05 cm; 0:000 rad: :00 cm; mirror equation: :00 cm