# Physics exercises_solution: Chapter 34

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## Physics exercises_solution: Chapter 34

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 34

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## Nội dung Text: Physics exercises_solution: Chapter 34

1. 34.1: If up is the  y -direction and right is the  x -direction, then the object is at ( x0 ,  y 0 ), P2 is at ( x0 ,  y 0 ), and mirror 1 flips the y -values, so the image is at ( x0 , y 0 ) which is P3. 34.2: Using similar triangles, htree d d 28.0 m  0.350 m  tree  htree  hmirror tree  0.040 m  3.24 m. hmirror d mirror d mirror 0.350 m 34.3: A plane mirror does not change the height of the object in the image, nor does the distance from the mirror change. So, the image is 39.2 cm to the right of the mirror, and its height is 4.85 cm. R 34.0 cm 34.4: a) f    17.0 cm. 2 2 b) If the spherical mirror is immersed in water, its focal length is unchanged—it just depends upon the physical geometry of the mirror. 34.5: a) 1 1 1 1 2 1 b)       s   33.0 cm, to the left of the s s f s  22.0 cm 16.5 cm mirror. s 33.0 cm y   y  (0.600 cm)  1.20 cm, and the image is inverted and real. s 16.5 cm
2. 34.6: a) 1 1 1 1 2 1 b)       s   6.60 cm, to the right of the s s f s 22.0 cm 16.5 cm mirror. s (6.60 cm) y   y  (0.600 cm)  0.240 cm, and the image is upright and s 16.5 cm virtual. 1 1 1 1 1 1 34.7:       s   1.75 m. s s f s  1.75 m 5.58  1010 m 1.75 m  3.14  10 11  y   my 5.58  1010  (3.14  10 11 )(6794  10 3 m)  2.13  10  4 m. 1 1 1 1 2 1 34.8: R  3.00 cm,       s   1.40 cm (in the s s f s 3.00 cm 21.0 cm s  1.40 cm ball). The magnification is m      0.0667. s 21.0 cm
3. 1 1 1 1 1 1 s f sf s f 34.9: a)        s  . Also m    . s s f s f s fs s f s f s b) For f  0, s  f  s   0, so the image is always on the outgoing side and is f real. The magnification is m   0, since f  s. f s f c) For s  2 f  m   1, which means the image is always smaller and f inverted since the magnification is negative. f For f  s  2 f  0  s  f  f  m   1. f d) Concave mirror: 0  s  f  s   0, and we have a virtual image to the right f of the mirror. m   1, so the image is upright and larger than the object. f sf sf 34.10: For a convex mirror, f  0  s     0. Therefore the image is s f s f f  f f always virtual. Also m     0, so the image is erect, and f s  f s f s m  1 since f  s  f , so the image is smaller.
4. 34.11: a) b) s   0 for s  f , s  0. c) s   0 for 0  s  f . d) If the object is just outside the focal point, then the image position approaches positive infinity. e) If the object is just inside the focal point, the image is at negative infinity, “behind” the mirror. f) If the object is at infinity, then the image is at the focal point. g) If the object is next to the mirror, then the image is also at the mirror. h) i) The image is erect if s  f . j) The image is inverted if s  f . k) The image is larger if 0  s  2 f . l) The image is smaller if s  2 f or s  0. m) As the object is moved closer and closer to the focal point, the magnification INCREASES to infinite values.
5. 34.12: a) a) s   0 for  f  s  0. b) s   0 for s   f and s  0. c) If the object is at infinity, the image is at the outward going focal point. d) If the object is next to the mirror, then the image is also at the mirror. For the answers to (e), (f), (g), and (h), refer to the graph on the next page. e) The image is erect (magnification greater than zero) for s   f . f) The image is inverted (magnification less than zero) for s   f . g) The image is larger than the object (magnification greater than one) for  2 f  s  0. h) The image is smaller than the object (magnification less than one) for s  0 and s   2 f .
6. 34.13: a) 1 1 1 1 2 1 b)       s   5.45 cm, to the right of the s s f s 20.0 cm 12.0 cm mirror. s  5.45 cm y   y  (0.9 cm)  0.409 cm, and the image is upright and virtual. s 12.0 cm s  48.0 34.14: a) m     4.00, where s comes from part (b). s 12.0 1 1 1 1 2 1 b)       s  48.0 cm. Since s  is negative, s s f s  32.0 cm 12.0 cm the image is virtual. c) n a nb 1.309 1.00 34.15:  0   0  s   2.67 cm. s s 3.50 cm s
7. n a nb 1.33 1.00 34.16: a)  0   0  s   5.26 cm, so the fish appears s s 7.00 cm s 5.26 cm below the surface. n n 1.33 1.00 b) a  b  0    0  s   24.8 cm, so the image of the fish s s 33.0 cm s appears 24.8 cm below the surface. 34.17: a) For R  0 and na  nb , with  a     and  b     , we have: n nb b  na b   b      a (   )  na  nb   (nb  na ) . nb h h h But   ,   , and   , so subbing them in one finds: s  s R n a nb ( nb  n a )   . s s R Also, the magnification calculation yields: y y n y n y y n s tan  a and tan  b  a  b m  a . s  s s s y nb s b) For R  0 and n a  nb , with  a     and  b     , we have : nb   n a  h h h n n (n  na ) (nb  na ) . But   ,   , and    a  b  b , so subbing them s s R s s R n n (n  na ) in one finds: a  b  b . Also, the magnification calculation yields: s s R n y n y y n s na tan  a  nb tan  b  a   b  m    a . s s y na s na nb nb  na 1 1.60 0.60 34.18: a)       s  8.00 cm. s s R  s 3.00 cm n n n  na 1 1.60 0.60 b) a  b  b     s   13.7 cm. s s R 12.0 cm s 3.00 cm n n n  na 1 1.60 0.60 c) a  b  b     s   5.33 cm. s s R 2.00 cm s 3.00 cm
8. n a nb nb  n a sR ( s   R) s ( s   R) 34.19:    na  nb  nb . s s R ( s  R) s R s  ( s  R) 90.0 cm (160 cm  3.00 cm)  na  (1.60)  1.52. 160 cm (90.0 cm  3.00 cm) n a nb nb  n a 1 1.60 0.60 34.20:       s   14.8 cm. s s R 24.0 cm s 4.00 cm   n s    14.8 cm  y   a  y    n s   (1.60)(24.0 cm)  1.50 mm  0.578 mm, so the image height   b    is 0.578 mm, and is inverted. n a nb nb  n a 1 1.60  0.60 34.21:       s   8.35 cm. s s R 24.0 cm s 4.00 cm   n s    (8.35 cm)  y   a  y    ns   (1.60)(24.0 cm)  1.50 mm  0.326 mm, so the image height is   b    0.326 mm, and is erect. na nb nb  na 1.33 1.00  0.33 34.22: a)       s  14.0 cm, so the fish s s R 14.0 cm s  14.0 cm appears to be at the center of the bowl.   n s     (1.33)(17.0 cm)  m a   n s   (1.00)(17.0 cm)   1.33.   b    n n n  na 1.00 1.33 0.33 b) a  b  b     s  56.4 cm, which is outside s s R  s  14.0 cm the bowl.
9. 34.23: For s  18 cm : 1 1 1 1 1 1 a)       s   63.0 cm. s s f s  14.0 cm 18.0 cm s 63.0 b) m      3.50. s 18.0 c) and d) From the magnification, we see that the image is real and inverted. For s  7.00 cm : 1 1 1 1 1 1 a)       s   14.0 cm. s s f s  14.0 cm 7.00 cm s  14.0 b) m      2.00. s 7.00 c) and d) From the magnification, we see that the image is virtual and erect.
10. 1 1 1 1 1 1 34.24: a)       f  48.0 cm, and the lens is s s f f 16.0 cm  12.0 cm diverging.  s   (12.0)  b) y   y    (0.850 cm)     0.638 cm, and is erect.  s  16.0  c) y  1.30 s 34.25: m    3.25   . Also: y 0.400 s 1 1 1 1 1 1 s s         1  3.25  s   15.75 cm s s f s 7.00 cm s  s 7.00 cm (to the left). 1 1 1     s  4.85 cm, and the image is virtual s 7.0 cm  15.75 cm (since s   0). y 4.50 s 34.26: m      s  0.711s . Also : y 3.20 s 1 1 1 1 1 1       s   217 cm (to the right). s s f 0.711 s  s  90.0  s  0.711(217 cm)  154 cm, and the image is real (since s   0). 1 1  1 1  1 1  1 1  34.27:   (n  1)    R R   18.0 cm  s   (0.48)  5.00 cm  3.50 cm     s s  1 2     s   10.3 cm (to the left of the lens).
11. 34.28: a) Given s   80.0s, and s  s   6.00 m  81.00s  6.00 m  s  0.0741 m and s   5.93 m. b) The image is inverted since both the image and object are real ( s   0, s  0). 1 1 1 1 1 c)      f  0.0732 m, and the lens is converging. f s s  0.0741 m 5.93 m 1  1 1   1 1  34.29:  (n  1)   R R    (0.60)    4.00 cm  8.00 cm    f  1 2     f  4.44 cm,  13.3 cm. f 1  13.3 cm; f 2  4.44 cm; f 3  4.44 cm; f 4  13.3 cm; f 5  13.3 cm; f 6  13.3 cm f 7  4.44 cm; f 8  4.44 cm. 34.30: We have a converging lens if the focal length is positive, which requires: 1  1 1   1 1   (n  1)    R R   0   R  R   0. This can occur in one of three ways:    f  1 2   1 2  (i) {R1  R2 }  {R1 , R2  0} (ii) R1  0, R2  0 (iii) {| R1 |  | R2 |}  {R1 , R2  0}. Hence the three lenses in Fig. (35.29a). We have a diverging lens if the focal length is negative, which requires: 1  1 1   1 1   (n  1)    R R   0   R  R   0. This can occur in one of three    f  1 2   1 2  ways: (i) {R1  R2 }  {R1 , R2  0} (ii) R1  R2  0 (iii) R1  0, R2  0. Hence the three lenses in Fig. (34.29b).
12. 34.31: a) The lens equation is the same for both thin lenses and spherical mirrors, so the derivation of the equations in Ex. (34.9) is identical and one gets: 1 1 1 1 1 1 s f sf s f        s  , and also m    . s s f s f s fs s f s f s b) Again, one gets exactly the same equations for a converging lens rather than a concave mirror because the equations are identical. The difference lies in the interpretation of the results. For a lens, the outgoing side is not that on which the object lies, unlike for a mirror. So for an object on the left side of the lens, a positive image distance means that the image is on the right of the lens, and a negative image distance means that the image is on the left side of the lens. c) Again, for Ex. (34.10) and (34.12), the change from a convex mirror to a diverging lens changes nothing in the exercises, except for the interpretation of the location of the images, as explained in part (b) above. 1 1 1 1 1 1 34.32:       s  7.0 cm. s s f s 12.0 cm  17.0 cm s (17.0) y  0.800 cm m   2.4  y    0.34 cm, so the object is s 7.2 m  2 .4 0.34 cm tall, erect, same side. 1 1 1 1 1 1 34.33:       s  26.3 cm. s s f s  48.0 cm 17.0 cm s  17.0 y  0.800 cm m   0.646  y    1.24 cm tall, erect, same side. s  26.3 m 0.646
13. 1 1 1 1 1 1 34.34: a)       f  11.1 cm, converging. s s f f 16.0 cm 36.0 cm  s   36  b) y  y      (0.80 cm)     1.8 cm, so the image is inverted.  s  16  c) y s y  0.024 m  34.35: a) m    s  s  600 m   240 m   0.0600 m  60 mm.  y s y   1 1 1 1 1       f  60 mm. So one should use the f s s  6.0  10 mm 60 mm 5 85-mm lens. y  0.036 m  b) s   s  40.0 m   9.6 m   0.15 m  150 mm.  y   1 1 1 1 1       f  149 mm. So one should use the f s s  40  10 mm 150 mm 3 135-mm lens. 1 1 1 1 1 1 34.36:       s   0.0869 m. s s f 3.90 m s  0.085 m s 0.0869 y   y   1750 mm  39.0 mm, so it will not fit on the 24-mm  s 3.90 36-mm film. 1 1 1 1 1 1 34.37:       s  1020 cm. s s f s 20.4 cm 20.0 cm s f 5.00 m 34.38: y    y y (70.7 m)  0.0372 m  37.2 mm. s s 9.50  10 3 m
14. s f 28 mm 34.39: a) m   m  1.4  10 4. s s 200,000 mm s f 105 mm b) m    m   5.3  10  4. s s 200,000 mm s f 300 mm c) m    m   1.5  10 3. s s 200,000 mm 34.40: a) s1    s  f1  12 cm. 1 b) s 2  4.0 cm  12 cm  8 cm. 1 1 1 1 1 1 c)        s 2  24 cm, to the right. s s f   8 cm s 2  12 cm d) s1    s  f1  12 cm. 1 s 2  8.0 cm  12 cm  4 cm. 1 1 1 1 1 1        s 2  6 cm. s s f   4 cm s 2  12 cm f f 300 mm 34.41: a) f 4  4  D   75 mm. D 4 4 f b) f 8  D  , so the diameter is 0.5 times smaller, and the area is 0.25 times 8 smaller. Therefore only a quarter of the light entered the aperture, and the film must be exposed four times as long for the correct exposure. 34.42: The square of the aperture diameter (~ the area) is proportional to the length of the 2  1   8 mm   1  exposure time required.  s   23.1 mm    250 s  .   30      34.43: a) A real image is formed at the film, so the lens must be convex. 1 1 1 1 s f sf b)   so  and s   , with f  50.0.0 mm s s f s sf s f For s  45 cm  450 mm, s  56 mm. For s  , s   f  50 mm. The range of distances between the lens and film is 50 mm to 56 mm.
15. 1 1 1 1 1 1 34.44: a)       s  0.153 m  15.3 cm. s s f s 9.00 m 0.150 m s  9.00 b) m    58.8  dimensions are (24 mm  36 mm) m  (1.41 m  2.12 m). s 0.153 1 1 34.45: a) f    0.364 m  36.4 cm. The near-point is normally at power 2.75 m 1 1 1 1 1 1 1 25 cm :       s   80 cm, in front of the eye. s s f 25 cm s  36.4 cm 1 1 b) f    0.769 m  76.9 cm. The far point is ideally at power  1.30 m 1 1 1 1 1 1 1 infinity, so:       s  76.9 cm. s s f  s  76.9 cm na nb nb  na 1 1.40 0.40 34.46:       R  0.710 cm. s s R 40.0 cm 2.60 cm R 1 1 1 1 1 1 34.47: a)      power   2.33 diopters. f s s  0.25 m  0.600 m f 1 1 1 1 1 1 b)      power   1.67 diopters. f s s    0.600 m f 25.0 cm 25.0 cm 34.48: a) Angular magnification M    4.17. f 6.00 cm 1 1 1 1 1 1 b)       s  4.84 cm. s s f s  25.0 cm 6.00 cm 1 1 1 1 1 1 34.49: a)       s  6.06 cm. s s f s  25.0 m 8.00 cm s  25.0 cm b) m    4.13  y   ym  (1.00 mm)(4.13)  4.13 mm s 6.06 cm
16. y y 2.00 mm 34.50:    f    80.0 mm  8.00 cm. f  0.025 rad s 1 1 1 1 1 34.51: m    6.50  s   6.50s      s s s  s  6.50s 4.00 cm 1 1  1  1    s  3.38 cm, s   6.50s  22.0 cm. s  6.50  4.00  (250 mm)s1 (250 mm)(160 mm  5.0 mm) 34.52: a) M    317. f1 f 2 (5.00 mm)(26.0 mm) y y  0.10 mm b) m   y    3.15  10  4 mm. y m 317 34.53: a) The image from the objective is at the focal point of the eyepiece, so  s1  d oe  f 2  19.7 cm  1.80 cm  17.9 cm 1 1 1 1 1 1        s  0.837 cm. s s f s 17.9 cm 0.800 cm s  17.9 cm b) m1    21.4. s 0.837 cm 25.0 cm 25.0 cm c) The overall magnification is M  m1  (21.4)  297. f2 1.80 cm
17.  s1 34.54: Using the approximation s1  f , and then m1  , we have: f1 f  16 mm : s   120 mm  16 mm  136 mm; s  16 mm s  136 mm  m1    8.5. s 16 mm s  124 mm f  4 mm : s   120 mm  4 mm  124 mm; s  4 mm  m1    31. s 4 mm f  1.9 mm : s   120 mm  1.9 mm  122 mm; s  1.9 mm s  122 mm  m1    64. s 1.9 mm The eyepiece magnifies by either 5 or 10, so: a) The maximum magnification occurs for the 1.9-mm objective and 10x eyepiece:  M  m1 me  (64)(10)  640. b) The minimum magnification occurs for the 16-mm objective and 5x eyepiece:  M  m1 me  (8.5)(5)  43. f1 95.0 cm 34.55: a) M     6.33. f2 15.0 cm 1 1 1 1 1 1 b)       s   0.950 m, so the height of an s s f 3000 m s  0.950 m s 0.950 image of a building is y   y  (60.0 m)  0.019 m. s 3000 c)    M  6.33arctan(60.0 3000)  6.33 (60.0) (3000)  0.127 rad. 34.56: f 1  f 2  d ss  f1  d ss  f 2  1.80 m  0.0900 m  1.71 m f 171 M  1   19.0. f2 9.00 f 34.57:  19.0  f  (19.0) D  (19.0)(1.02 m)  19.4 m. D s f    34.58: y  y  y  θf  (0.014) 3 (18 m)  4.40  10 m. s s  180 
18. R 34.59: a) f 1   0.650 m  d  f 1  f 2  0.661 m. 2 f 0.650 m b) M  1   59.1. f 2 0.011 m 1 1 1 1 1 1 34.60:       f  1.50 m  R  s s f 0.75 m  1.3 m 0.75 m  0.12 m f 2 f  3.0 m. So the smaller mirror must be convex (negative focal length) and have a radius of curvature equal to 3.0 m. 34.61: If you move away from the mirror at 2.40 m s, then your image moves away from the mirror at the same speed, but in the opposite direction. Therefore you see the image receding at 4.80 m s, the sum of your speed and that of the image in the mirror. 34.62: a) There are three images formed. 34.63: The minimum length mirror for a woman to see her full height h, is h 2 , as shown in the figure below.
19. s  s  4.00 m 34.64: m  2.25    1.25s  4.00 m  s  3.2 m. So the mirror is s s 7.20 m from the wall. Also: 1 1 2 2 1 1       R  4.43 m. s s R R 3.2 m 7.20 m y  360 s 8.00 m 34.65: a) m    60.0   s   0.133 m is where the filament y 6.00 s 60.0 should be placed. 1 1 2 2 1 1 b)       R  0.261 m. s s R R 0.133 m 8.00 m 1 1 2 1 1 2 34.66:       s   0.0894 m. s s R 13.0 m s   0.180 m  s    0.0894  y   y    1.50 m     0.0103 m.  s  13.0  b) The height of the image is less then 1% of the true height of the car, and is less than the image would appear in a plane mirror at the same location. This gives the illusion that the car is further away then “expected.” 34.67: a) R  0 and s  0, so a real image ( s   0) is produced for virtual object positions between the focal point and vertex of the mirror. So for a 24.0 cm radius mirror, the virtual object positions must be between the vertex and 12.0 cm to the right of the s s mirror. b) The image orientation is erect, since m      0. s s c)
20. 34.68: The derivations of Eqs. (34.6) and (34.7) are identical for convex mirrors, as long as one recalls that R and s  are negative. Consider the diagram below: 1 1 2 R 1 1 2 1 y s We have:    s  f      and m    , since s  s R 2 s s R f y s is not on the outgoing side of the mirror. 1 1 2 1 1 2 34.69: a)       s   46 cm, so the image is virtual. s s R 8.0 cm s  19.4 cm s  46 b) m      5.8, so the image is erect, and its height is: s 8 .0 y   (5.8) y  (5.8)(5.0 mm)  29 mm. c) When the filament is 8 cm from the mirror, there is no place where a real image can be formed. 5  s 2 5 1 2 3 34.70: m  s s   since m  0, s   0,    R  s 2 s 5 2s  s  R 4 3 3  s  R and s    R. 10 4