Upcoming SlideShare
×

# Anschp34

6,259 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
6,259
On SlideShare
0
From Embeds
0
Number of Embeds
9
Actions
Shares
0
116
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Anschp34

1. 1. Chapter 34. Reflection and Mirrors Physics, 6th Edition Chapter 34. Reflection and Mirrors Reflection from Plane Mirrors 34-1. A man 1.80 m tall stands 1.2 m from a large plane mirror. How tall is his image? How far is he from his image? ( Image symmetric, reversed.) 1.2 m Image distance q = object distance p: 1.8 m q = p = 1.2 m; y’ = y = 1.8 m 33-2. What is the shortest mirror length required to enable a 1.68 m woman to see her entire image? ( Virtual image is behind the mirror. Rays show that only half length is needed.) Mirror height = (1/2) object height 5 ft, 8 in. 1.68 m h= h = 2 ft, 10 in. 2 It doesn’t matter where she stands. *33-3. A plane mirror moves at a speed of 30 km/h away from a stationary person. How fast does this persons image appear to be moving in the opposite direction? Each time mirror moves, image also moves, so that speed is doubled: vi = 60 km/h *33-4. The optical lever is a sensitive measuring device that utilizes minute rotations of a plane mirror to measure small deflections. The device is illustrated in Fig. 34-19. When the mirror is in position 1, the light ray follows the path IVR1. If the mirror is rotated through an angle θ to position 2, the ray will follow the path IVR2. Show that the reflected beam turns through an angle 2θ, which is twice the angle through which the mirror itself turns. Remember the basic principles of ray tracing and that the angle of incidence always equals the angle of reflection. Refer to the figure drawn on the following page. 217
2. 2. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-4. (Cont.) Show that the deviated ray turns through twice the angle turned by mirror. In horizontal initial position, θi = θr. Now from figure: N1 N2 I R1 θ i1 = θ i1 + θ ; θr2 = θr1 + θ θ R2 θ i 2 = θ r 2 = (θ i 1 + θ) + (θ r 1 + θ) 2θ 2 ∠R1R2 = 2 θ 1 θ Images Formed by Spherical Mirrors 4-5. A light bulb 3 cm high is placed 20 cm in front of a concave mirror with a radius of curvature of 15 cm. Determine the nature, size, and location of the image formed. Sketch the ray-tracing diagram. [ y = 3 cm, p = 20 cm ] pf (20 cm)(7.5 cm) q= = ; q = 12 cm, real p− f 20 cm - 7.5 cm y ' −q −qy −(12 cm)(3 cm) M= = ; y' = = ; y’ = -1.80 cm y p p (20 cm) q = 12.0 cm, y’ = - 1.80 cm; Real, inverted, and diminished. 34-6. A spherical concave mirror has a focal length of 20 cm. What are the nature size and location of the image formed when a 6 cm tall object is located 15 cm from this mirror? pf (15 cm)(20 cm) q= = ; p− f 15 cm - 20 cm C F virtual image q = -60 cm, virtual and enlarged y ' −q −qy −(−60 cm)(6 cm) M= = ; y' = = ; y’ = +18 cm y p p (20 cm) q = -60 cm, y’ = 18.0 cm; virtual, erect, and enlarged. 218
3. 3. Chapter 34. Reflection and Mirrors Physics, 6th Edition 34-7. A 8-cm pencil is placed 10 cm from a diverging mirror of radius 30 cm. Determine the nature, size, and location of the image formed. Sketch the ray-tracing diagram. For diverging mirror: f = (R/2) = -15 cm virtual image pf (10 cm)(-15 cm) q= = ; q = -6.00 cm, virtual p − f 10 cm - (-15 cm) F C y ' −q −qy −(−6 cm)(8 cm) M= = ; y' = = ; y’ = +16 cm y p p (10 cm) q = -6.00 cm, y’ = 4.80 cm; virtual, erect, and diminished. 34-8. A spherical convex mirror has a focal length 25 cm. What are the nature, size, and location of the image formed of a 5-cm tall object located 30 cm from the mirror? For diverging mirror: f = -25 cm virtual image pf (30 cm)(-25 cm) q= = ; q = -13.6 cm, virtual p − f 30 cm - (-25 cm) F C y ' −q −qy −(−13.6 cm)(5 cm) M= = ; y' = = ; y’ = +2.27 cm y p p (30 cm) q = -13.6 cm, y’ = 2.27 cm; virtual, erect, and diminished. 34-9. An object 5 cm tall is place halfway between the focal point and the center of curvature of a concave spherical mirror of radius 30 cm. Determine the location and magnification the of the image? f = (R/2) = 15 cm. p = 22.5 cm pf (22.5 cm)(15 cm) C q= = ; q = 45 cm, real p− f 22.5 cm - 15 cm F y ' −q −(45 cm) M= = ; M= ; M = -2.00 y p (22.5 cm) q = 45.0 cm, M = - 2.00; Real, inverted, and enlarged. 219
4. 4. Chapter 34. Reflection and Mirrors Physics, 6th Edition 34-10. A 4-cm high source of light is placed in front of a spherical concave mirror whose radius is 40 cm. Determine the nature size and location of the images formed for the following object distances: (a) 60 cm, (b) 40 cm, (c) 30 cm, (d) 20 cm, and (e) 10 cm. Draw the appropriate ray-tracing diagrams. (In the interest of space, diagrams are not drawn) Givens: f = (40 cm/2) = 20 cm; y = 4 cm, p = 60, 40, 30, 20, and 10 cm. pf y ' −q −qy q= ; M= = ; y' = p− f y p p (60 cm)(20 cm) −(30 cm)(4 cm) (a) q = = 30 cm, real y'= = -2 cm, inverted 60 cm - 20 cm (60 cm) (40 cm)(20 cm) −(40 cm)(4 cm) (b) q = = 40 cm, real y'= = -4 cm, inverted 40 cm - 20 cm (40 cm) (30 cm)(20 cm) −(60 cm)(4 cm) (c) q = = 60 cm, real y'= = -8 cm, inverted 30 cm - 20 cm (30 cm) (20 cm)(20 cm) −(∞)(4 cm) (d) q = = ∞, no image y'= = ∞, no image 20 cm - 20 cm (20 cm) (10 cm)(20 cm) −(−20 cm)(4 cm) (e) q = = -20 cm, virtual y'= = 8 cm, erect 10 cm - 20 cm (10 cm) 34-11. At what distance from a concave spherical mirror of radius 30 cm must an object be placed to form an enlarged, inverted image located 60 cm from the mirror? Enlarged and inverted means object between F and C. C and the image is beyond the radius. F f = (R/2) = (30 cm/2); f = +15 cm q = +60 cm , positive since image is real. qf (60 cm)(15 cm) p= = ; p = 20 cm q− f 60 cm - 15 cm 220
5. 5. Chapter 34. Reflection and Mirrors Physics, 6th Edition Magnification 34-12. What is the magnification of an object if it is located 10 cm from a mirror and its image is erect and seems to be located 40 cm behind the mirror? Is this mirror diverging or converging? ( Erect image means diverging mirror.) q = -40 cm , p = 10 cm C F virtual image −q −(−40 cm) M= = ; M = -4.00 p 10 cm 34-13. A Christmas tree ornament has a silvered surface and a diameter of 3 in. What is the magnification of an object placed 6 in. from the surface of this ornament? [ R= D/2 ] R −1.5 in. virtual image f = = ; f = -0.75 in. (–) since converging 2 2 (6 in.)(-0.75 in.) F C q= ; q = -0.667 in., virtual 6 in. - (-0.75 in. R = 3 in. −q −(−0.667 in.) M= = ; M = +0.111 p 6.00 in. 34-14. What type of mirror is required to form an image on a screen 2 m away from the mirror when an object is placed 12 cm in front of the mirror? What is the magnification? It must be a converging mirror, since image is real. Also, from position and size, it must be enlarged and C inverted. Thus, q = +2.00 m, p = 12 cm. F −q −(2 m) M= = ; M = -16.7 p 0.12 m pq (0.12 m)(2 m) f = = ; f = +11.3 cm Positive means it is converging. p + q 0.12 m + 2 m 221
6. 6. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-15. A concave shaving mirror has a focal length of 520 mm. How far away from it should an object be placed for the image to be erect and twice its actual size? −q pf M= = +2; q = − 2 p; q= ; p p− f C F virtual image pf −2 p = ; − 2( p − f ) = f ; p− f f 520 mm −2 p + 2 f = f ; p= = ; p = 260 mm 2 2 *34-16. If a magnification of +3 is desired, how far should the mirror of Problem 34-15 be placed from the face? −q pf pf M= = +3; q = −3 p; q= ; −3 p = ; − 3( p − f ) = f ; p p− f p− f 2 f 2(520 mm) −3 p + 3 f = f ; p= = ; p = 357mm 3 3 *34-17. An object is placed 12 cm from the surface of a spherical mirror. If an erect image is formed that is one-third the size of the object, what is the radius of the mirror. Is it converging or diverging? (It is diverging since image is erect and diminished.) −q 1 − p −(12 cm) M= =+ ; q= = ; q = −4 cm p 3 3 3 pq (12 cm)(-4 cm) f = = ; f = -6.00 cm, diverging p + q 12 cm + (-4 cm) *34-18. A concave spherical mirror has a radius of 30 cm and forms an inverted image on a wall 90 cm away. What is the magnification? [ f = (30 cm/2) = 15 cm; q = +90 cm ] (90 cm)(15 cm) −90 cm p= = 18 cm; M= ; M = -5.00 90 cm - 15 cm 18 cm 222
7. 7. Chapter 34. Reflection and Mirrors Physics, 6th Edition Challenge Problems 34-19. What are the nature, size, and location of the image formed when a 6-cm tall object is located 15 cm from a spherical concave mirror of focal length 20 cm? (15 cm)(20 cm) q= = -60 cm, virtual C 15 cm - 20 cm F virtual image −(−60 cm)(6 cm) y'= = 24 cm, erect (15 cm) q = -60.0 cm, y’ = 24.0 cm; virtual, erect, and enlarged. 34-20. An erect image has a magnification of +0.6. Is the mirror diverging or converging? What is the object distance if the image distance is –12 cm? The mirror is diverging since image is diminished, erect, and virtual. −q − q −(−12 cm) M= = 0.6; p = = ; p = 20 cm p 0.6 0.6 34-21. An object is located 50 cm from a converging mirror whose radius is 40 cm. What is the image distance and the magnification? pf (50 cm)(20 cm) q= = ; q = 33.3 cm, real p− f 50 cm - 20 cm y ' − q −(33.3 cm) M= = = ; M = -0.667 y p (50 cm) q = 33.3 cm, M = - 0.667; Real, inverted, and diminished. 34-22. What is the focal length of a diverging mirror if the image of an object located 200 mm from the mirror appears to be a distance of 120 mm behind the mirror? pq (200 mm)(−120 mm) f = = ; f = -75.0 mm p + q 200 mm - (-120 mm) 223
8. 8. Chapter 34. Reflection and Mirrors Physics, 6th Edition 34-23. A silver ball is 4.0 cm in diameter. Locate the image of a 6-cm object located 9 cm from the surface of the ball. What is the magnification? [ R = 4 cm/2 = 2 cm. ] f = (R/2)=(-2 cm/2) = -1 cm; p = 9 cm; virtual image pf (9 cm)(-1 cm) q= = ; q = -0.9 cm p − f 9 cm - (-1 cm) F C −q −(−0.9 cm) R = 4 cm M= = ; M = +0.100 p 9.00 cm 34-24. An object 80 mm tall is placed 400 mm in front of a diverging mirror of radius –600 mm. Determine the nature size and location of the image virtual image f = (R/2)=(-600 mm/2) = -300 mm pf (400 mm)(-300 mm) F C q= = ; q = -171 cm p − f 400 mm - (-300 mm) R = 3 in. y ' −q −(−171 mm)(80 mm) M= = ; y' = ; y’ = +34.3 mm y p 400 mm *34-25. An object 10 cm tall is located 20 cm from a spherical mirror. If an erect image 5-cm tall is formed, what is the focal length of the mirror? y ' 5 cm −q M= = = +0.5; M= ; q = − Mp = −(0.5)(20 cm); q = −10 cm y 10 cm p pq (20 cm)(-10 cm) f = = ; f = -20 cm p + q 20 cm + (-10 cm) *34-26. What is the magnification if the image of an object is located 15 cm from a diverging mirror of focal length –20 cm? [ q = -15 cm; f = -20 cm ] qf (−15 cm)(-20 cm) −q −(−15 cm) p= = = 60 cm; M= = ; M = +0.250 q − f −15 cm − (−20 cm) p 60 cm 224
9. 9. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-27. An object is placed 200 mm from the vertex of a convex spherical mirror whose radius is 400 mm. What is the magnification of the mirror? virtual image f = R/2 = -400 mm/2 = -200 mm; p = 200 mm p pq (200 mm)(-200 mm) C F q= = ; q = - 100 mm R = 400 mm. p − q 200 mm - (-200 mm) −q −(−100 mm) Magnification: M= = ; M = +½ p 200 mm *34-28. A convex spherical mirror has a radius of –60 cm. How far away should an object be held if the image is to be one-third the size of the object? virtual image f = R/2 = -60m/2 = -30 mm; M = +1/3 −q 1 p pf M= = ; q=− ; q= F C p 3 3 p− f R = -60 cm −p pf = ; ( − 1)( p − f ) = 3 f ; − p+ f −3f = 0; 3 p− f p = -2f = -2(-30 cm); p = 60 cm *34-29. What should be the radius of curvature of a convex spherical mirror to produce an image one-fourth as large as the object which is located 40 in. from the mirror? −q 1 −p pf -1 f M= = ; q= = ; = ; p 4 4 p− f 4 40 in. − f −40 in. -40 in. + f = 4f f = ; f = -26.7 in. 3 *34-30. A convex mirror has a focal length of –500 mm. If an object is placed 400 mm from the vertex, what is the magnification? (400 mm)(-500 mm) −q −(−222 mm) q= = −222 mm; M = = ; M = +0.556 400 mm - (-500 mm) p 400 mm 225
10. 10. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-31. A spherical mirror forms a real image 18 cm from the surface. The image is twice as large as the object. Find the location of the object and the focal length of the mirror. Since the image is real and enlarged, we draw as shown: Note: q = -18 cm, M = -2 (inverted) C −q q −(−18 cm) F M= = −2; p= = ; p = 9.00 cm p 2 2 pq (9 cm)(-18 cm) f = = ; f = 6.00 cm p + q 9 cm +(-18 cm) *34-32. A certain mirror placed 2 m from an object produces an erect image enlarged three times. Is the mirror diverging or converging? What is the radius of the mirror? An erect, enlarged image is consistent C only for converging mirrors. virtual image F −q M= = +3; q = −3 p = −3(2 m); q = -6.00 m p pq (2 cm)(-6 cm) f = = ; f = +3.00 cm, converging p + q 2 cm +(-6 cm) *34-33. The magnification of a mirror is –0.333. Where is the object located if its image is formed on a card 540 mm from the mirror? What is the focal length? −q q 540 mm M= = −0.333; p= = ; p = 1.62 m p 0.333 0.333 pq (1.62 m)(0.540 m) f = = ; f = +405 mm p + q 1.62 m + 0.540 m 226
11. 11. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-34. What should be the radius of curvature of a concave mirror to produce an image one- fourth as large as an object 50 cm away from the mirror? The magnification must be negative if image is diminished by concave mirror: −q 1 p (50 cm) M= =− ; q= = ; q = 12.5 cm p 4 4 4 pq (50 cm)(12.5 cm) f = = ; f = +10 cm p + q 50 cm + 12.5 cm *34-35. A spherical shaving mirror as a magnification of +2.5 when an object is located 15 cm from the surface. What is the focal length of the mirror? −q C M= = +2.5; q = −2.5(15 cm); p F virtual image q = -37.5 cm pq (15 cm)(-37.5 cm) f = = ; f = +25.0 cm, converging p + q 15 cm +(-37.5 cm) Critical Thinking Problems 34-36. A baseball player is 6 ft tall and stands 30 ft in front of a plane mirror. The distance from the top of his cap to his eyes is 8 in. Draw a diagram showing location of the images formed of his feet and of the top of his cap? What is the minimum length of mirror required for him to see his entire image. If he walks 10 ft closer to the mirror what is the new separation of object and image? From the drawing of reflected rays, you see that: y = ½(8 in.) + ½(72 in. – 8 in.) ; ymin = 36 in. Since the image moves closer by the same amount that the ballplayer does a new object distance of 20 ft means a separation of 2(20 ft) = 40 ft 227
12. 12. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-37. The diameter of the moon is 3480 km and it is 3.84 x 108 m away from the earth. A telescope on the earth utilizes a spherical mirror, whose radius is 8.00 m to form an image of the moon. What is the diameter of the image formed. What is the magnification of the mirror? ( y = 3.48 x 10-6 m, p = 3.84 x 108 m. ) Moon pf (3.84 x 106 m)(4 m) q= = = 4.00 m (at F) p− f 3.84 x 108 m - 4 m y ' −q −4 m M= = ; M= ; M = -1.04 x 10-8 y p 3.84 x 108 m y ' = My = (−1.04 x 10-8 )(3.48 x 106 m); y = 3.62 cm *34-38. An image 60 mm long is formed on a wall located 2.3 m away from a source of light 20 mm high. What is the focal length of this mirror? Is it diverging or converging? What is the magnification? ( The image in inverted so that M is negative, image is inverted.) −q −( p + 2.3 m) q = p + 2.3 m; M= = = −3 p p p C p + 2.3 m = 3p; p = 1.15 m; q = p + 2.3 m = 3.45 m F pq (1.15 m)(3.45 m) 2.3 m f = = ; f = +0.862 m p + q 1.15 m + 3.45 m The positive focal length indicates that the mirror is converging. −q −(3.45 m) M= = ; M = -3.00 p 1.15 m 228
13. 13. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-39. Derive an expression for calculating the focal length of a mirror in terms of the object distance p and the magnification M. Apply it to Problem 34-35. Derive a similar relation for calculating the image distance q in terms of M and p. Apply it to Problem 34-33. −q pf pf M= ; q = − Mp; q= ; − Mp = p p− f p− f Mp -M(p – f) = f; -Mp + Mf = f; Mf – f = Mp f = ( M − 1) Mp (2.5)(15 cm) M = 2.5, p = 15 cm; f = = ; f = +25 cm ( M − 1) (2.5 − 1) −q −q qf −q qf M= ; p= ; p= ; = p M q− f M q− f −q -(q – f) =M f; -q + f = Mf Mf – f = -q f = ( M − 1) −q −540 mm M = -0.333, q = 540 mm; f = = f = +405 mm ( M − 1) −0.333 − 1 229
14. 14. Chapter 34. Reflection and Mirrors Physics, 6th Edition *34-40. A concave mirror of radius 800 mm is placed 600 mm from a plane mirror that faces it. A source of light placed midway between the mirrors is shielded so that the light is first reflected from the concave surface. What are the position and magnification of the image formed after reflection from the plane mirror? (Hint: treat the image formed by the first mirror as the object for the second mirror.) p C q 0.600 m R p= = 0.300 m; f = = 0.400 m 2 2 F virtual image 600 mm First find q formed by spherical mirror. pf (0.300 m)(0.400 m) q= = ; q = −1.20 m ; (behind mirror) p − f (0.300 m - 0.400 m) Since plane mirror is 0.600 m in left of spherical mirror, the final image q’= p’ is: q’ = 1.20 m + 0.600 m = 1.8 m; q’ = 1.8 m to left or behind the plane mirror. −q −( −1.2 m) M1 = = = +4; M 2 = +1; M 1 x M 2 = (+4)(+1); M = +4 p 0.300 m 230