Chapter 17 continued r i 42¡ r, mirror 90¡ r 90¡ 42¡ 48¡ C F

17.2 Curved Mirrors page 480 Level 1 61. A concave mirror has a focal length of Figure 17-21 10.0 cm. What is its radius of curvature? real; inverted; larger r 2f 2(10.0 cm) 20.0 cm Level 2 62. An object located 18 cm from a convex 65. Star Image Light from a star is collected by mirror produces a virtual image 9 cm from a concave mirror. How far from the mirror the mirror. What is the magnification of the is the image of the star if the radius of image? curvature is 150 cm? d m i Stars are far enough away that the do light coming into the mirror can be (9 cm) considered to be parallel and parallel 18 cm light will converge at the focal point. 0.5 Since r 2f, r 150 cm f 75 cm 63. Fun House A boy is standing near a 2 2 convex mirror in a fun house at a fair. He

notices that his image appears to be 66. Find the image position and height for the Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.60 m tall. If the magnification of the object shown in Figure 17-22. mirror is 1, what is the boy’s height? 3 h m i ho h h i 3.8 cm F o m 16 cm 0.60 m 1 31 cm 3 1.8 m

64. Describe the image produced by the object Figure 17-22 in Figure 17-21 as real or virtual, inverted or upright, and smaller or larger than the 1 1 1 f d d object. o i d f o di do f (31 cm)(16 cm) 31 cm 16 cm

366 Solutions Manual Physics: Principles and Problems Chapter 17 continued 33 cm what is the magnification of the image? r (40 mm) hi di f 20 mm m 2 2 ho do 1 1 1 d h d d f h i o o i i d o d f (16 mm)(20 mm) o (33 cm)(3.8 cm) di 80 mm do f 16 mm 20 mm 31 cm di (80 mm) 4.1 cm m 5 do 16 mm

67. Rearview Mirror How far does the image 70. A 3.0-cm-tall object is 22.4 cm from a of a car appear behind a convex mirror, concave mirror. If the mirror has a radius of with a focal length of 6.0 m, when the car curvature of 34.0 cm, what are the image is 10.0 m from the mirror? position and height? 1 1 1 r f d d f o i 2 d f o 34.0cm di do f 2 (10.0 m)(6.0 m) 17.0 cm 10.0 m (6.0 m) 1 1 1 3.8 m f do di d f o 68. An object is 30.0 cm from a concave mirror di do f of 15.0 cm focal length. The object is (22.4 cm)(17.0 cm) 1.8 cm tall. Use the mirror equation to 22.4 cm 17.0 cm find the image position. What is the image height? 70.5 cm

1 1 1 h d m i i do di f ho do d f o d h di i o do f hi do (30.0 cm)(15.0 cm) (70.5 cm)(3.0 cm) 30.0 cm 15.0 cm 22.4 cm 30.0 cm 9.4 cm h d m i i ho do Level 3 71. Jeweler’s Mirror A jeweler inspects a d h h i o watch with a diameter of 3.0 cm by placing i d o it 8.0 cm in front of a concave mirror of (30.0 cm)(1.8 cm) 12.0-cm focal length. (30.0 cm) a. Where will the image of the watch Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Companies, a division of The McGraw-Hill © Glencoe/McGraw-Hill, Copyright 1.8 cm appear? 1 1 1 69. Dental Mirror A dentist uses a small mir- do di f ror with a radius of 40 mm to locate a cavity in a patient’s tooth. If the mirror is concave and is held 16 mm from the tooth, Physics: Principles and Problems Solutions Manual 367 Chapter 17 continued d f (8.0 cm)(12.0 cm) o 1 1 1 di do f 8.0 cm 12.0 cm do di f 24 cm fd o di b. What will be the diameter of the image? do f h d (10.0 cm)(150 cm) i i 9.4 cm ho do 150 cm ( 10.0 cm) d h (24 cm)(3.0 cm) d (9.4 cm) i o m i 0.063 hi d 150 cm do 8.0 cm o 9.0 cm hi mho (0.063)(12 cm) 0.75 cm 72. Sunlight falls on a concave mirror and Mixed Review forms an image that is 3.0 cm from the pages 480–481 mirror. An object that is 24 mm tall is Level 1 placed 12.0 cm from the mirror. 74. A light ray strikes a plane mirror at an angle a. Sketch the ray diagram to show the of 28° to the normal. If the light source is location of the image. moved so that the angle of incidence increases by 34°, what is the new angle of O1 Ray 1 reflection? Ray 2 i i, initial 34¡ C F 28¡ 34¡ Horizontal scale: I 1 block 1.0 cm 1 62¡ Vertical scale: 1 block 4 mm r i 62¡ b. Use the mirror equation to calculate the image position. 75. Copy Figure 17-23 on a sheet of paper. Draw rays on the diagram to determine the Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 1 1 height and location of the image. do di f fd (3.0 cm)(12.0 cm) o di do f 12.0 cm 3.0 cm 4.0 cm c. How tall is the image? 3.0 cm F d m i 4.0cm 0.33 8.0 cm 4.0 cm do 12.0 cm hi mho ( 0.33)(24 mm) 8.0 mm

73. Shiny spheres that are placed on pedestals Figure 17-23 on a lawn are convex mirrors. One such sphere has a diameter of 40.0 cm. A 12-cm-tall robin sits in a tree that is 1.5 m from the sphere. Where is the image of the robin and how tall is the image? r 20.0 cm, f 10.0 cm

368 Solutions Manual Physics: Principles and Problems Chapter 17 continued b. What is the image height? O1 h I hi 1.0 cm i 1 m di 2.7 cm ho Horizontal scale: F diho 1 block 1.0 cm h i d Vertical scale: o 2 blocks 1.0 cm (22.9 cm)(2.4 cm) 30.0 cm The image height is 1.0 cm, and its 1.8 cm location is 2.7 cm from the mirror. 78. What is the radius of curvature of a concave Level 2 mirror that magnifies an object by a factor 76. An object is located 4.4 cm in front of a of 3.2 when the object is placed 20.0 cm concave mirror with a 24.0-cm radius. from the mirror? Locate the image using the mirror equation. hi r m f h 2 o d md 24.0cm i o 2 (3.2)(20.0 cm) 12.0 cm 64 cm 1 1 1 f do di 1 1 1 f d d d f o i d o i d f d d o f oi d d (4.4 cm)(12.0 cm) o i (20.0 cm)(64 cm) 4.4 cm 12.0 cm 6.9 cm 20.0 cm ( 64 cm) 29 cm 77. A concave mirror has a radius of curvature r 2f of 26.0 cm. An object that is 2.4 cm tall is (2)(29 cm) placed 30.0 cm from the mirror. 58 cm a. Where is the image position? r f 79. A convex mirror is needed to produce an 2 image one-half the size of an object and 26.0cm located 36 cm behind the mirror. What 2 focal length should the mirror have? 13.0 cm h d m i i 1 1 1 ho do f d d o i d h i o d f do d o hi i d f o ( 36 cm)ho Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Companies, a division of The McGraw-Hill © Glencoe/McGraw-Hill, Copyright (30.0 cm)(13.0 cm) ho 30.0 cm 13.0 cm 2 22.9 cm 72 cm 1 1 1 f do di

Physics: Principles and Problems Solutions Manual 369 Chapter 17 continued d d a. What kind of mirror would do this job? oi f do di An enlarged, upright image results (72 cm)(36 cm) only from a concave mirror, with the 72 cm (36 cm) object inside the focal length. 72 cm b. What is its radius of curvature? d i 80. Surveillance Mirror A convenience store m do uses a surveillance mirror to monitor the store’s aisles. Each mirror has a radius of di mdo (7.5)(14.0 mm) curvature of 3.8 m. 105 mm a. What is the image position of a customer 1 1 1 who stands 6.5 m in front of the mirror? do di f A mirror that is used for surveillance d d (14.0 mm)(105 mm) oi f is a convex mirror. So the focal di do 14.0 mm ( 105 mm) length is the negative of half the 16 mm radius of curvature. r r 2f (2)(16mm) f 2 32 mm 3.8m 2 82. The object in Figure 17-24 moves from position 1 to position 2. Copy the diagram 1.9 m onto a sheet of paper. Draw rays showing 1 1 1 how the image changes. f do di d f o di do f (6.5 m)(1.9 m) 6.5 m (1.9 m) 1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. C 2 F 1.5 m 1.0 m b. What is the image height of a customer 1.5 m who is 1.7 m tall? 2.0 m h d 2.5 m m i i ho do Figure 17-24 d h h i o i d o O1 Ray 1 O2 Ray 1

(1.5 m)(1.7 m) Ray 2 Ray 2 6.5 m 1 C 2 F I1 0.38 m I2 Horizontal scale: Level 3 1 block 10 cm 81. Inspection Mirror A production-line inspector wants a mirror that produces an 83. A ball is positioned 22 cm in front of a spheri- image that is upright with a magnification cal mirror and forms a virtual image. If the of 7.5 when it is located 14.0 mm from a spherical mirror is replaced with a plane mir- machine part. ror, the image appears 12 cm closer to the mir- ror. What kind of spherical mirror was used?

370 Solutions Manual Physics: Principles and Problems Chapter 17 continued The object position for both mirrors is O1 Ray 1 22 cm. So, the image position for the Horizontal scale: Ray 2 plane mirror is 22 cm. 1 block 1.0 m Vertical scale: Because the spherical mirror forms a C F hi 2.4 m 2 blocks 1.0 m virtual image, the image is located di 14 m behind the mirror. Thus, the image posi- I1 tion for the spherical mirror is negative. The image is 2.4 m tall, and it is 14 m di di, plane 12 cm from the mirror. 22 cm 12 cm 86. A 4.0-cm-tall object is placed 12.0 cm from a 34 cm convex mirror. If the image of the object is 1 1 1 2.0 cm tall, and the image is located at f do di 6.0 cm, what is the focal length of the mir- d d ror? Draw a ray diagram to answer the ques- oi f tion. Use the mirror equation and the mag- do di nification equation to verify your answer. (22 cm)(34 cm) 22 cm (34 cm) O1 Ray 1 I1 62 cm Ray 2 The focal length is positive, so the F spherical mirror is a concave mirror. Horizontal scale: 1 block 1.0 cm f 12 cm Vertical scale: 84. A 1.6-m-tall girl stands 3.2 m from a convex 3 blocks 2.0 cm mirror. What is the focal length of the mirror if her image appears to be 0.28 m tall? 1 1 1 h d f do di m i i h d d d o o oi f h d do di d i o i h (12.0 cm)(6.0 cm) o 12.0 cm (6.0 cm) (0.28 m)(3.2 m) 1.6 m 12 cm 0.56 m Thinking Critically 1 1 1 pages 481–482 f do di 87. Apply Concepts The ball in Figure 17-25 d d oi slowly rolls toward the concave mirror on the f do di right. Describe how the size of the ball’s image (3.2 m)(0.56 m) changes as it rolls along. 3.2 m (0.56 m) 0.68 m

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Companies, a division of The McGraw-Hill © Glencoe/McGraw-Hill, Copyright 85. Magic Trick A magician uses a concave C F mirror with a focal length of 8.0 m to make a 3.0-m-tall hidden object, located 18.0 m from the mirror, appear as a real image that is seen by his audience. Draw a scale ray diagram to find the height and location of the image. Figure 17-25 Physics: Principles and Problems Solutions Manual 371 Chapter 17 continued Beyond C, the image is smaller than the mirror, what is the focal length of the con- ball. As the ball rolls toward the mirror, cave mirror? the image size increases. The image is di, initial do, initial the same size as the ball when the ball is at C. The image size continues to 6.0 cm increase until there is no image when d d (8.0 cm) the ball is at F. Past F, the size of the i i, initial image decreases until it equals the 6.0 cm (8.0 cm) ball’s size when the ball touches the 14.0 cm mirror. 1 1 1 88. Analyze and Conclude The object in f do di Figure 17-26 is located 22 cm from a d d oi concave mirror. What is the focal length f do di of the mirror? (6.0 cm)(14.0 cm) f 6.0 cm (14.0 cm) f 1.0101 cm

91. Analyze and Conclude The layout of the two-mirror system shown in Figure 17-11 is that of a Gregorian telescope. For this 22 cm question, the larger concave mirror has a radius of curvature of 1.0 m, and the smaller mirror is located 0.75 m away. Why is the secondary mirror concave? The smaller mirror is concave to produce Figure 17-26 a real image at the eyepiece that is

r Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. f 2 upright. The light rays are inverted by the first concave mirror and then inverted d o again by the secondary concave mirror. 2

22 cm 92. Analyze and Conclude An optical arrange- 2 ment used in some telescopes is the 11 cm Cassegrain focus, shown in Figure 17-27. This telescope uses a convex secondary 89. Use Equations Show that as the radius of mirror that is positioned between the curvature of a concave mirror increases to primary mirror and the focal point of infinity, the mirror equation reduces to the the primary mirror. relationship between the object position Convex Concave and the image position for a plane mirror. secondary mirror primary mirror As f → ,1/f → 0. The mirror equation then becomes 1/do 1/di,or do di. F 90. Analyze and Conclude An object is located 6.0 cm from a plane mirror. If the plane mir- ror is replaced with a concave mirror, the Telescope tube Eyepiece resulting image is 8.0 cm farther behind the Figure 17-27 mirror. Assuming that the object is located a. A single convex mirror produces only between the focal point and the concave virtual images. Explain how the convex 372 Solutions Manual Physics: Principles and Problems Chapter 18 continued n 83. If an object is 10.0 cm from a converging 1 A c sin ng lens that has a focal length of 5.00 cm, how far from the lens will the image be? sin11.00 1.52 1 1 1 f d d 41.1¡ o i d f When the light ray in the glass strikes o di the surface at a 62¡ angle, total internal do f reflection occurs. (10.0 cm)(5.00 cm) 10.0 cm 5.00 cm 10.0 cm 62° 62° 28° 28°

45° 45° Level 2 84. A convex lens is needed to produce an image that is 0.75 times the size of the object and located 24 cm from the lens on the other side. What focal length should be specified? 18.2 Convex and Concave Lenses h d page 510 m i i ho do Level 1 d 81. The focal length of a convex lens is 17 cm. d i o m A candle is placed 34 cm in front of the lens. Make a ray diagram to locate the (24 cm) 0.75 image. 32 cm di 34 cm 1 1 1 Ray 1 O1 f do di d 34 cm d d Ray 2 i f o i d d F F o i (32 cm)(24 cm) Horizontal scale: 32 cm 24 cm I1 1 block 2 cm 14 cm

85. An object is located 14.0 cm from a convex 82. A converging lens has a focal length of lens that has a focal length of 6.0 cm. The 25.5 cm. If it is placed 72.5 cm from an object is 2.4 cm high. object, at what distance from the lens will a. Draw a ray diagram to determine the loca- the image be? tion, size, and orientation of the image. 1 1 1 f do di O1 Ray 1 do f hi 1.8 cm d Ray 2 i d f di 10.5 cm Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Companies, a division of The McGraw-Hill © Glencoe/McGraw-Hill, Copyright o F F (72.5 cm)(25.5 cm) 72.5 cm 25.5 cm I1 Horizontal scale: 39.3 cm 1 block 1.0 cm The image is 39.3 cm from the lens. Vertical scale: 1 block 0.4 cm

Physics: Principles and Problems Solutions Manual 389 Chapter 18 continued b. Solve the problem mathematically. b. If the original lens is replaced with a lens having twice the focal length, what are 1 1 1 f do di the image position, size, and orientation? d f f 2f o new di do f 2(6.00 cm) (14.0 cm)(6.0 cm) 14.0 cm 6.0 cm 12.0 cm 10.5 cm 1 1 1 f do di h d i i m d f ho do o new di, new do fnew d h h i o i d (15.0 cm)(12.0 cm) o 15.0 cm 12.0 cm (10.5 cm)(2.4 cm) 60.0 cm 14.0 cm h d 1.8 cm, so the image is m i i h d inverted o o d h i, new o hi, new 86. A 3.0-cm-tall object is placed 22 cm in front do of a converging lens. A real image is formed (60.0 cm)(3.0 cm) 11 cm from the lens. What is the size of the 15 cm image? 12 cm h d m i i ho do The image is inverted compared to the object. d h h i o i d o 88. A diverging lens has a focal length of

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (11 cm)(3.0 cm) 15.0 cm. An object placed near it forms a 22 cm 2.0-cm-high image at a distance of 5.0 cm 1.5 cm from the lens. The image is 1.5 cm tall. a. What are the object position and object height? 1 1 1 Level 3 f do di 87. A 3.0-cm-tall object is placed 15.0 cm in d f front of a converging lens. A real image is d i o d f formed 10.0 cm from the lens. i a. What is the focal length of the lens? ( 5.0 cm)( 15.0 cm) 5.0 cm (15.0 cm) 1 1 1 7.5 cm f do di d d hi di o i m f h d do di o o d h (15.0 cm)(10.0 cm) o i ho 15.0 cm 10.0 cm di 6.00 cm (7.5 cm)(2.0 cm) 5.0 cm 3.0 cm

390 Solutions Manual Physics: Principles and Problems Chapter 18 continued b. The diverging lens is now replaced by a b. A 1000.0-mm lens is focused on an converging lens with the same focal object 125 m away. What is the image length. What are the image position, position? height, and orientation? Is it a virtual 1 1 1 image or a real image? f di do f f d f new o So di (15.0 cm) do f (125 m)(1.0000 m) 15.0 cm 125 m 1.00 m 1 1 1 1.01 m 1.01103 mm fnew do di, new d f 90. Eyeglasses To clearly read a book 25 cm o new di, new do fnew away, a farsighted girl needs the image to be 45 cm from her eyes. What focal length is (7.5 cm)(15 cm) 7.5 cm 15 cm needed for the lenses in her eyeglasses? 15 cm 1 1 1 f di do h d i i d d m o i ho do So f do di d h i, new o (25 cm)(45 cm) hi, new do 25 cm (45 cm) ( 15 cm)(3.0 cm) 56 cm 7.5 cm 6.0 cm Level 2 This is a virtual image that is upright 91. Copy Machine The convex lens of a copy compared to the object. machine has a focal length of 25.0 cm. A letter to be copied is placed 40.0 cm from the lens. 18.3 Applications of Lenses a. How far from the lens is the copy pages 510–511 paper? Level 1 1 1 1 89. Camera Lenses Camera lenses are f di do described in terms of their focal length. d f A 50.0-mm lens has a focal length of o di 50.0 mm. do f a. A camera with a 50.0-mm lens is (40.0 cm)(25.0 cm) focused on an object 3.0 m away. What 40.0 cm 25.0 cm is the image position? 66.7 cm 1 1 1 b. How much larger will the copy be? f di do h d i i d f h d o o o So di Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Companies, a division of The McGraw-Hill © Glencoe/McGraw-Hill, Copyright d f o d h (66.7 cm)(h ) h i o o 3 i (3.0 10 mm)(50.0 mm) do 40.0 cm 3 3.0 10 mm 50.0 mm 1.67ho 51 mm The copy is enlarged and inverted.

Physics: Principles and Problems Solutions Manual 391 Chapter 18 continued 92. Camera A camera lens with a focal length 94. Telescope The optical system of a toy of 35 mm is used to photograph a distant refracting telescope consists of a converging object. How far from the lens is the real objective lens with a focal length of image of the object? Explain. 20.0 cm, located 25.0 cm from a converging eyepiece lens with a focal length of 35 mm; for a distant object, do can be 4.05 cm. The telescope is used to view a considered at , thus 1/do is zero. According to the thin lens equation, 10.0-cm-high object, located 425 cm from the objective lens. di f. a. What are the image position, height, Level 3 and orientation as formed by the objec- 93. Microscope A slide of an onion cell is tive lens? Is this a real or virtual image? placed 12 mm from the objective lens of 1 1 1 a microscope. The focal length of the f do di objective lens is 10.0 mm. d f o di a. How far from the lens is the image do f formed? (425 cm)(20.0 cm) 1 1 1 425 cm 20.0 cm f d d i o 21.0 cm d f So d o h d i i i do f m ho do (12 mm)(10.0 mm) d h 12 mm 10.0 mm h i o i d 6.0101 mm o (21.0 cm)(10.0 cm) b. What is the magnification of this image? 425 cm d 1 i 6.0 10 mm mo 5.0 0.494 cm do 12 mm

This is a real image that is inverted Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. The real image formed is located compared to the object. 10.0 mm beneath the eyepiece lens. If the focal length of the eyepiece is b. The objective lens image becomes the 20.0 mm, where does the final image object for the eyepiece lens. What are appear? the image position, height, and orienta- tion that a person sees when looking 1 1 1 into the telescope? Is this a real or f di do virtual image? d f d o i do, new 25.0 cm di do f 25.0 cm 21.0 cm (10.0 mm)(20.0 mm) 10.0 mm 20.0 mm 4.0 cm 20.0 mm, or 20.0 mm beneath the eyepiece 1 1 1 fnew do, new di, new d. What is the final magnification of this d f compound system? d o, new new i, new d f d o, new new i ( 20.0 mm) me 2.00 do 10.0 mm (4.0 cm)(4.05 cm) 4.0 cm 4.05 cm m m m (5.0)(2.00) total o e 3.2102 cm 1.0101

392 Solutions Manual Physics: Principles and Problems Chapter 18 continued ho, new hi 97. A 3.0-cm-tall object is placed 20 cm in front of a converging lens. A real image is formed 0.494 cm 10 cm from the lens. What is the focal h d length of the lens? m i i h d o o 1 1 1 d d h f o di h i, new o, new i, new d d d o, new o i f 2 do di ( 3.2 10 cm)( 0.494 cm) 4.0 cm (20 cm)(10 cm) 4.0101 cm 20 cm 10 cm 7cm This is a virtual image that is invert- ed compared to the object. c. What is the magnification of the Level 2 telescope? 98. Derive n sin 1/sin 2 from the general form of Snell’s law of refraction, h m i, new n sin n sin . State any assumptions h 1 1 2 2 o and restrictions. 1 4.0 10 cm The angle of incidence must be in air. If 10.0 cm we let substance 1 be air, then n1 4.0 1.000. Let n2 n. Therefore, Mixed Review n1 sin 1 n2 sin 2 pages 511—512 sin 1 n sin 2 Level 1 sin 1 95. n A block of glass has a critical angle of 45.0°. sin 2 What is its index of refraction? 99. Astronomy How many more minutes n sin 2 would it take light from the Sun to reach c n 1 Earth if the space between them were filled n 2 with water rather than a vacuum? The Sun n1 , n2 1.00 for air 8 sin c is 1.510 km from Earth.

1.00 Time through vacuum n1 sin 45.0¡ 8 t d (1.5 10 km)(1000 m/1 km) 1.41 c 3.00108 m/s 5.0102 s 96. Find the speed of light in antimony trioxide if it has an index of refraction of 2.35. Speed through water c 3.00108 m/s n c v v n 1.33 2.26108 m/s v c n Time through water 3.00108 m/s d (1.5108 km)(1000 m/1 km) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Companies, a division of The McGraw-Hill © Glencoe/McGraw-Hill, Copyright 2.35 t v 2.26108 m/s 8 1.28 10 m/s 660 s t 660 s 500 s 160 s (160 s)(1 min/60 s) 2.7 min

Physics: Principles and Problems Solutions Manual 393