## Solutions to Physics: Principles with Applications, 5/E, Giancoli ...

Solutions to Physics: Principles with Applications, 5/E, Giancoli. Chapter 23. Page 23 – 1. CHAPTER 23. 1. For a flat mirror the image is as far behind the mirror ...

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

CHAPTER 23 1.

For a flat mirror the image is as far behind the mirror as the object is in front, so the distance from object to image is do + di = 1.5 m + 1.5 m = 3.0 m.

2.

Because the angle of incidence must equal the angle of reflection, we see from the ray diagrams that the ray that reflects to the top of the object must be as far below the horizontal line to the reflection point on the mirror as the top is above the line, regardless of object position.

Mirror

Position 1

3.

We show some of the rays from the tip of the arrow that form the three images. Single reflections form the two side images. Double reflections form the third image. The two reflections have reversed the orientation of the image.

Position 2

I 3

I 2

I 1

Object

4.

The angle of incidence is the angle of reflection. Thus we have tan q = (H – h)/L = h/x; (1.62 m – 0.43 m)/(2.10 m) = (0.43 m)/x, which gives x = 0.76 m = 76 cm.

Mirror

q h

q x  L

5.

f

From the triangle formed by the mirrors and the first reflected ray, we have q + a + f = 180°; 40° + 135° + f = 180°, which gives f = 5°.

q

Page 23 – 1

a

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

6.

Because the rays entering your eye are diverging from the image position behind the mirror, the diameter of the area on the mirror and the diameter of your pupil must subtend the same angle from the image: Dmirror/di = Dpupil/(do + di); Dmirror/(70 cm) = (5.5 mm)/(70 cm + 70 cm), which gives Dmirror = 2.75 mm. Thus the area on the mirror is Amirror = (πDmirror2 = (π(2.75 ´ 10–3 m)2 = 5.9 ´ 10–6 m2.

7.

For the first reflection at A the angle of incidence q1 is the angle of reflection. For the second reflection at B the angle of incidence q2 is the angle of reflection. We can relate these angles to the angle at which the mirrors meet, f, by using the sum of the angles of the triangle ABC: f + (90° – q1) + (90° – q2) = 180°, or f = q1 + q2 . In the same way, for the triangle ABD, we have a + 2q1 + 2q2 = 180°, or a = 180° – 2(q1 + q2) = 180° – 2f. At point D we see that the deflection is b = 180° – a = 180° – (180° – 2f) = 2f.

D q2

f C

q2 q

1

a q1

b

8.

(a) The velocity of the wave, which specifies the direction of the light wave, is in the direction of the ray. If we consider a single reflection, because the angle of incidence is equal to the angle of reflection, the component of the velocity perpendicular to the mirror is reversed, while the component parallel to the mirror is unchanged. When we have three mirrors at right angles, we can choose the orientation of the mirrors for our axes. In each of the three reflections, the component of velocity perpendicular to the mirror will reverse. Thus after three reflections, all three components of the velocity will have been reversed, so the wave will return along the line of the original direction. (b) Assuming the mirrors are large enough, the incident ray will make two reflections only if its direction is parallel to one of the mirrors. In that case, it has only two components, both of which will be reversed, so the ray will be reversed along the line of the original direction.

9.

The rays from the Sun will be parallel, so the image will be at the focal point. The radius is r = 2f = 2(17.0 cm) = 34.0 cm.

10. To produce an image at infinity, the object will be at the focal point: do = f = r/2 = (27.0 cm)/2 = 13.5 cm.

11.  Object  Object  C

Image  Image

Page 23 – 2

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

12. The ball is a convex mirror with a focal length f = r/2 = (– 4.5 cm)/2 = – 2.25 cm. We locate the image from (1/do) + (1/di) = 1/f; [1/(30.0 cm)] + (1/di) = 1/(– 2.25 cm), which gives di = – 2.09 cm. The image is 2.09 cm behind the surface, virtual. The magnification is m = – di/do = – (– 2.09 cm)/(30.0 cm) = + 0.070. Because the magnification is positive, the image is upright. 13. We find the image distance from the magnification: m = hi/ho = – di/do ; + 3 = – di/(1.3 m), which gives di = – 3.9 m. We find the focal length from (1/do) + (1/di) = 1/f; [1/(1.3 m)] + [1/(– 3.9 m)] = 1/f, which gives f = 1.95 m. The radius of the concave mirror is r = 2f = 2(1.95 m) = 3.9 m.

14. We find the image distance from the magnification: m = hi/ho = – di/do ; + 4.5 = – di/(2.20 cm), which gives di = – 9.90 cm. We find the focal length from (1/do) + (1/di) = 1/f; [1/(2.20 cm)] + [1/(– 9.90 cm)] = 1/f, which gives f = 2.83 cm. Because the focal length is positive, the mirror is concave with a radius of r = 2f = 2(2.83 cm) = 5.7 cm.

15. To produce an upright image, we have di < 0. A smaller image means |di | < do, so f < 0, which means the mirror is convex. The focal length of the mirror is f = r/2 = (– 3.2 m)/2 = – 1.6 m. We locate the image from (1/do) + (1/di) = 1/f; [1/(15.0 m)] + (1/di) = 1/(– 1.6 m), which gives di = – 1.45 m. Because the image distance is negative, the image is virtual. We find the image height from the magnification: m = hi/ho = – di/do ; hi/(1.3 m) = – (– 1.45 m)/(15.0 m), which gives hi = 0.13 m.

Page 23 – 3

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

16. (a) We see from the ray diagram that the image is behind the mirror, so it is virtual. We estimate the image distance as – 7 cm. O

(b) If we use a focal length of – 10 cm, we locate the image from (1/do) + (1/di) = 1/f; [1/(20 cm)] + (1/di) = 1/(– 10 cm), which gives di = – 6.7 cm. (c) We find the image size from the magnification: m = hi/ho = – di/do ; hi/(3.00 mm) = – (– 6.7 cm)/(20 cm), which gives hi = 1.0 mm.

17. (a) With di = do , we locate the object from (1/do) + (1/di) = 1/f; (1/do) + (1/do) = 1/f, which gives do = 2f = r. The object should be placed at the center of curvature. (b) Because the image is in front of the mirror, di > 0, it is real. (c) The magnification is m = – di/do = – do/do = – 1. Because the magnification is negative, the image is inverted. (d) As found in part (c), m = – 1.

18. We take the object distance to be ∞, and find the focal length from (1/do) + (1/di) = 1/f; (1/∞) + [1/(– 14.0 cm)] = 1/f, which gives f = – 14.0 cm. Because the focal length is negative, the mirror is convex. The radius is r = 2f = 2(– 14.0 cm) = – 28.0 cm.

19. We find the image distance from (1/do) + (1/di) = 1/f = 2/r, which we can write as di = rdo/(2do – r). The magnification is m = – di/do = – r/(2do – r). If do > r, then (2do – r) > r, so |m | = r/(2do – r) = r/(> r) < 1. If do < r, then (2do – r) < r, so |m | = r/(2do – r) = r/(< r) > 1.

Page 23 – 4

C

Solutions to Physics: Principles with Applications, 5/E, Giancoli

20. From the ray that reflects from the center of the mirror, we have tan q = ho/do = hi/di ; |m | = ho/hi = di/do . Because the image distance on the ray diagram is negative, we get m = – di/do .

h o  d o

21. From the ray diagram, we see that tan q = ho/do = hi/di ; tan a = ho/(do + r) = hi/(r – di ). When we divide the two equations. we get (do + r)/do = (r – di )/di ; 1 + (r/do) = (r/di) – 1, or (r/do) – (r /di) = – 2; (1/do) – (1/di) = – 2/r = – 1/f, with f = r/2. From the ray diagram, we see that di < 0. If we consider f to be negative, we have (1/do) + (1/di) = 1/f.

h o O

22. We find the image distance from the magnification: m = hi/ho = – di/do ; + 0.80 = – di/(2.2 m), which gives di = – 1.76 m. We find the focal length from (1/do) + (1/di) = 1/f; [1/(2.2 m)] + [1/(– 1.76 m)] = 1/f, which gives f =

d o

Chapter 23

q q

h i

d i

q q

h i a

d i

C

C

– 8.8 m.

23. (a) As the radius of a sphere gets larger, the surface is flatter. The plane mirror can be considered a spherical mirror with an infinite radius, and thus f = ∞. (b) When we use the mirror equation, we get (1/do) + (1/di) = 1/f; (1/do) + (1/di) = 1/∞ = 0, or di = – do . (c) For the magnification, we have m = – di/do = – (– do)/do = + 1. (d) Yes, these are consistent with the discussion on plane mirrors.

24. (a) To produce a smaller image located behind the surface of the mirror requires a (b) We find the image distance from the magnification: m = hi/ho = – di/do ; (3.5 cm)/(4.5 cm) = – di/(28 cm), which gives di = – 21.8 cm. As expected, di < 0. The image is located 22 cm behind the surface. (c) We find the focal length from (1/do) + (1/di) = 1/f; [1/(28 cm)] + [1/(– 21.8 cm)] = 1/f, which gives f = – 98 cm. (d) The radius of curvature is  Page 23 – 5

convex mirror.

Solutions to Physics: Principles with Applications, 5/E, Giancoli

r = 2f = 2(– 98 cm) =

– 196 cm.

Page 23 – 6

Chapter 23

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

25. (a) To produce a larger image requires a concave mirror. (b) The image will be erect and virtual. (c) We find the image distance from the magnification: m = hi/ho = – di/do ; 1.3 = – di/(20.0 cm), which gives di = – 26.0 cm. We find the focal length from (1/do) + (1/di) = 1/f; [1/(20.0 cm)] + [1/(– 26.0 cm)] = 1/f, which gives f = 86.7 cm. The radius of curvature is r = 2f = 2(86.7 cm) = 173 cm.

26. (a) The speed in crown glass is v = c/n = (3.00 ´ 108 m/s)/(1.52) = (b) The speed in Lucite is v = c/n = (3.00 ´ 108 m/s)/(1.51) =

1.97 ´ 108 m/s. 1.99 ´ 108 m/s.

27. We find the index of refraction from v = c/n; 2.29 ´ 108 m/s = (3.00 ´ 108 m/s)/n, which gives n =

28. We find the index of refraction from v = c/n; 0.85vwater = 0.85c/1.33 = c/n, which gives n =

1.31.

1.56.

29. The % uncertainty in the index is ± [(0.000010)/(1.00030)](100) = ± 9.997 ´ 10–4 %, which will be the % uncertainty in the speed. Thus we have v = c/n = [(2.997925 ´ 108 m/s)/(1.00030)] ± 9.997 ´ 10–4 % = 2.997026 ´ 108 m/s ± 9.997 ´ 10–4 % = (2.997026 ± 0.000030) ´ 108 m/s. 30. We find the angle of refraction in the glass from n1 sin q1 = n2 sin q2 ; (1.00) sin 63° = (1.50) sin q2 , which gives q2 =

36°.

31. We find the angle of refraction in the water from n1 sin q1 = n2 sin q2 ; (1.33) sin 42.5° = (1.00) sin q2 , which gives q2 =

64.0°.

32. We find the incident angle in the water from n1 sin q1 = n2 sin q2 ; (1.33) sin q1 = (1.00) sin 60°, which gives q1 =

Page 23 – 7

41°.

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

33. We find the incident angle in the air from n1 sin q1 = n2 sin q2 ; (1.00) sin q1 = (1.33) sin 21.0°, which gives q1 = 28.5°. Thus the angle above the horizon is 90.0° – q1 = 90.0° – 28.5° = 61.5°.

34. For the refraction at the first surface, we have nair sin q1 = n sin q2 ; (1.00) sin 45.0° = (1.52) sin q2 , which gives q2 = 27.7°. We find the angle of incidence at the second surface from (90° – q2) + (90° – q3) + A = 180°, which gives q3 = A – q2 = 60° – 27.7° = 32.3°. For the refraction at the second surface, we have n sin q3 = nair sin q4 ; (1.52) sin 32.3° = (1.00) sin q4 , which gives

q4 =

A

q 1 q 2

q 4

q 3

54.3° from the normal.

35. We find the angle of incidence from the distances: tan q1 = L1/h1 = (2.7 m)/(1.3 m) = 2.076, so q1 = 64.3°. For the refraction from air into water, we have nair sin q1 = nwater sin q2 ; (1.00) sin 64.3° = (1.33) sin q2 , which gives q2 = 42.6°. We find the horizontal distance from the edge of the pool from L = L1 + L2 = L1 + h2 tan q2 = 2.7 m + (2.1 m) tan 42.6° = 4.6 m.

h 2

36. The angle of reflection is equal to the angle of incidence: qrefl = q1 = 2q2 . For the refraction we have nair sin q1 = nglass sin q2 ; (1.00) sin 2q2 = (1.52) sin q2 . We use a trigonometric identity for the left­hand side: sin 2q2 = 2 sin q2 cos q2 = (1.52) sin q2 , or cos q2 = 0.760, so q2 = 40.5°. Thus the angle of incidence is q1 = 2q2 =

81.0°.

Page 23 – 8

n air

q 1

h 1

q 2

L 1  L 2

n w ater

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

37. (a) We find the angle in the glass from the refraction at the air–glass surface: n1 sin q1 = n2 sin q2 ; n 2  n 3 n 1  (1.00) sin 43.5° = (1.52) sin q2 , which gives q2 = 26.9°. q 3  (b) Because the surfaces are parallel, the refraction angle from the first surface is the incident angle at the second q 2 q 2 surface. We find the angle in the water from the refraction at the glass–water surface: q 1 n2 sin q2 = n3 sin q3 ; (1.52) sin 26.9° = (1.33) sin q3 , which gives q3 = 31.2°. (c) If there were no glass, we would have n1 sin q1 = n3 sin q3¢; (1.00) sin 43.5° = (1.33) sin q3¢, which gives q3¢ = 31.2°. Note that, because the sides are parallel, q3 is independent of the presence of the glass.

38. Because the surfaces are parallel, the angle of refraction from the first surface is the angle of incidence at the second, Thus for the refractions, we have n1 sin q1 = n2 sin q2 ;

n 1 q 3

n2 sin q2 = n3 sin q3 . When we add the two equations, we get n1 sin q1 = n1 sin q3 , which gives q3 = q1 . Because the ray emerges in the same index of refraction, it is undeviated.

39. Because the glass surfaces are parallel, the exit beam will be traveling in the same direction as the original beam. We find the angle inside the glass from nair sin q = n sin f . If the angles are small, we use cos f » 1, and sin f » f, where f is in radians. (1.00) q = nf, or f = q/n. We find the distance along the ray in the glass from L = t/cos f » t. We find the perpendicular displacement from the original direction from d = L sin (q – f) » t(q – f) = t[q – (q/n)] = tq(n – 1)/n.

n 2

n 1

q 2

q 2

q 1

n

q

q–f

f  f

L

n air  = 1.00

n air  = 1.00

40. When the light in the material with a higher index is incident at the critical angle, the refracted angle is 90°: nLucite sin q1 = nwater sin q2 ; (1.51) sin q1 = (1.33) sin 90°, which gives q1 = 61.7°. Because Lucite has the higher index, the light must start in Lucite. 41. When the light in the liquid is incident at the critical angle, the refracted angle is 90°: nliquid sin q1 = nair sin q2 ; nliquid sin 44.7° = (1.00) sin 90°, which gives nliquid = 1.42. Page 23 – 9

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Page 23 – 10

Chapter 23

Solutions to Physics: Principles with Applications, 5/E, Giancoli

42. We find the critical angle for light leaving the water: n sin q1 = sin q2 ; (1.33) sin qC = sin 90°, which gives qC = 48.8°. If the light is incident at a greater angle than this, it will totally reflect. We see from the diagram that R > H tan qC = (62.0 cm) tan 48.8° = 70.7 cm.

Chapter 23

air

n H

q 1  q 1

43. We find the distance L between reflections from d = L sin q; 10–4 m = L sin 15°, which gives L = 3.9 ´ 10–4 m.

44. We find the angle of incidence from the distances: tan q1 = L/h = (7.0 cm)/(8.0 cm) = 0.875, so q1 = 41.2°. For the maximum incident angle for the refraction from liquid into air, we have nliquid sin q1 = nair sin q2 ; nliquid sin q1max = (1.00) sin 90°, which gives sin q1max = 1/nliquid . Thus we have sin q1 ≥ sin q1max = 1/nliquid ; sin 41.2° = 0.659 ≥ 1/nliquid , or nliquid ≥ 1.5.

45. For the refraction at the first surface, we have nair sin q1 = n sin q2 ; A (1.00) sin q1 = n sin q2 , which gives sin q2 = sin q1/n. We find the angle of incidence at the second surface from (90° – q2) + (90° – q3) + A = 180°, which gives q 1 q3 = A – q2 = 75° – q2 . q 2 For the refraction at the second surface, we have n sin q3 = nair sin q4 = (1.00) sin q4. n  The maximum value of q4 before internal reflection takes place at the second surface is 90°. Thus for internal reflection to occur, we have n sin q3 = n sin (A – q2) ≥ 1. When we expand the left­hand side, we get n(sin A cos q2 – cos A sin q2) ≥ 1. If we use the result from the first surface to eliminate n, we get sin q1 (sin A cos q2 – cos A sin q2)/(sin q2) = sin q1(sin A/tan q2 – cos A) ≥ 1, or 1/tan q2 ≥ [(1/sin q1) + cos A]/sin A = [(1/sin 45°) + cos 75°]/sin 75° = 1.732, which gives tan q2 ≤ 0.577, so q2 ≤ 30°. From the result for the first surface, we have nmin = sin q1/sin q2max = sin 45°/sin 30° = 1.414, so n ≥ 1.414.

Page 23 – 11

q 3

q 4

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

46. For the refraction at the side of the rod, we have n2 sin g = n1 sin d. d  n 1  The minimum angle for total reflection gmin occurs when d = 90°: n2 sin gmin = (1.00)(1) = 1, or sin gmin = 1/n2 . 90°  g We find the maximum angle of refraction at the end of the b rod from bmax = 90° – gmin . a n 2  Because the sine function increases with angle, for the refraction at the end of the rod, we have n1 sin amax = n2 sin bmax ; (1.00) sin amax = n2 sin (90° – gmin) = n2 cos gmin . If we want total internal reflection to occur for any incident angle at the end of the fiber, the maximum value of a is 90°, so n2 cos gmin = 1. When we divide this by the result for the refraction at the side, we get tan gmin = 1, or gmin = 45°. Thus we have n2 ≥ 1/sin gmin = 1/sin 45° = 1.414.

47. (a) The ray enters normal to the first surface, so there is no deviation there. The angle of incidence is 45° at the second surface. When there is air outside the surface, we have n1 sin q1 = n2 sin q2 ; n1 sin 45° = (1.00) sin q2 . For total internal reflection to occur, sin q2 ≥ 1, so we have n1 ≥ 1/sin 45° = 1.414. (b) When there is water outside the surface, we have n1 sin q1 = n2 sin q2 ; (1.50) sin 45° = (1.33) sin q2 , which gives sin q2 = 0.80.

n 2  45°

n 1 45°

Because sin q2 < 1, the prism will not be totally reflecting. (c) For total reflection when there is water outside the surface, we have n1 sin q1 = n2 sin q2 ; n1 sin 45° = (1.33) sin q2 . For total internal reflection to occur, sin q2 ≥ 1, so we have n1 ≥ 1.33/sin 45° = 1.88.

48. (a) From the ray diagram, the object distance is about six focal lengths, or

F I  O

(b) We find the object distance from Page 23 – 12

390 mm.

90°

Solutions to Physics: Principles with Applications, 5/E, Giancoli

(1/do) + (1/di) = 1/f; (1/do) + (1/78.0 mm) = 1/65.0 mm, which gives do = 390 mm =

Page 23 – 13

Chapter 23

39.0 cm.

Solutions to Physics: Principles with Applications, 5/E, Giancoli

49. (a) To form a real image from parallel rays requires a (b) We find the power of the lens from (1/do) + (1/di) = 1/f = P, when f is in meters; (1/∞) + (1/0.185 m) = P = 5.41 D.

Chapter 23

converging lens.

50. To form a real image from a real object requires a converging lens. We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/225 cm) + (1/48.3 cm) = 1/f, which gives f = + 39.8 cm. Because di > 0, the image is real.

51. (a) The power of the lens is P = 1/f = 1/0.295 m = 3.39 D, converging. (b) We find the focal length of the lens from P = 1/f; – 6.25 D = 1/f, which gives f = – 0.160 m = – 16.0 cm, diverging.

52. (a) We locate the image from (1/do) + (1/di) = 1/f; (1/18 cm) + (1/di) = 1/24 cm, which gives di = – 72 cm. The negative sign means the image is 72 cm behind the lens (virtual). (b) We find the magnification from m = – di/do = – (– 72 cm)/(18 cm) = + 4.0.

53. (a) Because the Sun is very far away, the image will be at the focal point. We find the size of the image from m = hi/ho = – di/do ; hi/(1.4 ´ 106 km) = – (28 mm)/(1.5 ´ 108 km), which gives hi = – 0.26 mm. (b) For a 50 mm lens, we have hi/(1.4 ´ 106 km) = – (50 mm)/(1.5 ´ 108 km), which gives hi = – 0.47 mm. (c) For a 200 mm lens, we have hi/(1.4 ´ 106 km) = – (200 mm)/(1.5 ´ 108 km), which gives hi = – 1.9 mm. (d) The 28­mm lens simulates being farther away, so it would be a wide­angle lens. The 200 mm lens simulates being closer, so it would be a telephoto lens.

54. (a) We find the image distance from (1/do) + (1/di) = 1/f; (1/10.0 ´ 103 mm) + (1/di) = 1/80 mm, which gives di = 81 mm. (b) For an object distance of 3.0 m, we have (1/3.0 ´ 103 mm) + (1/di) = 1/80 mm, which gives di = 82 mm. (c) For an object distance of 1.0 m, we have (1/1.0 ´ 103 mm) + (1/di) = 1/80 mm, which gives di = 87 mm. (d) We find the smallest object distance from (1/domin) + (1/120 mm) = 1/80 mm, which gives di = 240 mm = 24 cm.

Page 23 – 14

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

55. We find the image distance from (1/do) + (1/di) = 1/f; (1/0.140 m) + (1/di) = – 6.0 D, which gives di = – 0.076 m = – 7.6 cm (virtual image behind the lens). We find the height of the image from m = hi/ho = – di/do ; hi/(1.0 mm) = – (– 7.6 cm)/(14.0 cm), which gives hi = 0.54 mm (upright).

56. (a) We see that the image is behind the lens, so it is virtual. (b) From the ray diagram we see that we need a converging lens. (c) We find the image distance from the magnification: m = – di/do ; + 2.5 = – di/(8.0 cm), which gives di = – 20 cm. We find power of the lens from (1/do) + (1/di) = 1/f = P, when f is in meters; (1/0.080 m) + [1/(– 0.20 m)] = P = 7.5 D.

F

57. We can relate the image and object distance from the magnification: m = – di/do , or do = – di/m. We use this in the lens equation: (1/do) + (1/di) = 1/f; – (m/di) + (1/di) = 1/f, which gives di = (1 – m)f. (a) If the image is real, di > 0. With f > 0, we see that m < 1; thus m = – 2.00. The image distance is di = [1 – (– 2.00)](50.0 mm) = 150 mm. The object distance is do = – di/m = – (150 mm)/(– 2.00) = 75.0 mm. (b) If the image is virtual, di < 0. With f > 0, we see that m > 1; thus m = + 2.00. The image distance is di = [1 – (+ 2.00)](50.0 mm) = – 50 mm. The object distance is do = – di/m = – (– 50 mm)/(+ 2.00) = 25.0 mm.

Page 23 – 15

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

58. We can relate the image and object distance from the magnification: m = – di/do , or do = – di/m. We use this in the lens equation: (1/do) + (1/di) = 1/f; – (m/di) + (1/di) = 1/f, which gives di = (1 – m)f. (a) If the image is real, di > 0. With f < 0, we see that m > 1; thus m = + 2.00. The image distance is di = [1 – (+ 2.00)](– 50.0 mm) = 50.0 mm. The object distance is do = – di/m = – (50.0 mm)/(+ 2.00) = – 25.0 mm. The negative sign means the object is beyond the lens, so it would have to be an object formed by a preceding optical device. (b) If the image is virtual, di < 0. With f < 0, we see that m < 1; thus m = – 2.00. The image distance is di = [1 – (– 2.00)](– 50.0 mm) = – 150 mm. The object distance is do = – di/m = – (– 150 mm)/(– 2.00) = – 75.0 mm. The negative sign means the object is beyond the lens, so it would have to be an object formed by a preceding optical device.

59. (a) We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/31.5 cm) + [1/(– 8.20 cm)] = 1/f, which gives f = The image is in front of the lens, so it is virtual. (b) We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/31.5 cm) + [1/(– 38.0 cm)] = 1/f, which gives f = The image is in front of the lens, so it is virtual.

– 11.1 cm (diverging).

+ 184 cm (converging).

60. (a) We find the image distance from (1/do) + (1/di) = 1/f; (1/1.20 ´ 103 mm) + (1/di) = 1/135 mm, which gives di = 152 mm (real, behind the lens). We find the height of the image from m = hi/ho = – di/do ; hi/(2.20 cm) = – (152 mm)/(1.20 ´ 103 mm), which gives hi = – 0.279 cm (inverted). (b) We find the image distance from (1/do) + (1/di) = 1/f; (1/1.20 ´ 103 mm) + (1/di) = 1/(– 135 mm), which gives di = – 121 mm (virtual, in front of the lens). We find the height of the image from m = hi/ho = – di/do ; hi/(2.20 cm) = – (– 121 mm)/(1.20 ´ 103 mm), which gives hi = + 0.222 cm (upright).

Page 23 – 16

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

61. The sum of the object and image distances must be the Screen  distance between object and screen: do + di = L = 60 cm. For the lens we have F  I  (1/do) + (1/di) = 1/f; (1/do) + [1/(60 cm – do)] = 1/(15 cm), O  F  which gives a quadratic equation: do2 – (60 cm)do + 900 cm2 = 0, or (do – 30 cm)2 = 0. Thus the two answer are the same: 30 cm, so the lens should be placed midway between the object and screen. Note that in general the screen must be at least 4f from the object for an image to be formed on the screen.

62. For a real object and image, both do and di must be positive, so the magnification will be negative: m = – di/do ; – 2.75 = – di/do , or di = 2.75do . We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/2.75do) = 1/(+ 75 mm), which gives do = 102 cm. The image distance is di = 2.75do = 2.75(102 cm) = 281 cm. The distance between object and image is L = do + di = 102 cm + 281 cm = 382 cm.

63. (a) For the thin lens we have O  (1/do) + (1/di) = 1/f; [1/(f + x)] + [1/(f + x¢)] = 1/f, F  which can be written as 2f + x + x¢ = (f + x)(f + x¢)/f x  = f + (x + x¢) + (xx¢/f ), or xx¢ = f 2. (b) For the standard form we have (1/do) + (1/di) = 1/f; (1/45.0 cm) + (1/di) = 1/32.0 cm, which gives di = + 110.8 cm. (c) For the Newtonian form we have xx¢ = f 2; (45.0 cm – 32.0 cm)x¢ = (32.0 cm)2, which gives x¢ = 78.7 cm. Thus the distance from the lens is di = x¢ + f = 78.7 cm + 32.0 cm = 110.8 cm.

x¢ F  I

64. We find the image formed by the refraction of the first lens: (1/do1) + (1/di1) = 1/f1 ; (1/35.0 cm) + (1/di1) = 1/27.0 cm, which gives di1 = + 118.1 cm. This image is the object for the second lens. Because it is beyond the second lens, it has a negative object distance: do2 = 16.5 cm – 118.1 cm = – 101.5 cm. We find the image formed by the refraction of the second lens: (1/do2) + (1/di2) = 1/f2 ; [1/(– 101.5 cm)] + (1/di2) = 1/27.0 cm, which gives di2 = + 21.3 cm. Thus the final image is real, 21.3 cm beyond second lens. The total magnification is the product of the magnifications for the two lenses: m = m1m2 = (– di1/do1)(– di2/do2) = di1di2/do1do2  Page 23 – 17

Solutions to Physics: Principles with Applications, 5/E, Giancoli

= (+ 118.1 cm)(+ 21.3 cm)/(+ 35.0 cm)(– 101.5 cm) =

Page 23 – 18

Chapter 23

– 0.708 (inverted).

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

65. The image of an infinite object formed by the refraction of the first lens will be at the focal point: di1 = f1 = + 20.0 cm. This image is the object for the second lens. Because it is beyond the second lens, it has a negative object distance: do2 = 14.0 cm – 20.0 cm = – 6.0 cm. We find the image formed by the refraction of the second lens: (1/do2) + (1/di2) = 1/f2 ; [1/(– 6.0 cm)] + (1/di2) = 1/(– 31.5 cm), which gives di2 = + 7.4 cm. Thus the final image is real 7.4 cm beyond second lens. 66. We see from the ray diagram that the image from the first lens will be a virtual image at its focal point. This is a real object for the second lens, and must be at the focal point of the second lens. If L is the separation of the lenses, the focal length of the first lens is f1 = L – f2 = 21.0 cm – 31.0 cm = – 10.0 cm.

f 1   0

f 1  f 2

67. We find the focal length by finding the image distance for an object very far away. For the first converging lens, we have (1/do1) + (1/di1) = 1/fC ; (1/∞) + (1/di1) = 1/fC , or, as expected, di1 = fC . The first image is the object for the second lens. If the first image is real, the second object distance is negative: do2 = – di1 = – fC . For the second diverging lens, we have (1/do2) + (1/di2) = 1/fD ; [1/(– fC)] + (1/di2) = 1/fD . Because the second image must be at the focal point of the combination, we have (– 1/fC) + (1/fT ) = 1/fD , which gives 1/fD = (1/fT) – (1/fC).

68. We find the focal length of the lens from 1/f = (n – 1)[(1/R1) + (1/R2)] = (1.52 – 1){[1/(– 31.2 cm)] + [1/(– 23.8 cm)]}, which gives 69. We find the index from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)]; 1/28.9 cm = (n – 1)[(1/31.0 cm) + (1/31.0 cm)], which gives n =

70. When the surfaces are reversed, we get 1/f = (n – 1)[(1/R1) + (1/R2)] = (1.51 – 1){[1/(– 18.4 cm)] + (1/∞)}, which gives f = – 36.0 cm, which is the result from Example 23–17.

Page 23 – 19

f = – 26.0 cm.

1.54.

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

71. We find the radius from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)]; 1/28.5 cm = (1.46 – 1)[(1/∞) + (1/R2)], which gives R2 =

13.1 cm.

72. We find the radius from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)]; 1/(– 25.4 cm) = (1.50 – 1)[(1/∞) + (1/R2)], which gives R2 = The negative sign indicates concave. 73. We find the radius from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)]; + 1.50 D = (1.56 – 1)[(1/20.0 cm) + (1/R2)], which gives R2 =

– 12.7 cm.

– 43.1 cm (concave).

74. We find the focal length of the lens from 1/f = (n – 1)[(1/R1) + (1/R2)] = (1.56 – 1){[1/(– 21.0 cm)] + [1/(+18.5 cm)]}, which gives f = 277.5 cm. We find the image distance from (1/do) + (1/di) = 1/f; (1/100 cm) + (1/di) = 1/277.5 cm, which gives di = – 156 cm (in front of the lens). The magnification is m = – di/do = – (– 156 cm)/(100 cm) = + 1.56 (upright).

75. The refraction equations are all based on n1 sin q1 = n2 sin q2 . The lensmaker’s equation is derived assuming air (n = 1.00) on the left­hand side. If we have some other material with n different from 1.00, we can make the equation equivalent to this by using an effective index: sin q1 = (n2/n1) sin q2 = neff sin q2 , where neff = n2/n1 . Thus we have Pair = 1/fair = (nglass – 1)[(1/R1) + (1/R2)]; Pwater = 1/fwater = [(nglass/nwater) – 1][(1/R1) + (1/R2)]. If we divide the two equations, we get Pwater/Pair = [(nglass/nwater) – 1]/(nglass – 1); Pwater/(+ 5.2 D) = [(1.50/1.33) – 1]/(1.50 – 1), which gives Pwater = + 1.3 D.

Page 23 – 20

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

76. For a plane mirror each image is as far behind the mirror as the object is in front. Each reflection produces a front­to­back reversal. We show the three images and the two intermediate images that are not seen.  d

I 3

I 2

I 1

I 1¢

I 2¢

(a) The first image is from a single reflection, so it is d1 = 2D = 2(1.5 m) = 3.0 m away. The second image is from two reflections, so it is d2 = L + d + D = 2.0 m + 0.5 m + 1.5 m = 4.0 m away. The third image is from three reflections, so it is d3 = 2L + D + D = 2(2.0 m) + 1.5 m + 1.5 m = 7.0 m away. (b) We see from the diagram that the first image is facing toward you; the second image is facing away from you; the third image is facing toward you. 77. We find the angle of incidence for the refraction from water into air: nwater sin q1 = nair sin q2 ;

q 2

(1.33) sin q1 = (1.00) sin (90° – 14°),

which gives q1 = 47°. We find the depth of the pool from tan q1 = x/h; tan 47° = (5.50 m)/h =, which gives h =

n air  n w ater

q 1

x

5.2 m.

78. At the critical angle, the refracted angle is 90°. For the refraction from plastic to air, we have nplastic sin qplastic = nair sin qair ; nplastic sin 37.3° = (1.00) sin 90°, which gives nplastic = 1.65. For the refraction from plastic to water, we have nplastic sin qplastic¢ = nwater sin qwater ; (1.65) sin qplastic¢ = (1.33) sin 90°, which gives qplastic¢ =

53.7°.

79. We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/7.50 ´ 103 mm) = 1/100 mm, which gives do = 101 mm = We find the size of the image from m = hi/ho = – di/do ; hi/(0.036 m) = – (7.50 m)/(0.101 m), which gives hi = – 2.7 m.

Page 23 – 21

0.101 m.

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

80.

ª 30°  30°

30°

30°

ª

I 1

I 4

30°  30°  30°

ª

ª I 2

ª

I 3

I 5

ª Object 81. We find the object distance from the required magnification (which is negative for a real object and a real image): m = hi/ho = – di/do ; – (2.7 ´ 103 mm)/(36 mm) = – (9.00 m)/do , which gives do = 0.120 m. We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/0.120 m) + (1/9.00 m) = 1/f, which gives f = 0.118 m = + 12 cm.

82. We get an expression for the image distance from the lens equation: (1/do) + (1/di) = 1/f; 1/di = (1/f ) – (1/do), or di = fdo/(do – f ). The magnification is m = – di/do = – f/(do – f ). If the lens is converging, f > 0. For a real object, do > 0. When do > f, we have (do – f ) > 0, so all factors in the expressions for di and m are positive; thus di > 0 (real), and m < 0 (inverted). When do < f, we have (do – f ) < 0, so the denominator in the expressions for di and m are negative; thus di < 0 (virtual), and m > 0 (upright). For an object beyond the lens, do < 0. When – do > f, we have (do – f ) < 0, so both numerator and denominator in the expression for di are negative; thus di > 0, so the image is real. The numerator in the expression for m is negative; thus m > 0, so the image is upright. When 0 < – do < f, we have (do – f ) < 0, so we get the same result: real and upright.

Page 23 – 22

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

83. We get an expression for the image distance from the lens equation: (1/do) + (1/di) = 1/f; 1/di = (1/f) – (1/do), or di = fdo/(do – f ). If the lens is diverging, f < 0. If we write f = – | f |, we get di = – | f |do/(do + | f |). For a real object, do > 0. All factors in the expression for di are positive, thus di < 0, so the image is always virtual. We can have a real image, di > 0, if do < 0, and | do || < | f |, so the denominator is still positive. Thus to have a real image from a diverging lens, the condition is 0 < – do < – f.

84. For the refraction at the second surface, we have n sin q3 = nair sin q4; (1.52) sin q3 = (1.00) sin q4.

A

The maximum value of q4 before internal reflection takes place at the second surface is 90°. Thus for internal reflection not to occur, we have (1.52) sin q3 ≤ 1.00; sin q3 ≤ 0.658, so q3 ≤ 41.1°. We find the refraction angle at the second surface from (90° – q2) + (90° – q3) + A = 180°, which gives q2 = A – q3 = 72° – q3 . Thus q2 ≥ 72° – 41.1° = 30.9°. For the refraction at the first surface, we have nair sin q1 = n sin q2 ; (1.00) sin q1 = (1.52) sin q2 ; which gives sin q1 = (1.52) sin q2 . For the limiting condition, we have sin q1 ≥ (1.52) sin 30.9° = 0.781, so q1 ≥ 51.3°.

q 1 q 2

q 3

q 4

85. (a)  f 1

f 2  F 1

I 1

F 1

F 2  F 2

I 2

(b) We see that the mage is real and upright, and estimate that it is 20 cm beyond the second lens. We find the image formed by the refraction of the first lens: (1/do1) + (1/di1) = 1/f1 ; (1/30 cm) + (1/di1) = 1/15 cm, which gives di1 = + 30 cm. This image is the object for the second lens. Because it is in front of the second lens, it is a real object, with an object distance of do2 = 50 cm – 30 cm = 20 cm. We find the image formed by the refraction of the second lens: (1/do2) + (1/di2) = 1/f2 ; (1/20 cm) + (1/di2) = 1/10 cm, which gives di2 = + 20 cm. Thus the final image is real, 20 cm beyond second lens. The total magnification is the product of the magnifications for the two lenses: m = m1m2 = (– di1/do1)(– di2/do2) = di1di2/do1do2 = (+ 30 cm)(+ 30 cm)/(+ 20 cm)(+ 20 cm) = + 1.0. Page 23 – 23

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Thus the final image is

upright, same size as object.

Page 23 – 24

Chapter 23

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 23

86. (a) We find the focal length by finding the image distance for an object very far away. For the first lens, we have (1/do1) + (1/di1) = 1/f1 ; (1/∞) + (1/di1) = 1/f1 , or, as expected, di1 = f1 . The first image is the object for the second lens. If the first image is real, the second object is virtual: do2 = – di1 = – f1 . For the second lens, we have (1/do2) + (1/di2) = 1/f2 ; [1/(– f1)] + (1/di2) = 1/f2 . Because the second image must be at the focal point of the combination, we have (– 1/f1) + (1/fT ) = 1/f2 , which gives 1/fT = (1/f1) + (1/f2). When we solve for fT , we get fT = f1f2/(f1 + f2). (b) If we use the intermediate result 1/fT = (1/f1) + (1/f2), we see that P = P1 + P1 .

87. (a) We use the lens equation with do + di = dT : (1/do) + (1/di) = 1/f ; (1/do) + [1/(dT – do) = 1/f . When we rearrange this, we get a quadratic equation for do : do2 – dTdo + dT f = 0, which has the solution do = ! [dT ± (dT2 – 4dT f )1/2]. If dT > 4f, we see that the term inside the square root dT2 – 4dT f > 0, and (dT2 – 4dT f )1/2 < dT , so we get two real, positive solutions for do . (b) If dT < 4f, we see that the term inside the square root dT2 – 4dT f < 0 , so there are no real solutions for do . (c) When there are two solutions, the distance between them is ∆d = do1 – do2 = ! [dT + (dT2 – 4dT f )1/2] – ! [dT – (dT2 – 4dT f )1/2] = (dT2 – 4dT f )1/2. The image positions are given by di = dT – do = ! [dT — (dT2 – 4dT f )1/2]. The ratio of image sizes is the ratio of magnifications: m = m2/m1 = (di2/do2)/(di1/do1) = (di2/do2)(do1/di1) = {! [dT + (dT2 – 4dT f )1/2]/! [dT – (dT2 – 4dT f )1/2]}2 =

{[dT + (dT2 – 4dT f )1/2]/[dT – (dT2 – 4dT f )1/2]}2.

88. We find the focal length by finding the image distance for an object very far away. For the first lens, we have (1/do1) + (1/di1) = 1/f1 ; (1/∞) + (1/di1) = 1/(10.0 cm), or, as expected, di1 = 10.0 cm. The first image is the object for the second lens. The first image is real, so the second object is virtual: do2 = – di1 = – 10.0 cm. For the second lens, we have (1/do2) + (1/di2) = 1/f2 ; [1/(– 10.0 cm)] + (1/di2) = 1/(– 20.0 cm), which gives di2 = + 20.0 cm. Because the second image must be at the focal point of the combination, we have f = + 20.0 cm (converging).

Page 23 – 25

Solutions to Physics: Principles with Applications, 5/E, Giancoli

89. For both mirrors the image is virtual (behind the mirror), so the image distances are negative. The image distance for the plane mirror is di1 = – do , and the image is upright and the same size. Because the angle subtended by the image is small, it is q1 = h1/(do – di1) = h/2do . The image distance for the convex mirror is di2 , and the image is upright and smaller. Because the angle subtended by the image is small, it is q2 = h2/(do – di2) = q1/2 = h/4do , or h2/h = (do – di2)/4do . We find the image distance from the magnification: m2 = h2/h = – di2/do = (do – di2)/4do , which gives di2 = – do/3. We find the required focal length of the convex mirror from (1/do) + (1/di2) = 1/f2 ; (1/do) + [1/(– do/3)] = 1/f2 , which gives f2 = – do/2. Thus the radius of curvature is R2 = 2f2 = 2(– do/2) = – do = – 3.25 m.

Chapter 23

q 1

h 1

h O

I 1

d o  q 2

h 2 I 2

90. The two students chose different signs for the magnification, i. e., one upright and one inverted. The focal length of the concave mirror is f = R/2 = (40 cm)/2 = 20 cm. We relate the object and image distances from the magnification: m = – di/do ; ± 3 = – di/do , which gives di = — 3do . When we use this in the mirror equation, we get (1/do) + (1/di) = 1/f ; (1/do) + [1/(— 3do)] = 1/f , which gives do = 2f/3, 4f/3 = 13.3 cm, 26.7 cm. The image distances are = – 40 cm (virtual, upright), and + 80 cm (real, inverted). 91. The refraction equations for the two surfaces are all based on n¢ sin q1 = n sin q2 , and n sin q3 = n¢ sin q4 . Eq. 23–10 was derived based on having air (n¢ = 1) outside the lens. We can make these equivalent to having air outside the lens, if we write them as sin q1 = (n/n¢) sin q2 , and (n/n¢) sin q3 = sin q4 . Thus we see that we have an effective index, neff = n/n¢, to use in Eq. 23–10: 1/f ¢ = [(n/n¢) – 1][(1/R1) + (1/R2)]. The focal length in air is 1/f = (n – 1)[(1/R1) + (1/R2)]; so we have 1/f ¢ = [(n/n¢) – 1]/f(n – 1). Eq. 23–8 is derived using the focal points. Thus, if we use the above focal length, the derivation is the same, so we have the same equation: (1/do) + (1/di) = 1/f ¢, where 1/f ¢ = [(n/n¢) – 1]/f(n – 1). Eq. 23–9 is derived by comparing heights and distances, so it is unchanged: m = hi/ho = – di/do .

Page 23 – 26