24 Geometrical Optics &...

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1 804 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 24 Geometrical Optics &... Answers to Discussion Questions 24. The ocal length increases because the rays are not bent as strongly at the water-glasnterace It will be positive. It s a diverging lens The ocal length dependn the index o reraction and that dependn the wavelength They provide an air space in ront o the cornea so that the eye can work with its normal amount o reraction at that irst interace Form a real image o a very distant object; the image-distance then approaches the ocal length. Place the two in contact, shine in parallel light and measure, knowing that / / + +/, where + is given and is to be ound The rump in the real, the head in the virtual.

2 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT The gas provides a medium with an index greater than one and so lessens the bending o the rays. At is pumped out, the rays are brought to convergence more strongly and the ocal point moves toward the lens The rays must have been converging to begin with The radiu curvature is and so is. That means the object- and image-distances must have equal magnitudes. Thus, the magniication is Apoint source emits spherical waves which strike the surace and appear to come rom a point source at the other ocus. Hence the relected waves are spherical too. 24. I the soldiers lined up along a parabolic arc with the ship s sails at the ocu the parabola, parallel rays rom the Sun relected o their shields would converge on the ship The ocal length is shorter and so the dioptric power is greater The target is at one o the two ocii o the hyperboloid and rays relected rom it appear to come rom the other ocus, F (H), but this also a ocus, F (E), othe ellipsoidal mirror. Rays appearing to come rom one ocu the ellipsoid, ater relecting o it, converge to the other ocus, F 2 (E), atthe ilm plane The rays will be more converging than necessary so objects at ininity (and ar away) will be blurry because the eye cannot handle converging rays. Nearer objects with more strongly diverging rays will be seen clearly The object has a diameter d, where d(200) m, d m, which is the same as the wavelength o violet light. We cannot hope to see objects that are smaller than the

3 806 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT probe being used; namely, the wavelength o light. The amount o diraction will then obscure the image completely. So the details we wish to observe cannot be iner than λ, which puts a practical limit on the magniication. A 2000 microscope will only make larger blurred images showing no more detail The special lat ilament o the lamp is at the center o the curvature o the spherical mirror so as to relect light back toward the ilament which is at the ocu the lens. Thus most o the light emerges as a parallel beam. By shiting the location o the lamp with respect to the lens, the shape o the beam can be changed The ilament is located at one ocu the ellipsoid and the relected light is made to converge at the other ocus, which also corresponds to the ront ocal point o the lens combination geometrical optics An idealized domain o optics, where the diraction o light is negligible and all the light travel in straight linen accordance with the Law Relection and Reraction. lens A ground or molded piece o glass, plastic, or other transparent material with opposite suraces either or both o which are curved, by mean which light rays are reracted so that they converge or diverge to orm an image. aspherical suraces Suraces whose shape do not resemble part o a sphere. converging lens A lens which causes the incoming rays to converge to some extent, to bend towards the central axis. diverging lens A lens which turns the incoming rayutward away rom the central axis. concave Curved like the inner surace o a sphere. convex Having a surace or boundary that curver bulgeutward, as the exterior o a sphere. paraxial Rays that are not ar rom the central axis and enter only at shallow angles are known to be paraxial. thin lens Lenses or which the radii o curvature o the suraces are large compared to the thickness. object distance The distance between the object and the optical instrument (such as a lenr a mirror). image distance The distance between the optical instrument (such as a lenr a mirror) and the image it orms. Thin-Lens Equation / +/ (n l )(/R /R 2 ). Lensmaker s Formula / (n l )(/R /R 2 ).

4 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 807 ocal point A point at which ray light or other radiation converge or rom which they appear to diverge, as ater reraction or relection in an optical system. ocal plane The plane containing the ocal point, on which parallel rays ocus. positive lens Converging lens. negative lens Diverging lens. optical center A special point at the center o any thin lens, such that any paraxial ray heading toward it will pass through undeviated and may be drawn as a straight line. real image An image ormed by converging light rays. virtual image An image rom which ray relected or reracted light appear to diverge. transverse magniication The ratio o any transverse (i.e., perpendicular to the optical axis) dimension o the image ormed by an optical system to the corresponding dimension o the object. longitudinal magniication The ratio o any longitudinal (i.e., parallel to the optical axis) dimension o the image ormed by an optical system to the corresponding dimension o the object. cornea The transparent convex anterior portion o the outer ibrous coat o the eyeball that covers the iris and the pupil and is continuous with the sclera. crystalline lens A transparent, biconvex body o the eye between the iris and the vitreous humor that ocuses light rays entering through the pupil to orm an image on the retina. retina A delicate, multilayered, light-sensitive membrane lining the inner eyeball and connected by the optic nerve to the brain. accommodation The ine ocusing perormed by the human eye. ar-point The point that is seen clearly by the unaccomodated eye. near-point The closest point that can be clearly seen with maximum accommodation. angular magniication The ratio o the size o the retinal image ormed by the device to the size o the retinal image ormed by the unaided eye at normal viewing distance. magniying glass A lenr combination o lenses that enlarges the image o an object. It is usually a convex lens placed within one ocal length o the object. intermediate image An image ormed by one o a serie optical instrument and one which serves as the eective object o the next instrument. dioptic power A measure o the ocusing power o a lens, deined as the reciprocal o the ocal length. arsightedness An abnormal condition o the eye in which vision is better or distant objects than or near objects. It results rom the eyeball being too short rom ront to back, causing images to be ocused behind the retina. nearsightedness A visual deect in which distant objects appear blurred because their images are ocused in ront o the retina rather than on it objective eyepiece The lenr lens group closest to the eye in an optical instrument. mirror ormula / +/ 2/R.

5 808 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Answers to Multiple Choice Questions. b 2. c 3. b 4. c 5. d 6. d 7. b 8. b 9. a 0. c. b 2. d 3. c 4. e 5. c 6. e 7. e 8. b 9. a 20. d 2. d 22. a 23. d 24. c 25. b Solutions to Problems 24. First, use the Lensmaker s Formula, Eq. (24.2), to ind [ ( (n l ) )] [ ( )] (.50 ) R R m 0.50 m m Now locate the image rom Eq. (24.4), / +/ /, or (.00 m)(0.50 m).00 m 0.50 m.0 m, i.e., the image is located.0 mtothe right o the lens With the previous problem in mind, we irst ind [ ( (n l ) )] [ ( )] (.50 ) R R 2.00 m.5m m Now ind rom Eq. (24.4), / +/ / (2.00 m)(.5 m) 2.00 m.5 m 6.0 m, i.e., the light source should be positioned 6.0 minront o the lens Let R R 2 R or the lens. Then Eq. (24.) gives / +/ (n l )(/R +/R) 2(n l )/R. Plug in 24.0 cm, and n l.50, and solve or R R 2(n l ) / +/ 2(.50 ) /(24.0 cm)+/(24.0 cm) 2cm.

6 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT The lens convex since it is thicker in the middle. The ocal length is given by Eq. (24.2), with n l.58, R.00 m, and R 2 (or the lat surace) [ ( (n l ) )] [ ( (.50 ) R R 2.00 m )] 2.0m. So the sunlight (with ) will be ocused a distance 2.0 mbehind the lens Use Eq. (24.2) or the ocal length, with n l.58, R +.00 m, and R 2.00 m [ ( (n l ) )] [ ( )] (.58 ) R R 2.00 m 0.86 m..00 m 24.6 Forabi-convex lenne o the two radii o curvature is negative. Since R 2.0m < R mwemusttake R 2.0mand R mtoensure that > 0. Plug these into Eq. (24.2), along with n l.5, toind [ ( (n l ) )] [ ( )] (.5 ) R R m 2.9m. 5.0 m 24.7 From the Lensmaker s Formula [Eq. (24.2)] [ ( (n l ) )] [ ( )] (.50 ) R R m.0 m m Note that <0 since it s a concave lens Similar to the previous problem, this time R m and R m, so [ ( (n l ) )] [ ( )] (.50 ) R R m +.0m m Note that >0 since it s a convex lens.

7 80 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 24.9 The ocal length o the lens given by the Lensmaker s ormula, / (n l )(/R /R 2 ). In this case.60 m, R 2.00 m, R 2 (or the planar surace), which we plug into the ormula to solve or n l n l + (/R /R 2 ) + (.60 m)(/2.00 m / ) It is clear that 2.50 m or the lens, since 2.50 m when. Nowset 4.00 m and ind rom Eq. (24.4), / +/ / (4.00 m)(2.50 m) 4.00 m 2.50 m 6.67 m, i.e., the screen must be positioned at 6.67 mbehind the lens. Since > 0 the image is real. 24. From the problem statement we know that when 25.0 cm. Plug these into Eq. (24.4) to obtain 25.0cm. Now set 50 cm and ind the corresponding value o rom Eq. (24.4) (50 cm)(25.0 cm) 50 cm 25.0 cm 30.0 cm, i.e., the image is at 30.0 cm behind the lens. Again, since > 0 the image is real Foraplane wave (with parallel light rays), so 00 cm. Now replace the value o with 200 cm and ind the new value o rom Eq. (24.4) (200 cm)(00 cm) 200 cm 00 cm 200 cm. Again, since > 0 the image is real Use the Lensmaker s Formula / (n l )(/R /R 2 ). In this case, R, and R 2 are all positive, so R 2 >R > 0. Set 00 cm, n l 3/2, and R 2 00 cm and solve or R, the smaller o the two radii [ R (n l ) + ] [ R 2 (00 cm)(3/2 ) + ] 33.3cm. 00 cm

8 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Apply Eq. (24.) / +/ (n l )(/R /R 2 ). In this case R 2, n.50, 50.0cm, and. Plug these date into the equation above and solve or R R ( /so +/ n l + ) ( /50.0 cm +/ R ) 25.0 cm. Thus the radii o the lens are 25.0 cm and In deriving Eq. (24.2), / a (n l )(/R /R 2 ),wemade use o the Law o Reraction or a light ray entering the lens rom air n i sin θ i n t sin θ t,orsin θ t (n t /n i ) sin θ i. Here i stands or air and t or the glass. Taking n i n air and n t n l, this becomes In the current case i stands or water (w), so sin θ t sin θ t n l sin θ i. ( nt n i ) ( ) nl sin θ i sin θ n i. w Comparing this expression with that in the previous case, we ind that, to obtain the new ocal length w,wemust replace n l with n l /n w when the lens submerged in water in stead o air. Thus Eq. (24.2) becomes ( )( nl ). w n w R R 2 Numerically, the actor on the RHS o the equation is now n l /n w.5/(4/3) 0.25, which is to be compared with the old actor o n l 0.5 when the lens submerged in air. Thus or w 4 a. a n /n l w w n l , 24.6 (a) is the distance between the lens and the Sun, and is virtually ininity. (b) > 2. (c) Real, since >. (d) Inverted, which is the case or all real images ormed by a positive lens. (e) miniied, since > 2.

9 82 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT () (g) Use Eq. (24.4), the Gaussian Lens Equation, with and cm, to ind / / +/,or/ / +/ / +/ /, and so 80.0 cm (a) 3.00 m (the distance between the object and the lens). (b) 3.00 m and 00.0 cm.000 m, so > 2. (c) Since > the image is real and can be projected onto a screen. (d) Inverted, which is the case or all real images ormed by a positive (convex) lens. (e) miniied, since > 2. () O I (g) Use Eq. (24.4), the Gaussian Lens Equation, with 3.00 m and.000 m, to ind / / +/, (3.00 m)(.000 m) 3.00 m.000 m +.50 m, i.e., the image o the person is located.50 mbehind the lens. It is real (since magniication o M T.50 m 3.00 m 0.500, > 0), with a i.e., it is hal-sized and inverted (since M T magniication. < 0). Here we used Eq. (24.7) or the transverse

10 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT The object distance (300 cm) is much greater than the ocal length (5.00 cm), and much greater than 2, hence the image is real, inverted, miniied, and located near the ocal plane ( ). Now use Eq. (24.4), with 300 cm and cm, to ind / / +/,or (300 cm)(5.00 cm) 300 cm 5.00 cm +5. cm, i.e., the image is located 5. cm behind the lens, with a magniication o M T 5. cm 300 cm 0.07, meaning that the image is much lesn size than the object (a) 2.0cm (the distance between the object and the lens). (b) 4.8 cm, which is the distance between the image and the lens it is negative since it is to the let o the lens, the same side as the object. (c) That would be Eq. (24.4), the Gaussian Lens Equation / / +/. (d) s s (2.0 cm)( 4.8 cm) o i 8.0 cm. / +/ cm 4.8 cm (e) <0means that it is a negative (concave) lens. () The image is virtual and thereore cannot be projected onto a screen. (g) See Fig in the text. (h) Upright, as the case or any image ormed by a concave lens. (i) See Fig in the text Use Eq. (24.4), the Gaussian Lens Equation, with 30.0 cm and +0.0 cm, to ind / / +/,or (30.0 cm)(0.0 cm) 30.0 cm 0.0 cm +5.0 cm, i.e., the image o the cat is located 5.0 cm behind the lens. It is real (since > 0), with a magniication o M T 5.0 cm 30.0 cm 0.500, i.e., it is hal-sized and inverted (since M T < 0). Here we used Eq. (24.7) or the transverse magniication.

11 84 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 24.2 With the previous problem in mind, we use Eq. (24.4), with 00.0 cm and 60.0cm, to ind (00.0 cm)(60.0 cm) +50 cm, 00.0 cm 60.0 cm i.e., the image o the rog is located 50 cm behind the lens. It is real (since > 0), with a magniication o M T s 50 cm i 00.0 cm.50, so the image is 50% larger than the rog itsel and inverted (since M T < 0) Similar to the previous two problems, we again apply Eq. (24.4), / +/ /, with 00 cm and 50.0 cm, to ind (00 cm)(50.0 cm) 00 cm 50.0 cm +00 cm, i.e., the image o the lower is located 00 cm behind the lens. It is real (since magniication o M T 00 cm 00 cm.00, so the image is lie-sized (since M T.00) and inverted (since M T < 0). > 0), with a The location o the image is a distance rom the lens, where is (see the previous three problems) (80.0 mm)(20 mm) 80.0 mm 20 mm 240 mm, where we plugged in 80.0 mm and 20 mm. Note that here < 0, soimage is virtual and a distance 240 mm in ront o the lens (i.e., on the same side o the lens as the object) The image produced by a concave lens always virtual, miniied, and right-side-up. To locate the image o the beacon light, plug in 50m and 0 minto Eq. (24.4) and solve or s o (50 m)( 0 m) 50 m ( 0 m) 8.3 m, so the image appears to be located 8.3 m outside the window (on the same side as the object). You can easily veriy that 0 <M T / <, meaning that the image is miniied and right-side-up.

12 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT (a) First, ind the location o the image by solving or (0.75 m)(0.25 m) 0.75 m 0.25 m 0.38 m, rom Eq. (24.4), which gives so the image is located a distance 0.38 mbehind the lens. It is real (since > 0), with a magniication o M T / (0.375 m)/(0.75 cm) 0.50, so the image is hal-sized (since M T 0.50) and inverted (since M T < 0). (b) The ray diagram is shown below The eyeglass a positive (convex) lens with +2.0 m. The location o the image o the bus at 2.5 mbehind the glass, so +2.5 m. Now solve or rom Eq.(24.4), / / +/ (2.5 m)(2.0 m) 2.5 m 2.0 m 0m, i.e., the bus a distance 0 maway rom the eyeglass The lens positive (convex) since it has to ocus a real image onto the ilm, so mm. Also, to ocus the image o the bug on the ilm the distance between the lens and the ilm should be,so 00 mm. It ollows that (see the previous problem) (00 mm)(60.0 mm) 00 mm 60 mm 50 mm, so the bug should be a distance 50 mm away rom the camera lens With the previous problem in mind, we plug mm, along with 5.00 m, into Eq. (24.4) and solve or, the desired distance between the lens and the ilm (5.00 m)( m) 5.00 m m m 52.5 mm.

13 86 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Since the camera iriginally ocused on a distance object (i.e., ) the original distance between the lens and the ilm is the same as the ocal length, at 52.0 mm. So to re-ocus when the object is 5.00 m rom the lens the lens has to be advanced by 52.5 mm 52.0 mm 0.5mm When the eye in the relaxed state we may use Eq. (24.9) to ind the magniication o the lens to be M A d n /, where d n misthe normal viewing distance (to the near-point) or a standard observer. In this case M A 2.0, sothe ocal length o the lens d n M A m m Similar to the previous problem, we apply Eq.(24.9) or the magniication M A, with d n and 0cm M A d n 25 cm 0 cm cm 24.3 The ocal length o the magniier is related to its magniication M A via M A d n /, where d n m or a standard observer. Since M A 8, d n M A m 8 so the lens should be placed 3 cm above the lat surace m 3cm, (a) 0.0cm (the distance between the object and the lens). while 20.0cm, so <. (b) Virtual, since <. (c) Erect, since <. (d) Magniied, since <. (e) I O

14 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 87 () Apply Eq. (24.4), / / +/ (0.0 cm)(20.0 cm) 0.0 cm 20.0 cm 20.0 cm, i.e., the image o the candle is located 20.0 cm to the let o the lens. (g) M T 20.0 cm 0.0 cm (h) M T > 0 means that the image is erect. (i) Since M T 2.00 the image is twice the size o the object, or (2.00)(.0 cm) 2.0cm tall (a) 2.0cm (the distance between the object and the lens). 4.0 cm, which is the distance between the image and the lens it is negative since it in the same side as the object. (b) Use Eq. (24.4), the Gaussian Lens Equation / / +/,toobtain / +/ + (2.0 cm)( 4.0 cm) 2.0 cm 4.0 cm 6.0 cm. (c) <0means that it is a negative (concave) lens. (d) The image is virtual and thereore cannot be projected onto a screen. (e) I O () Upright, as the case or any image ormed by a concave lens. (g) M T s 4.0 cm i 2.0 cm (h) M T > 0 means that the image is erect. (i) Since M T the image o the butterly is timetriginal size, or (0.333)(3.0 cm).0cm.

15 88 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT First, compute the ocal length o the lens rom Eq. (24.2) [ ( (n l ) )] [ ( )] (.5 ) R R m 0.60 m m (a) For a lie-sized, real image M T /.0, so, and Eq. (24.4) now becomes / / +/ 2/, which gives 2 2(0.60 m).2 m, i.e., the object is to be placed.2 minront o the lens. Here we noted that the image must be real since it can be projected onto a screen, and so > 0. (b) Since (see the previous part),.2maswell, meaning that the screen is to be placed.2 mbehind the lens or a sharp image to be ocused on it. (c) When an object is placed a distance in ront o a convex len ocal length a real image is ormed a distance behind the lens (i >). Here, and are related via the Gaussian Lens Equation [Eq. (24.4)] / / +/.Inour case the object (grasshopper) is located 0 cm to the let o the lens while itmage is 30 cm to the right o the lens, so 0cm and 30cm. Plug these data into Eq. (24.4) to ind, the ocal length o the lens + 0 cm + 30 cm 4 30 cm, and so 30cm/4 7.5cm. (a) Now the grasshopper jumps 7.5 cm towards the lens so decreases by 7.5 cm, to 0 cm 7.5 cm 2.5cm. Plug this new value o, along with 7.5cm, into Eq. (24.4) again and solve or, the new location o the image (2.5 cm)(7.5 cm) 2.5 cm 7.5 cm 3.8 cm,

16 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 89 which means that the image is now 3.8 cm to the let o the lens (i.e., on the same side o the lens as the grasshopper itsel), since < 0. (b) At irst 30cm and 0cm, so the transverse magniication is M T / 30 cm/0 cm 3.0. The image is real (as > 0), inverted (as M T < 0), and magniied to three times the original size o the grasshopper (as M T 3.0). Similarly, or the new image 3.8 cm < 0,soitisnow virtual. The transverse magniication is now M T / 3.8 cm/2.5 cm +.5 > 0, sothe new image is right-side-up and is magniied to.5 times the original size. (c) The two ray diagrams below depict the situation beore and ater the grasshopper makes the jump Foranimage to be projected onto a screen it must be real, so +0 m. Also, it is enlarged 00 times so M T / 00, which gives /00 0 m/ m. Plug the value and into Eq. (24.4) to ind the desired ocal length o the projector lens ( + ) ( 0.0 m + ) m 9.9cm. 0 m The size o Fred-the-chicken s ace is y o 5.0cm, while the size o the corresponding image on the ilm is y o 24mm. Thus the transverse magniication M T provided by the camera lens

17 820 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT [see Eqs (24.8) and (24.9)] M T / y i /y o 0.24 cm/5.0 cm. Now plug in 2.0m into this equation to ind 0.96 m. The ocal length o the camera then ollows rom Eq. (24.4) ( + ) ( 2.0 m + ) 0.65 m mm m Since the Moon is very ar away rom us we may take m, which yields 50mm by virtue o Eq. (24.4). The diameter o the Moon is y o 0.273( m) m. Plug these valuento Eqs. (24.8) and (24.9), M T / y i /y o, to ind the size (diameter) o the image o the Moon on the ilm y i s y i o (50 mm)( m) mm. m (I you let out the absolute value sign you would get y i only means that the image o the Moon inverted.) 0.45 mm, where the minus sign Consider the two similar trianglen Fig. P39, one with a base y o and corresponding height, and the other with a base y i and corresponding height. By similarity y o / y i /,or y i /y o /,sothe transverse magniication o the pinhole camera is, by deinition, M T y i y o, which identical with the expression or M T or a lens. Note the negative sign we added here to relect the act that the image inverted. This ormula tells us that the magniication o the pinhole camera is proportional to, the pinhole-ilm plane distance, which is about the same as the length o the camera box. To obtain a larger image, just extend the length o the camera body accordingly. (Many commercial pinhole cameras do come with an adjustable body length.) First, use Eqs.(24.8) and (24.9) to ind M T y i /y o /, and so y i y o (.50 cm)(60.0 cm) 3.00 cm cm. (Note that > 0 since the image is real.) Now ind rom Eq. (24.4) ( + ) ( ) 60.0 cm cm cm

18 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Let the magnitude o each radiu the equi-convex lens be R, then R R 2 R, and the ocal length o the lens satisies Eq. (24.2) ( (n ) ) ( (n l R R l ) 2 R ) 2(n ) l. R R To ind R we need to know, which can be computed rom Eq.(24.4) with the value and. Since any real image projected by a convex lens always behind the lens the separation between the object and the image is d +, which we know to be 0.60 m. Also, the magniication is M T y i /y o /, which gives (y i /y o ) ( 25 cm/5.0 cm) 5.0. Note that y i is negative, since any real image projected by a single convex lens alwaynverted. The two last equations above give 0.0 m and 0.50 m, which we plug into Eq. (24.4), along with the expression we obtained above or / / 2(n l )/R / +/, and solve or R R 2(n l ) + 2(.50 )(0.0 m)(0.50 m) 0.0 m m m 8.3 cm First, use the result o Problem (24.5) to ind the ocal length w o the lens underwater ( ) ( ) nl.56 w n l /n w a (0 cm) cm..56/.33 Now apply Eq. (24.4), with 00 cm and w, or the ish, to ind the corresponding value o w (00 cm)( cm) +48 cm, w 00 cm cm i.e., the image o the ish is a distance 48 cm behind the lens The positive lens (with ocal length 60cm) projects the object, which is the picture on the TV screen, onto the wall, where the image o the TV screen is located. So the distance d between the TV screen and the wall is that between the object and the image d +. Now let s ind and. Since the picture is enlarged 3 times, the transverse magniication is [see Eq. (24.7)] M T / 3. Here the minus sign corresponds to the act that, being real, the image on the wall must inverted. The last equation can be rewritten as 3, which we plug into Eq. (24.4), along with 60cm 0.60 m ,

19 822 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT which gives 4(0.60 m)/ m. Thus 3 3(0.80 m) 2.4 m. The distance between the TV screen and the wall is then d m +2.4 m 3.2 m. The primary beneit o using a large lens that more light rom the TV screen can be collected by the lens and projected onto the wall, resulting in a brighter image. Since the image on the wall inverted (with respect to the picture on the TV screen), the TV set has to be placed upside down, i the inal image on the wall is to appear normal (i.e., right-side-up) Apply Eq. (24.4), with 00 cm, to obtain ( + ) ( 00 cm + ) 50.0cm. 00 cm Thus by deinition the power o the lens D 50.0 cm 2.00 m 2.00 D. Note that the unit or the power is the diopter (D) D m For the irst lens 20.0 m and 0.50 m, so Eq. (24.4) gives the location o o the irst image (denoted as I in the ray diagram) / / +/,or (20.0 m)(0.50 m) 20.0 m 0.50 m m, which putt a distance m m.487 minront o the second len ocal length 2 (.00 m). The location o the inal image (I 2 )isthen (.487 m)(.00 m).487 m.00 m 3. m, i.e., it is 3. mbehind (i.e., to the right o) the second lens. It is real, right-side-up, and miniied, as shown in the ray diagram below. L L 2 y to object (height y) 2 I 2 I 2 (inal image)

20 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT With the previous problem in mind, this time 50.0cm and 25.0cm, hence or the irst image (I ) (50.0 cm)(25.0 cm) 50.0 cm 25.0 cm 50.0 cm, and so 2 30 cm 50.0 cm 80.0 cm or the second lens, with cm. Thus (80.0 cm)(40.0 cm) 80.0 cm 40.0 cm 80.0 cm, meaning that the inal image (I 2 )islocated a distance 80.0 cm behind the second lens. The ray diagram is shown below. L L 2 I 2 2 I 2 (inal image) Sunlight comen parallel, so or the irst lens, and 50.0cm; which means that, or the second lens (with cm), 2 70 cm 50.0 cm 20 cm. Thus or the inal image (I 2 ) (20 cm)(60.0 cm) 20 cm 60.0 cm 20 cm, meaning that the inal image o the Sun is at 20 cm behind the second lens. This where we should place the screen. The ray diagram is shown below. L L 2 parallel light rays rom the Sun I I 2 (inal image) 2 2

21 824 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT According to Eq. (24.6) the total power o two lensen contact is D D + D 2.Inthis case D.0 D (or the negative lens) and D 0.75 D, so the power o the liquid lens D 2 D D 0.75 D (.0 D) D The power D is deined as /, which in turn is given by Eq. (24.4). In this case n 3/2, R +00 cm R 2 (or an equiconvex lens); so D (n l ) ( R R 2 ) ( )( ) m +.00 D..00 m The eective ocal length o two lensen contact with each other is given by Eq. (24.4) / / +/ 2. Plug in 5.0 cm and cm and solve or ( + ) ( ) cm cm cm The combination is thereore equivalent to a negative lens with a ocal length o 40.0 cm Since both lens are positive, adding the second lens will ocus the laser beam even aster than beore, resulting in a shorter ocal length. So the parallel beam will be ocused closer to the lenses and allowed to spread out more beore reaching the screen, causing the size o the blotch to increase. Since 2 20cm the ocal length o the two-lens combination is [see Eq. (24.4)] ( + ) ( 2 20 cm + ) 0cm. 20 cm Suppose that the three lenses are lined up rom let to right, starting rom the irst lens. Now consider a parallel beam incident rom let through the irst lens, which ocuses the beam a distance 0cm behind it. Since the second lens placed a distance 30 cm rom the irst one this ocused light happens to be a distance 30 cm 0 cm 20cm, or one ocal length, in ront o the second lens, whose ocal length is 2 20cm. Thus the second lens re-collimates the light beam, which is subsequently ocused again by the third lens, a distance 3 5.0cm behind it.

22 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 825 [In general, we need to apply Eq. (24.4) successively three times, once or each lens, to locate the inal image. This particular case is simple since we are dealing with a parallel beam o light, and the distance between the irst two lenses happens to be + 2.] First, combine the irst and the second lenses to ind the equivalent ocal length 2 o the two-lens combination by applying Eq. (24.4) / 2 / +/ 2. Now apply Eq. (24.4) again to combine this with the remaining (third) lens to ind the ocal length o the three-lens system cm + 20 cm cm 0.25 cm, so (/0.25) cm +4.0 cm. Since >0 the lens system is equivalent to a single positive lens and can orm real images The reractive power D is deined as /, which is given by Eq. (24.4) D +.00 m m 4.0 m +4.0 D We want the corrective lens to make an object positioned in the normal near-point (25.4 cm) to appear at 00 cm in ront o the lens so it can be viewed clearly. This means that, or 25.4cm m, we must have 00 cm.00 m. Note the negative sign here, which corresponds to the act that the image in the same side as the object. The desired value o the reractive power o the lens to be prescribed is then D m +.00 m 2.94 m D This problem is similar to the previoune. In this case the corrective lens should bring the image o an object at ininity to the ar-point, where it can be seen clearly. So or + we must have 00 cm, or.00 m, where the negative sign again corresponds to the act that the image in the same side as the object. The lens to be prescribed should then have a reractive power o D m.00 m.00 D.

23 826 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Without the corrective lens the person can only clearly see as ar as the ar-point. What the lens does that it brings the image o an object at ininity to the ar point, where it can be viewed clearly. Thus or we must have L, where L is the distance between the lens and the ar-point. Now apply Eq. (24.4), / / +/ / /L, and solve or L L i.e., the ar-point is 5.0 minront o the lens. / / / /( 5.0 m) 5.0 m, According to the discussion in the text on the compound microscope the total angular magniication M A is the product o the magniication the objective (M TO ) and the eyepiece (M AE ) M A M TO M AE By the deinition o the tube length, the total separation L total between the objective lens and the eyepiece is the sum o o the ocal lengths O and E o the two lenses plus the tube length L L total O + E +L. Inthis case the total separation is L total 0.0cm, O 0mm.0cm, and E 30mm 3.0cm; so the tube length is L L total O E 0.0 cm.0 cm 3.0 cm 6.0 cm The angular magniication o the telescope is M A O / E. Let M A 0 (negative, since the inal image inverted) and E 2.5cm, and solve or O O M A E ( 0)(2.5 cm) 25cm. The length o the scope is then E + O 2.5 cm +25cm 28cm A telescope consist an objective lens (O) and an eyepiece (E), with O > E. The two lenses at our disposal have ocal lengths 3.0 cm and /.00 m.00 m; so we put the the old eyeglass lens up ront as the objective lens, with O.00 m, and take the magniying glass as the eyepiece, with E 3.0cm. The tube length we ll need is L O + E.00 m m.03 m,

24 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 827 and the magniication that can be achieved is M A O E.00 m m (a) Apply the Thin Lens Equation to image, the intermediate image / +/ /, to obtain (5.0 cm)(0.0 cm) 30.0cm. 5.0 cm 0.0 cm (b) The magniication is M T / (30.0 cm)/5.0 cm 2.0. (c) The image is real (as > 0), inverted (as M T < 0), and magniied size (as M T > ). (d) Image is located at (30.0 cm 25.0 cm 5.0cm behind the second lens and is thereore a virtual object or that lens cm < 0. (e) For the inal image (image 2, produced by the second lens) /2 +/2 / 2, and so ( 5.0 cm)( 7.5 cm) 5.0 cm ( 7.5 cm) +5 cm. () The inal image is real (as 2 (g) > 0), so it can be projected onto a screen. I 2 (inal image) O 2 2 L I L 2 (h) The irst lens produces an inverted image (as > ) while the second one keeps the orientation unchanged rom the irst image (since it is a concave lens). So the inal image is inverted (relative to the object).

25 828 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT (i) The magniication o the second lens M T2 2 /2 (5.0 cm)/( 5.0 cm) 3.0. (j) M T M T M T2 ( 2.0)(3.0) 6.0. (k) Since M T 6.0 the inal image is 6.0 times the size o the object, or 6.0(.0 cm) 6.0 cm tall (a) Apply the Thin Lens Equation to image, the intermediate image / +/ /, to obtain s (0.0 cm)(5.0 cm) o 5.0 cm 0.0 cm 0cm. The magniication is M T / (0.0 cm)/0.0 cm.0. The image is real (as > 0), inverted (as M T < 0), and lie size (as M T.0). (b) and (c) Image is located at (0.0 cm 4.0 cm 6.0cm behind the second lens and is a virtual object or that lens cm. Now /2 +/2 / 2, and so or the inal image (image 2) ( 6.0 cm)( 3.0 cm) 6.0 cm ( 3.0 cm) 6.0 cm. The magniication o image 2 (measured relative to image, which serves as the object or image 2) is M T2 2 /2 ( 6.0 cm)/( 6.0 cm).0. The inal image is virtual (as 2 < 0), with a total magniication o M T M T M T2 (.0)(.0) +.0. Itisupright (as M T > 0) and lie size (as M T.0). (d) Since M T.0 the inal image has the same size as the object, or 0.30 cm tall. (e) See Fig. P63 in the textbook First, locate the image projected by the irst lens. We have 30cm and m +30 cm, so rom Eq. (24.4) 30 cm 30 cm 0, or, which means that the light rays rom the object becomes parallel ater passing through the irst lens 2. The location o the inal image can then be obtained rom /2 /2 / 2,or 2 / 2 /2 /( 0.20 m) /( ) 0.20 m, i.e., the inal image is located a distance 0.20 minront o the second lens, coinciding with its let ocal point (assuming that lens is to the let o lens 2). Since the two lens are 0 cm apart the inal image is also 0.20 m 0 cm 0cm in ront o (i.e., to the let o) the irst

26 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 829 lens. It is virtual, with a inal transverse magniication M T which equals the product o the magniication the two lenses ( M T M T M T2 s )( i s ) ( i2 ) ( 0.20 m ) cm 3, so it is right-side-up and miniied to 2/3, or 67%, o the original size o the object The image produced by the irst lens real, so it is.0 mbehind that lens. Now, the second lens is 90 cm behind the irst one, so the image projected by the irst lens at.00 m 90 cm 0cm behind the second lens 2 0 cm. Also, the inal image is located 0 cm beyond the irst image, so 2 0cm +0cm 20cm. Plug the value both 2 and 2 into Eq. (24.4) to ind 2, the ocal length o the second lens ( 0 cm)(20 cm) 0 cm +20cm 20 cm. Note that 2 < 0, since the second lens negative. The magniication o the second lens M T2 y i2 /y o2 2 /,sothe diameter o the inal image o the clock is y i2 y 2 i2 (20 cm)(20 cm) 40cm. 0 cm For the top portion o the glasses which corrects his near-sightedness, a negative len ocal length must be used to cast the image o an object at ininity to the ar-point, a distance 3.00 minront o the cornea, or 3.00 m 2.0 cm 2.98 minront o the glasses. This means that or, 2.98 m. Note the negative sign here, since the image in the same side o the lens as the object. Plug the value and into Eq. (24.4) to ind the desired power D o the (negative) lens D m 0.34 m 0.34 D. Now consider the lower portion o the glasses which corrects the ar-sightedness by bringing the image o an object placed at a distance d n (25.4cm) in ront o the eyeut to the near-point, which is a distance 45 cm rom the eyes, or 45 cm 2.0 cm 43cm rom the lens. So or 25.4cm 2.0 cm 23.4cm we must have 43 cm, where the negative sign is again due to the act that the image in the same side o the lens as the object. The desired power D 2 o the (positive) lens then D m m +.9 m +.9 D.

27 830 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT (a) Suppose the person wears a pair o glasses with negative lense ocal length to correct his near-sightedness. The lenses help bring the image o an object ininitely ar away to the ar-point in ront o the eyes so it can be viewed clearly, at.00 m 5.0 mm m rom the lenses. So or we must have m, where the negative sign is again due to the act that the image in the same side o the lens as the object. The desired power D o the lenses then D m.02 m.02 D. (b) Suppose that the new near-point is a distance d in ront o the eyes, or d 5.0 mm in n n ront o the glasses. An object placed at that point will have itmage ormed by the glasses a distance d n 25.0cm in ront o the eyes the closest distance rom the eyes where an image can be seen clearly. The distance between thimage and the glasses d n 5.0 mm. So or d 5.0 mm we must have s n i (d n 5.0 mm) (25.0 cm 5.0 mm) 23.5 cm. Plug the value, and into Eq. (24.4) D.02 m + d 5.0 mm cm. n Solve or d n to obtain d n m 32.4 cm, which is not terribly inconvenient or reading A microscope consist an objective lens (O) and an eyepiece (E), with E > O > 0. So O 2.0mm and E 2.0cm. Now locate the position o the image ormed by the objective lens. Plug 2.5mm and O 2.0mm into Eq. (24.4) and solve or O O (2.0 mm)(2.5 mm) 2.5 mm 2.0 mm 0mm. The separation o the lenses should then be + E Now the magniications. For the objective 0mm +2.0 cm 3.0 cm. M TO O O 0 mm 2.0 mm 2.0 mm 4 (reer to the discussion in the text ollowing the ormula M A M TO M AE ), and or the eyepiece M AE d n E 25.4 cm 2.0 cm 2.7 ; so the combined angular magniication o the microscope is M A M TO M AE ( 4)(2.7) 5.

28 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT (a) The total magniication is the product o M AE and M TO M A M TO M AE ( 20)(0) 200, or, to two signiicant igures, ( ). (b) For the eyepiece M AE d n / E,so E d n M AE 254 mm mm. For the objective M TO L/ O, where L 60 mm is the standard tube length, so O L M TO 60 mm mm. (c) For the image ormed by the objective L + O, and so rom / O / +/ / +/(L + O ) we may solve or O(L + O ) O(L + O ) (L + O ) O L (8.0 mm)(60 mm +8.0 mm) 60 mm 8.4 mm, i.e., the object is located 8.4 mm in ront o the objective lens Reer to the igure to the right, which shows the our suraces o the two lenses. Here R 2 R 2 in accordance with the problem statement. We are given, the ocal length o the two-lens system, and 2, that o L 2. This gives an equation or, the ocal length or L,per Eq.(24.4) / / +/ 2, so m 0.50 m 4.0 D. L L 2 R R 2 R 22 R 2 Now, i R R then R 2 R, asl is equi-convex. Eq. (24.2) then gives ( (n l ) R ) 2(n ) l. R R Equate the two expressions above or / and plug in n l.50 to obtain 4.0 D 2(.50 )/R, orr 0.25 m.

29 832 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Now ind the remaining unknown radius, R 22,byapplying / 2 (n l2 )(/R 2 /R 22 ). Plug in m, n l2.55, and R 2 R 2 R 0.25 m, and solve or R 22 R 22 R 2 ( n l2 ) R 2 ( n l2 ) (0.25 m)( 0.50 m)(.55) 0.25 m ( 0.50 m)(.55) 2.8 m. So inally R 0.25 m, R 2 R m, and R m Beore prescribing a new pair o glasses or her we need to know the location o her new nearpoint, which is right at the image ormed by the old glasses held at 80 cm rom her eyes. When she holds the book 80 cm away rom her eyet is at 80 cm 2.0 cm 78cm in ront o the glasses. Set D / +2.0Dand 78cm 0.78 mtoind / D / / +/, or D 2.0D 0.78 m 0.78 D. The new near-point is a distance rom the eyeglasses. Now, the new glasses should make the image o any object placed at the new near point appear to be a distance d n m rom the eyes, or m m minront o the eyeglasses. Thus or s s we must o i have s m. The desired power D o the new lens then i D s o + s i m 0.78 D ( 4.27 D) +5.0 D To correct the person s near-sightedness, a negative len ocal length must be used to cast the image o an object at ininity to the ar-point, a distance 200 cm in ront o the cornea, or 2.00 m 2.0 cm.98 minront o the glasses. This means that or,.98 m, where the negative relects the act that the image in the same side o the lens as the object. Plug the value and into Eq. (24.4) to ind the desired power D o the lens D m D. The dierence in power between the eyeglasses and the contact lens stems rom the act that the contact is worn right on the cornea, so the 2.0-cm distance between the cornea and the glasses eliminated, and the value o should just be 200 cm instead o 98 cm or. The power o the contact should then be D contact m D, which corresponds to a ocal length o 2.00 m. From the discussion above we see that the ar point in either case is 200 cm in ront o the cornea that s how the powers are chosen.

30 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT According to Eq. (24.8) the ocal length o a spherical mirror is related to its radiu curvature R via R/2. Plug in R 66.0 cm to obtain R 2 ( 66.0 cm) cm. 2 Note that >0, as the mirror is concave The radiu curvature o a convex mirror is positive, so R m. Thus rom Eq. (24.8) the ocal length o the mirror is Note that <0, as the mirror is convex. R 2 (0.50 m) 0.25 m The polished steel ball acts as a spherical convex mirror with R 2 (50 cm) 25cm. Thuts ocal length is R 2 (25 cm) 3 cm Use a concave mirror o radius R, and place the candle a distance R /2, or30 cm, in ront o its vertex. This gives R 2(30 cm) 60cm. The location o the image satisies Eq. (24.9), / +/ 2/R. Note that here R R 60 cm < 0 since it is a concave mirror. Set R /2 R/2 and solve or 2 R 2 R R/2 0, or, which means that the image will be ocused at ininity, so the relected beam must be parallel which is what we wanted. The ray diagram is shown below. R

31 834 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT We are given 200 cm and +400 cm, and we wish to ind the value o or the mirror. This calls or Eqs. (24.20) and (24.2), / +/ 2/R /, which we solve or / +/ Note that >0 since the mirror is concave. /200 cm +/400 cm +33 cm First, locate the image by inding rom Eq. (24.9) 2/R / 2/60 cm /60 cm +60 cm, where the positive sign indicates that the image is at 60 cm in ront o the vertex o the mirror, on the same side as the object itsel. It is real, lie-sized, and inverted, as shown in the ray diagram below (a) We are given 30cm 0.30 m or the location o the object and 9.0m or the location o the image. Note that here > 0 since the image is projected onto a screen, which must be on the same side o the mirror as the object no light rays rom the object can reach behind the mirror. These data lead to R, the radiu curvature o the mirror, via Eq. (24.9) / +/ 2/R, or 2 2 R / +/ /0.30 m +/9.0 m 0.58 m. Note that R<0 since the mirror is concave. (b) The transverse magniication is given by Eqs. (24.8) and (24.9) as M T y i /y o /, where y i and y o are the transverse size the image and the object, respectively. To ind y i, plug y o 5.0cm m, along with 9.0m and 0.30 m, into the ormula or M T above and solve or y i y i y o (0.050 m)(9.0 m) 0.30 m.5 m.

32 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 835 The image is thereore.5 m tall and inverted (andicated by the minus sign in ront o y i ). It i courses real, since it can be projected onto a screen To cast a real image onto a screen, the mirror must be concave; so R 0.20 m and R/2 ( 0.20 m)/ m +0 cm. Also, the screen is.0 m rom the vertex o the mirror so +.0 m. Now use Eq. (24.9) to ind 2/R +/ 2/( 0.20 m)+/.0 m 0. m cm, i.e., the candle is to be placed cm away rom the vertex o the mirror. The image is real (as >>0), with a transverse magniication o M T /.0 m/0. m 0, soitisinverted and enlarged to 0 times the original size o the candle, as shown in the ray diagram to the right From the hint given in the problem statement, we take R +20 cm and obtain R/2 00 cm/2 50 cm or the concave mirror. Plug this value or the ocal length, along with 20cm, into / +/ / to ind (20 cm)( 50 cm) 20 cm ( 50 cm) 4 cm, where the negative sign indicates that the image appears to be 4 cm behind the vertex o the mirror. It is virtual (as no light rom your nose can bypass the mirror and reach behind its vertex), with a transverse magniication o M T 4.3 cm 20 cm +0.7, so it is right-side-up (as M T > 0) and miniied to about 7% the original size o your nose, as shown in the ray diagram in the next page.

33 836 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Apply Eq. (24.9), / +/ 2/R, tolocate the image o the mouse. Here 00 cm and R 2 (80.0 cm) 40.0cm, so / 2/R /00 cm 2/40.0 cm 6.7 cm, meaning that the image is located at 6.7 cm behind the vertex o the mirror. It is virtual, right-side-up, and miniied see the ray diagram below With the previous problem in mind, this time / 2/R /0.0 m 2/.00 m 0.0 m and R.00 m, so m 8.3 cm,

34 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT 837 i.e., your image appears to be 8.3 cm behind the glasses, whose outer surace acts like a relecting spherical mirror. The image is virtual, right-side-up, and miniied, as shown in the ray diagram below The ocal length o the convex mirror in question is R/2 ( 8.0 mm)/2 4.0 mm. Since 20cm, with the previous problem in mind we have (20 cm)( 0.40 cm) 20 cm ( 0.40 cm) 0.39 cm 3.9 mm, where the negative sign indicates that your image appears to be 3.9 mm behind the cornea. It is virtual, with a transverse magniication o M T / ( cm)/20 cm , so it is right-side-up (as M T > 0) and miniied to about 2% o your original size (a) Since the mirror is concave >0. Now, 90.0cm and 2 (90.0 cm) 45.0cm. This gives, the ocal length, via Eq. (24.9) / +/ /, or 2 / +/ /(90.0 cm)+/(45.0 cm) 30.0 cm. (b) The image has to projected onto the screen so it must be real. (c) Inverted, as the case or all real images produced by a concave mirror. (d) Since > (or, since 2 > >) the image is magniied.

35 838 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT (e) O I () M T / 90.0 cm/45.0 cm 2.0. (g) Since M T 2.0 the image is 2.0 times the size o the object, or 2.0(4.0 mm) 8.0mm tall. (h) R 2 2(30.0 cm) 60.0 cm, it is negative since the mirror is concave (a) Since the mirror is convex R>0 and 2 R 2 (50.0 cm) 25.0 cm. (b) Virtual, since it is produced by a convex mirror. (c) Upright, since it is produced by a convex mirror. (d) Miniied, since it is produced by a convex mirror. As the object approaches the mirror the image size increases. To get a near-lie size image the object has to be essentially right in ront o the mirror. (e) I O 2 () Apply Eq. (24.9), / +/ 2/R, and solve or 2/R +/ 2/( 50.0 cm)+/(200 cm) 28.6 cm, i.e., the image is located 28.6 cm behind the mirror. (g) M T / ( 28.6 cm)/200 cm (h) Since M T 0.43 the image is 0.43 times the size o the object, or 0.43(00 cm) 4.3 cm tall.

36 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT First we need to locate the image o the satellite ormed by the mirror, by solving or rom Eq. (24.9), / +/ 2/R. Assuming that the satellite is directly above in the sky, then is approximately the altitude o itrbital 500 km m. Plug this value or, along with R.0 m (negative as the mirror is concave), into the equation above and solve or 2/R +/ 2/(.0 m)+/( m) m, so the (real) image o the satellite is located 0.50 minront o the vertex o the mirror. The size o the image can then be obtained rom the transverse magniication M T in Eqs. (24.8) and (24.9) (which are identical or both thin lenses and spherical mirrors) M T y o /y i /. Here y o 2.0misthe size o the satellite, 0.50 m, m, which we plug into the equation and solve or y i, the size o the image o the satellite in the telescope y i y o (2.0 m)(0.50 m) m m 2.0 µm. (I you let out the absolute value sign you would obtain y i indicates that the image inverted.) 2.0 µm, where minus sign only The magniication o the mirror is y i /y o /. Here y o.0m, y i.0 cm (note the negative sign here, since the image is miniied and, thereore, real, which means that it must be inverted), and 0m. Thus y i y o (0 m)(.0 cm).0 m +0 cm +0.0 m, so the detector should be located a distance 0 cm away rom the vertex o the mirror. Plug this value o into / / +/, along with 0m, to ind, the ocal length o the mirror + The ray diagram is shown below. (0 m)(0.0 m) 0 m +0.0 m m 9.9 cm. ray to object, where rays and 2 originate ray 2 image

37 840 CHAPTER 24 GEOMETRICAL OPTICS & OPTICAL EQUIPMEMT Since a convex mirror can only produce miniied images, we should select a concave one or the task, and put the object (the tooth) within one ocal length rom the vertex o the mirror to orm a virtual, enlarged image that is right-side-up. All that remains to be speciied is the radius R o the mirror. To ind R, Apply Eq. (24.9), / +/ 2/R, with.5cm, along with Eq. (24.7) or the magniication M T / +2(positive, as the image is right-side-up), or 2 2(.5 cm) 3.0 cm. Thus 2 2 R / +/ /.5 cm /3.0 cm 6.0 cm First, ind the value o, given R and,byapplying Eq. (24.9) / +/ The magniication is thereore [see Eq. (24.7)] M T R 2/R +/ 2 + R. R/(2 + R) R 2 + R. 2/R, or 24.9 Use the result o the previous problem, with R 60 cm (negative because the mirror is concave) and 2.4m, to ind the magniication M T R 2 + R 0.60 m 2(2.4 m)+( 0.60 m) 0.4. It is real, inverted, and miniied, as shown in the ray diagram below The magniication M T was computed in Problem (24.90) as a unction o and R M T R/(2 + R), which we may solve or R R 2 M T /( M T ). In this case M T and 00 mm, so