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

Chapter 25

CHAPTER 25 1.

From the definition of the f-stop, we have f-stop = f/D; 1.4 = (55 mm)/D1.4 , which gives D1.4 = 39 mm; 22 = (55 mm)/D22 , which gives D22 = 2.5 mm. Thus the range of diameters is 2.5 mm ? D ? 39 mm.

2.

We find the f-number from f-stop = f/D = (14 cm)/(6.0 cm) =

f/2.3.

3.

The exposure is proportional to the area and the time: Exposure ∝ At ∝ D2t ∝ t/( f-stop)2. Because we want the exposure to be the same, we have t1/( f-stop1)2 = t2/( f-stop2 )2; (1/60) s. [(1/250) s]/(5.6)2 = t2/(11)2, which gives t2 =

4.

The exposure is proportional to the area and the time: Exposure ∝ At ∝ D2t ∝ t/( f-stop)2. Because we want the exposure to be the same, we have t1/( f-stop1)2 = t2/( f-stop2 )2; [(1/125) s]/(11)2 = [(1/1000) s]/( f-stop2)2, which gives f-stop2 =

f/4.

5.

From the similar triangles on the ray diagram, we see that H/L1 = h/L2 ; (2.0 cm)/(100 cm) = h/(7.0 cm), which gives h = 0.014 cm = 1.4 mm. h H To find the radius of each spot of the image, we consider the light going through a slit and find the distance from the L1 L2 central bright spot to the first dark spot. For destructive interference, the path-length difference to the first dark spot is d sin θ = !λ; We find the locations on the screen from y = L2 tan θ. For small angles, we have sin θ ≈ tan θ, which gives y = L2(!λ/d) = !L2λ/d = (7.0 × 10–2 m)(550 × 10–9 m)/(1.0 × 10–3 m) = 1.9 × 10–5 m = 0.019 mm. Thus the diameter of the image spot is about 0.04 mm, which is much smaller than the separation of the spots, so they are easily resolvable.

6.

We find the effective f-number for the pinhole: f-stop2 = f/D = (70 mm)/(1.0 mm) = f/70. The exposure is proportional to the area and the time: Exposure ∝ At ∝ D2t ∝ t/( f-stop)2. Because we want the exposure to be the same, we have t1/( f-stop1)2 = t2/( f-stop2 )2; (1/6) s. [(1/250) s]/(11)2 = t2/(70)2, which gives t2 = Page 25 – 1

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Chapter 25

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

Chapter 25

7.

For an object at infinity, the image will be in the focal plane, so we have d1 = f = 135 mm. When the object is at 1.2 m, we locate the image from (1/do) + (1/di) = 1/f; (1/1200 mm) + (1/di) = 1/135 cm, which gives di = 152 mm. Thus the distance from the lens to the film must change by di – f = 152 mm – 135 mm = 17 mm.

8.

We find the object distances from (1/do) + (1/di) = 1/f; (1/do1) + (1/200.0 mm) = 1/200 mm, which gives do1 = ¡; (1/do2) + (1/200.6 mm) = 1/200 mm, which gives do1 = 6.87 × 103 mm = 6.87 m. Thus the range of object distances is

9.

6.87 m ? do ? ¡.

The converging camera lens will form a real, inverted image. For the magnification, we have m = hi/ho = – di/do ; – (24 × 10–3 m)/(22 m) = – di/(50 m), or di = 5.45 × 10–2 m. We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/50 m) + (1/5.45 × 10–2 m) = 1/f, which gives f = 5.4 × 10–2 m = 54 mm.

10. The length of the eyeball is the image distance for a far object, i. e., the focal length of the lens. We find the f-number from f/4.0. f-stop = f/D = (20 mm)/(5.0 mm) = 11. The actual near point of the person is 50 cm. With the lens, an object placed at the normal near point, 25 cm, or 23 cm from the lens, is to produce a virtual image 50 cm from the eye, or 48 cm from the lens. We find the power of the lens from (1/do) + (1/di) = 1/f = P, when distances are in m; + 2.3 D. (1/0.23 m) + (1/– 0.48 m) = P = 12. With the lens, the screen placed 50 cm from the eye, or 48.2 cm from the lens, is to produce a virtual image 120 cm from the eye, or 118.2 cm from the lens. We find the power of the lens from (1/do) + (1/di) = 1/f = P, when distances are in m; + 1.2 D. (1/0.482 m) + (1/– 1.182 m) = P = 13. With the contact lens, an object at infinity would have a virtual image at the far point of the eye. We find the power of the lens from (1/do) + (1/di) = 1/f = P, when distances are in m; (1/¡) + (1/– 0.17 m) = P = – 5.9 D. To find the new near point, we have (1/do) + (1/di) = 1/f = P; (1/do) + (1/– 0.12 m) = – 5.9 D, which gives do = 0.41 m. Glasses would be better, because they give a near point of 32 cm from the eye.

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14. (a) The lens is diverging, so it produces images closer than the object, thus the person is (b) We find the far point by finding the image distance for an object at infinity: (1/do) + (1/di) = 1/f = P;

nearsighted.

(1/¡) + (1/di) = – 5.0 D, which gives di = – 0.20 m = – 20 cm. Because this is the distance from the lens, the far point without glasses is 20 cm + 2.0 cm =

22 cm.

15. (a) We find the power of the lens for an object at infinity: (1/do) + (1/di) = 1/f = P; (1/¡) + (1/– 0.70 m) = P = – 1.43 D. (b) To find the near point with the lens, we have (1/do) + (1/di) = 1/f = P; (1/do) + (1/– 0.20 m) = – 1.43 D, which gives do = 0.28 m =

28 cm.

16. The 2.0 cm of a normal eye is the image distance for an object at infinity, thus it is the focal length of the lens of the eye. To find the length of the nearsighted eye, we find the image distance (distance from lens to retina) for an object at the far point of the eye: (1/do) + (1/di) = 1/f; (1/17 cm) + (1/di) = 1/20 cm, which gives di = 2.27 cm. Thus the difference is 2.27 cm – 2.0 cm = 0.3 cm. 17. We find the far point of the eye by finding the image distance with the lens for an object at infinity: (1/do1) + (1/di1) = 1/f1 ; (1/¡) + (1/di1) = 1/– 25.0 cm, which gives di1 = – 25.0 cm from the lens, or 26.8 cm from the eye. We find the focal length of the contact lens from (1/do2) + (1/di2) = 1/f2 ; (1/¡) + (1/– 26.8 cm) = 1/f2 , which gives f2 =

– 26.8 cm.

18. (a) We find the focal length of the lens for an object at infinity and the image on the retina: (1/do1) + (1/di1) = 1/f1 ; (1/¡) + (1/2.0 cm) = 1/f1 , which gives f1 = 2.0 cm. (b) With the object 30 cm from the eye, we have (1/do2) + (1/di2) = 1/f2 ; (1/30 cm) + (1/2.0 cm) = 1/f2 , which gives f2 = 1.9 cm. 19. The magnification with the image at infinity is M = N/f = (25 cm)/(12 cm) = 2.1×. 20. We find the focal length from M = N/f; 3.0 = (25 cm)/f, which gives f =

8.3 cm.

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21. (a) We find the focal length with the image at the near point from M = 1 + N/f1 ; 2.5 = 1 + (25 cm)/f1 , which gives f1 = 17 cm. (b) If the eye is relaxed, the image is at infinity, so we have M = N/f2 ; 2.5 = (25 cm)/f2 , which gives f2 = 10 cm. 22. Maximum magnification is obtained with the image at the near point. We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/– 25.0 cm) = 1/10.0 cm, which gives do = 7.1 cm from the lens. The magnification is M = 1 + N/f = 1 + (25.0 cm)/(10.0 cm) = 3.5×. 23. (a) The angular magnification with the image at the near point is 3.78×. M = 1 + N/f = 1 + (25.0 cm)/(9.00 cm) = (b) Because the object without the lens and the image with the lens are at the near point, the angular magnification is also the ratio of widths: M = hi/ho; 11.7 mm. 3.78 = hi/(3.10 mm), which gives hi = (c) We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/– 25.0 cm) = 1/9.00 cm, which gives do = 6.62 cm from the lens. 24. (a) We find the image distance from (1/do) + (1/di) = 1/f; (1/5.35 cm) + (1/di) = 1/6.00 cm, which gives di = – 49.4 cm. (b) From the diagram we see that the angular magnification is M = θ ′/θ = (ho/do)/(ho/N) = N/do 4.67×. = (25 cm)/(5.35 cm) =

hi

hi

θ′

ho O

I N ho

θ O

θ′

ho O

I

N

θ

ho O

25. (a) We find the image distance from (1/do) + (1/di) = 1/f; (1/8.5 cm) + (1/di) = 1/9.5 cm, which gives di = – 81 cm. (b) The linear magnification is m = – di/do = – (– 81 cm)/(8.5 cm) = 9.5×. (c) The angular magnification is M = θ ′/θ = (ho/do)/(ho/N) = N/do = (25 cm)/(8.5 cm) = 2.9×. Note that this is less than the linear magnification because the image is farther away.

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26. We find the focal length of the lens from M = N/f; 3.0 = (25 cm)/f, which gives f = 8.3 cm. (a) The magnification with the image at infinity is M1 = N1/f = (50 cm)/(8.3 cm) = 6.0×. (b) The magnification with the image at infinity is M2 = N2/f = (16 cm)/(8.3 cm) = 1.9×. Without the lens, the closest an object can be placed is the near point. A farther near point means a smaller angle subtended by the object without the lens, and thus greater magnification. 27. We find the object distance for an image at her eye’s near point: (1/do) + (1/di) = 1/f = P; 17 cm. (1/do) + (1/– 0.100 m) = – 4.0 D, which gives do = 1.7 × 10–2 m = We find the object distance for an image at her eye’s far point: (1/do) + (1/di) = 1/f = P; (1/do) + (1/– 0.200 m) = – 4.0 D, which gives do = 1.0 m = 100 cm. 28. The magnification of the telescope is given by M = – fo/fe = – (80 cm)/(2.8 cm) = – 29×. For both object and image far away, the separation of the lenses is L = fo + fe = 80 cm + 2.8 cm = 83 cm. 29. We find the focal length of the eyepiece from the magnification: M = – fo/fe ; – 25 = – (80 cm)/fe , which gives fe = 3.2 cm. For both object and image far away, the separation of the lenses is 83 cm. L = fo + fe = 80 cm + 3.2 cm = 30. We find the focal length of the objective from the magnification: M = fo/fe ; 8.0 = fo/(3.0 cm) , which gives fo = 24 cm. 31. We find the focal length of the eyepiece from the power: fe = 1/P = 1/40 D = 0.025 m = 2.5 cm. The magnification of the telescope is given by M = – fo/fe = – (85 cm)/(2.5 cm) = – 34×. 32. For both object and image far away, we find the focal length of the eyepiece from the separation of the lenses: L = fo + fe ; 76 cm = 74.5 cm + fe , which gives fe = 1.5 cm. The magnification of the telescope is given by M = – fo/fe = – (74.5 cm)/(1.5 cm) = – 50×.

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33. For both object and image far away, we find the (negative) focal length of the eyepiece from the separation of the lenses: L = fo + fe ; 33 cm = 36 cm + fe , which gives fe = – 3.0 cm. The magnification of the telescope is given by 12×. M = – fo/fe = – (36 cm)/(– 3.0 cm) = 34. The reflecting mirror acts as the objective, with a focal length fo = r/2 = (5.0 m)/2 = 2.5 m. The magnification of the telescope is given by M = – fo/fe = – (250 cm)/(3.2 cm) = – 78×. 35. We find the focal length of the mirror from M = – fo/fe ; – 120 = – fo/(3.5 cm), which gives fo = 4.2 × 102 cm = The radius is r = 2fo = 2(4.2 m) = 8.4 m. 36. For the magnification we have M = – fo/fe = – 180, or fo = 180fe . For both object and image far away, we have L = fo + fe ; 1.25 m = 180fe + fe , which gives fe = 6.91 × 10–3 m = The focal length of the objective is fo = 180fe = 180(6.91 × 10–3 m) = 1.24 m.

4.2 m.

6.91 mm.

37. We assume a prism binocular so the magnification is positive, but simplify the diagram by ignoring the prisms. We find θ the focal length of the eyepiece from the design magnification: M = fo/fe ; 6.0 = (26 cm)/fe , which gives fe = 4.33 cm. We find the intermediate image formed by the objective: (1/doo) + (1/doi) = 1/fo ; (1/400 cm) + (1/doi) = 1/26 cm, which gives doi = 27.8 cm. With the final image at infinity (relaxed eye), the intermediate image will be at the focal point of the eyepiece lens. From the diagram the angle subtended by the object is θ = h/doi , while the angle subtended by the image is θ ′ = h/fe . Thus the angular magnification is M = θ ′/θ = (h/fe)/(h/doi) = doi/fe = (27.8 cm)/(4.33 cm) = 6.4×. 38. The magnification of the microscope is M = N¬/fo fe = (25 cm)(17.5 cm)/(0.65 cm)(1.5 cm) =

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450×.

doi

θ

fe

θ′

h N

I

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

Chapter 25

39. We find the focal length of the eyepiece from the magnification of the microscope: M = N¬/fo fe ; 720 = (25 cm)(17.5 cm)/(0.40 cm) fe , which gives fe = 1.5 cm. 40. (a) The total magnification is M = MoMe = (62.0)(12.0) = 744×. (b) With the final image at infinity, we find the focal length of the eyepiece from Me = N/fe ; 12.0 = (25.0 cm)/fe , which gives fe = 2.08 cm. Because the image from the objective is at the focal point of the eyepiece, the image distance for the objective is di = ¬ – fe = 20.0 cm – 2.08 cm = 17.9 cm. We find the object distance from the magnification: Mo = di/do ; 62.0 = (17.9 cm)/do , which gives do = 0.289 cm. We find the focal length of the objective from the lens equation: (1/do) + (1/di) = 1/fo ; 0.284 cm. (1/0.289 cm) + (1/17.9 cm) = 1/fo , which gives fo = (c) We found the object distance: do = 0.289 cm. 41. (a) Because the image from the objective is at the focal point of the eyepiece, the image distance for the objective is dio = ¬ – fe = 16.0 cm – 1.8 cm = 14.2 cm. We find the object distance from the lens equation for the objective: (1/doo) + (1/dio) = 1/fo ; (1/doo) + (1/14.2 cm) = 1/0.80 cm, which gives doo = 0.85 cm. (b) With the final image at infinity, the magnification of the eyepiece is Me = N/f = (25.0 cm)/(1.8 cm) = 13.9×. The magnification of the objective is Mo = dio/doo = (14.2 cm)/(0.85 cm) = 16.7×. The total magnification is 230×. M = MoMe = (16.7)(13.9) = 42. (a) We find the object distance from the lens equation for the eyepiece: (1/doe) + (1/die) = 1/fe ; (1/doe) + (1/– 25 cm) = 1/1.8 cm, which gives doe = 1.7 cm. The image distance for the objective is dio = ¬ – doe = 16.0 cm – 1.7 cm = 14.3 cm. We find the object distance from the lens equation for the objective: (1/doo) + (1/dio) = 1/fo ; (1/doo) + (1/14.3 cm) = 1/0.80 cm, which gives doo = 0.85 cm. (b) With the final image at the near point, the magnification of the eyepiece is Me = 1 + N/f = 1 + (25.0 cm)/(1.8 cm) = 14.9×. The magnification of the objective is Mo = dio/doo = (14.3 cm)/(0.85 cm) = 16.9×. The total magnification is 250×. M = MoMe = (16.9)(14.9) =

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43. (a) We find the image distance from the lens equation for the objective: (1/doo) + (1/dio) = 1/fo ; (1/0.790 cm) + (1/dio) = 1/0.740 cm, which gives dio = 11.7 cm. For the relaxed eye, the image from the objective is at the focal point of the eyepiece: doe = 2.70 cm. The distance between lenses is ¬ = dio + doe = 11.7 cm + 2.70 cm = 14.4 cm. (b) With the final image at infinity, the magnification of the eyepiece is Me = N/f = (25.0 cm)/(2.70 cm) = 9.26×. The magnification of the objective is Mo = dio/doo = (11.7 cm)/(0.790 cm) = 14.8×. The total magnification is M = MoMe = (14.8)(9.26) = 137×. 44. (a) When lenses are in contact, the powers add: P = P1 + P2 = (1/– 0.28 m) + (1/0.23 m) = + 0.776 D. converging. It is a positive lens, and thus (b) The focal length is f = 1/P = 1/0.776 D = 1.3 m.

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45. (a) We find the incident angle from sin θ1 = h1/R = (1.0 cm)/(12.0 cm) = 0.0833, so θ1 = 4.78¡. For the refraction at the curved surface, we have sin θ1 = n sin θ2 ; sin 4.78¡ = (1.50) sin θ2 , θ1 which gives sin θ2 = 0.0556, so θ2 = 3.18¡.

θ1 – θ 2 = θ3

We see from the diagram that θ2 h θ3 = θ1 – θ2 = 4.78¡ – 3.18¡ = 1.60¡. θ4 θ3 h′ θ1 θ4 For the refraction at the flat face, we have d n sin θ3 = sin θ4 ; n (1.50) sin 1.60¡ = sin θ4 , R which gives sin θ4 = 0.0419, so θ4 = 2.40¡. We see from the diagram that h1′ = h1 – R sin θ3 = 1.0 cm – (12.0 cm) sin 1.60¡ = 0.665 cm, so the distance from the flat face to the point where the ray crosses the axis is d1 = h1′/tan θ4 = (0.665 cm)/tan 2.40¡ = 15.9 cm. (b) We find the incident angle from sin θ1 = h2/R = (4.0 cm)/(12.0 cm) = 0.333, so θ1 = 19.47¡. For the refraction at the curved surface, we have sin θ1 = n sin θ2 ; sin 19.47¡ = (1.50) sin θ2 , which gives sin θ2 = 0.222, so θ2 = 12.84¡. We see from the diagram that θ3 = θ1 – θ2 = 19.47¡ – 12.84¡ = 6.63¡. For the refraction at the flat face, we have n sin θ3 = sin θ4 ; (1.50) sin 6.63¡ = sin θ4 , which gives sin θ4 = 0.173, so θ4 = 9.97¡. We see from the diagram that h2′ = h2 – R sin θ3 = 4.0 cm – (12.0 cm) sin 6.63¡ = 2.61 cm, so the distance from the flat face to the point where the ray crosses the axis is d2 = h2′/tan θ4 = (2.61 cm)/tan 9.97¡ = 14.8 cm. (c) The separation of the Òfocal pointsÓ is ®d = d1 – d2 = 15.9 cm – 14.8 cm = 1.1 cm. (d) When h2 = 4.0 cm, the rays focus closer to the lens, so they will form a circle at the Òfocal pointÓ for h1 = 1.0 cm. We find the radius of this circle from similar triangles: h2′/d2 = r/(d1 – d2); (2.61 cm)/(14.8 cm) = r/(1.1 cm), whcih gives r = 0.19 cm. 46. The minimum angular resolution is θ = 1.22λ/D = (1.22)(550 × 10–9 m)/(100 in)(0.0254 m/in) = 47. We find the angle of acceptance from NA = n sin α; 1.41 = (1.80) sin α, which gives sin α = 0.783, so α = The resolving power is RP = 0.61λ/NA = 0.61(500 nm)/(1.41) = 216 nm.

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51.6¡.

2.64 × 10–7 rad = (1.51 × 10–5)¡.

Solutions to Physics: Principles with Applications, 5/E, Giancoli 48. (a) The numerical aperture is NA = n sin α = (1.60) sin 60¡ = 1.39. (b) The resolving power is RP = 0.61λ/NA = 0.61(550 nm)/(1.39) =

Chapter 25

240 nm.

49. The resolution of the telescope is θ = 1.22λ/D = (1.22)(500 × 10–9 m)/(12 in)(0.0254 m/in) = 2.00 × 10–6 rad. The separation of the stars is d = Lθ = (20 ly)(9.46 × 1015 m/ly)(2.00 × 10–6 rad) = 3.8 × 1011 m. 50. If the lines are barely resolved, their separation is the resolving power: RP = 0.61λ/NA = 0.61(540 nm)/(0.95) = 347 nm. Because RP ∝ λ, 400 nm requires a lower RP, so they will be resolved by violet light. 700 nm requires a higher RP, so they will not be resolved by red light. 51. The maximum resolving power is the spacing, so we find the minimum NA from RPmax = 0.61λ/NAmin ; 0.63 m = (0.61)(480 m)/NAmin , which gives NAmin = 0.46.

52. The normal human eye can resolve about 10–4 m, so the required minimum magnification is 160×. M = (10–4 m)/(0.63 × 10–6 m) = 53. (a) The resolution of the eye is θ = 1.22λ/D = (1.22)(500 × 10–9 m)/(5.0 × 10–3 m) = 1.22 × 10–4 rad. We find the maximum distance from d = Lθ ; 2.0 m = L(1.22 × 10–4 rad), which gives L = 1.6 × 104 m = 16 km. (b) The angular separation is the resolution: θ = 1.22 × 10–4 rad = (6.99 × 10–3)¡ = 0.42'. Our answer is less than the real resolution because of the presence of aberrations. 54. A path difference of one wavelength corresponds to a phase difference of 2?. The actual distance has not changed, but the number of wavelengths is different for each path. If the object has a thickness t, the number of wavelengths in the medium is t/λ1 , and the number in the object is t/λ2 . Thus the phase difference is δ = 2?[(t/λ2) – (t/λ1)] = 2?[(n2t/λ) – (n1t/λ)] = (2?/λ)(n2 – n1)t. 55. For the diffraction from the crystal, we have mλ = 2d sin φ; m = 1, 2, 3, É . For the first maximum, we get (1)(0.135 nm) = 2(0.280 nm) sin φ, which gives φ = 56. We find the spacing from mλ = 2d sin φ; m = 1, 2, 3, É . (2)(0.0973 nm) = 2d sin 23.4¡, which gives d = Page 25 – 11

14.0¡.

0.245 nm.

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

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

Chapter 25

57. (a) For the diffraction from the crystal, we have mλ = 2d sin φ; m = 1, 2, 3, É . When we form the ratio for the two orders, we get m2/m1 = (sin φ2)/(sin φ1); 52.6¡. 2/1 = (sin φ2)/(sin 23.4¡), which gives φ2 = (b) We find the wavelength from m1λ = 2d sin φ1 ; 0.19 nm. (1)λ = 2(0.24 nm) sin 23.4¡, which gives λ = 58. For the diffraction from the crystal, we have mλ = 2d sin φ; m = 1, 2, 3, É ; (1)λ = 2d sin φ1 ; (2)λ = 2d sin φ2 ; (3)λ = 2d sin φ3 . We see that each equation contains the ratio λ/d, so the wavelength and lattice spacing cannot be separately determined. 59. (a) Because X-ray images are shadows, the image will be the same size as the object, so the magnification is 1. (b) The rays from the point source will not refract, so we can use similar triangles to compare the image size to the object size for the front of the body: m1 = hi/ho1 = (d1 + d2)/d1 = (15 cm + 25 cm)/(15 cm) = 2.7. For the back of the body, the image and object have the same 1. size, so the magnification is

∗

ho1

d1

60. (a) With a positive (converging) lens, Sam is farsighted. (b) The focal length of the lens is f = 1/P = 1/3.2 D = 0.313 m = 31 cm. (c) The lens produces a virtual image at his near point: (1/do1) + (1/di1) = 1/f = P; (1/0.23 m) + (1/di1) = + 3.2 D, which gives di1 = – 0.87 m, so his near point is (d) For Pam, we find the object distance that will have an image at her near point: (1/do2) + (1/di2) = 1/f = P; (1/do2) + (1/– 0.23 m) = + 3.2 D, which gives do2 = 0.13 m, which is 15 cm 61. The exposure is proportional to the intensity, the area and the time: Exposure ∝ IAt ∝ ID2t ∝ It/( f-stop)2. With the same shutter speed, the time is constant. Because we want the exposure to be the same, we have I1/( f-stop1)2 = I2/( f-stop2 )2; I1/(5.6)2 = I2/(22)2, which gives I2 = 16I1. Note that we have followed convention to use multiples of 2.

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ho2

d2

89 cm.

from her eyes.

hi

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Chapter 25

62. When an object is very far away, the image will be at the focal point di = f. Thus the magnification is m = – di/do = – f/do , that is, proportional to f.

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Solutions to Physics: Principles with Applications, 5/E, Giancoli 63. For the magnification, we have m = – hi/ho = – di/do = – 1, so di = do . We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/do) = 1/50 mm, which gives do = The distance between the object and the film is d = do + di = 100 mm + 100 mm = 200 mm.

Chapter 25

100 mm.

64. The actual far point of the person is 180 cm. With the lens, an object far away is to produce a virtual image 180 cm from the eye, or 178 cm from the lens. We find the power of the lens from (1/do1) + (1/di1) = 1/f1 = P1 , when distances are in m; – 0.56 D (upper part). (1/¡) + (1/– 1.78 m) = P1 = The actual near point of the person is 40 cm. With the lens, an object placed at the normal near point, 25 cm, or 23 cm from the lens, is to produce a virtual image 40 cm from the eye, or 38 cm from the lens. We find the power of the lens from (1/do2) + (1/di2) = 1/f2 = P2 ; (1/0.23 m) + (1/– 0.38 m) = P2 = + 1.7 D (lower part). 65. The maximum magnification is achieved with the image at the near point: M1 = 1 + N1/f = 1 + (15.0 cm)/(8.0 cm) = 2.9×. For an adult we have M2 = 1 + N2/f = 1 + (25.0 cm)/(8.0 cm) = 4.1×. 66. The magnification for a relaxed eye is M = N/f = NP = (0.25 m)(+ 4.0 D) =

1.0×.

67. (a) The magnification of the telescope is given by – 2.25×. M = – fo/fe = – Pe/Po = – (4.5 D)/(2.0 D) = (b) To get a magnification greater than 1, for the eyepiece we use the lens with the smaller focal length, or greater power: 4.5 D. 68. The minimum angular resolution is θ = 1.22λ/D. The distance between lines is the resolving power: RP = fθ = 1.22λf/D= 1.22λ(f-stop). For f/2 we have RP2 = (1.22)(500 × 10–9 m)(2) = 1.22 × 10–6 m = 1.22 × 10–3 mm, so the resolution is 1/RP2 = 1/(1.22 × 10–3 mm) = 820 lines/mm. For f/16 we have RP2 = (1.22)(500 × 10–9 m)(16.7) = 1.02 × 10–5 m = 1.02 × 10–2 mm, so the resolution is 1/RP2 = 1/(1.02 × 10–2 mm) = 98 lines/mm.

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69. To find the new near point, we have (1/do1) + (1/di1) = 1/f1 = P1 , when distances are in m; (1/0.35 m) + (1/di1) = + 2.5 D, which gives di1 = – 2.8 m. To give him a normal near point, we have (1/do2) + (1/di2) = 1/f2 = P2 ; (1/0.25 m) + (1/– 2.8 m) = P2 = + 3.6 D. 70. For the minimum aperture the angle subtended at the lens by the smallest feature is the angular resolution: θ = d/L = 1.22λ/D ; 34 cm. (5 × 10–2 m)/(25 × 103 m) = (1.22)(550 × 10–9 m)/D, which gives D = 0.34 m = 71. The angular resolution of the eye, which is the required resolution using the telescope, is θeye = deye/Leye = (0.10 × 10–3 m)/(25 × 10–2 m) = 4.0 × 10–4 rad. The resolution without the telescope is θ = d/L = (10 km)/(3.84 × 105 km) = 2.6 × 10–5 rad. If we ignore the inverted image, the magnification is M = θeye/θ = fo/fe ; 14 cm. (4.0 × 10–4 rad)/(2.6 × 10–5 rad) = (2.2 m)/fe , which gives fe = 0.14 m = The resolution limit is θ = 1.22λ/D = (1.22)(500 × 10–9 m)/(0.12 m) = 5.1 × 10–6 rad. This is a distance of 2 km on the surface of the Moon. 72. (a) When an object is very far away, the image will be at the focal point di1 = f. From the magnification, we have m = – hi/ho = – di/do , so we see that hi is proportional to di . When the object and image are the same size, we get – hi/ho = – 1 = – di2/do2 , so do2 = di2 . From the lens equation, we get (1/do2) + (1/di2) = 1/f ; (1/di2) + (1/di2) = 1/f, which gives di2 = 2f. The required exposure time is proportional to the area of the image on the film: t ∝ A ∝ (hi)2 ∝ (di)2. When we form the ratio, w get t2/t1 = (di2/di1)2 = (2f/f)2 = 4. (b) For the increased object distances, we have (1/do3) + (1/di3) = 1/f ; (1/4f) + (1/di3) = 1/f, which gives di3 = 4f/3; and t3/t1 = (4/3)2 = 1.78. (1/do4) + (1/di4) = 1/f ; (1/5f) + (1/di4) = 1/f, which gives di4 = 5f/4; and t4/t1 = (5/4)2 = 1.56. These increased exposures are less than the minimal adjustment on a typical camera, so they are negligible.

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73. The focal length of the eyepiece is fe = 1/Pe = 1/20 D = 5.0 × 10–2 m = 5.0 cm. For both object and image far away, we find the focal length of the objective from the separation of the lenses: L = fo + fe ; 85 cm = fo + 5.0 cm, which gives fo = 80 cm. The magnification of the telescope is given by M = – fo/fe = – (80 cm)/(5.0 cm) = – 16×. 74. We find the focal lengths of the lens for the two colors: 1/fred = (nred – 1)(1/R1 + 1/R2) = (1.5106 – 1)[(1/18.4 cm) + (1/¡)] which gives fred = 36.04 cm. 1/forange = (norange – 1)(1/R1 + 1/R2) = (1.5226 – 1)[(1/18.4 cm) + (1/¡)] which gives forange = 35.21 cm. We find the image distances from (1/do) + (1/dired) = 1/fred ; (1/66.0 cm) + (1/dired) = 1/36.04 cm, which gives dired = 79.4 cm. (1/do) + (1/diorange) = 1/forange ; (1/66.0 cm) + (1/diorange) = 1/35.21 cm, which gives diorange = 75.5 cm. The images are 3.9 cm apart, an example of chromatic aberration. 75. We can relate the object and image distances from the magnification: m = – hi/ho = – di/do ; (8.25 × 10–3 m)/(1.75 m) = di/do , or di = (4.71 × 10–3)do . We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + [1/(4.71 × 10–3)do] = 1/200 × 10–3 m, which gives do = 42.7 m. Thus the reporter was standing 42.7 m from his subject.

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