Chapter 34 Solutions_8

# Chapter 34 Solutions_8 - Chapter 34 2. The bird is a...

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Unformatted text preview: Chapter 34 2. The bird is a distance d2 in front of the mirror; the plane of its image is that same distance d2 behind the mirror. The lateral distance between you and the bird is d3 = 5.00 m. We denote the distance from the camera to the mirror as d1, and we construct a right triangle out of d3 and the distance between the camera and the image plane (d1 + d2). Thus, the focus distance is d= ( d1 + d 2 ) 2 + d32 = ( 4.30m+3.30m ) + ( 5.00m ) 2 2 = 9.10m. 3. When S is barely able to see B the light rays from B must reflect to S off the edge of the mirror. The angle of reflection in this case is 45°, since a line drawn from S to the mirror’s edge makes a 45° angle relative to the wall. By the law of reflection, we find x d 3.0 m = tan 45° = 1 ⇒ x = = = 1.5 m. d /2 2 2 7. We use Eqs. 34‐3 and 34‐4, and note that m = –i/p. Thus, 1 1 12 − = = . p pm f r We solve for p: p= r 1 35.0 cm 1 1− = 1− = 10.5 cm. 2 m 2 2.50 FG H IJ K FG H IJ K 9. A concave mirror has a positive value of focal length. (a) Then (with f = +18 cm and p = +12 cm) , the radius of curvature is r = 2f = + 36 cm. (b) Eq. 34‐9 yields i = pf /( p − f ) = –36 cm. (c) Then, by Eq. 34‐7, m = −i/p = +3.0. (d) Since the image distance is negative, the image is virtual (V). (e) The magnification computation produced a positive value, so it is upright [not inverted] (NI). (f) For a mirror, the side where a virtual image forms is opposite from the side where the object is. 36. In addition to n1 =1.0, we are given (a) n2=1.5, (b) p = +10 and (c) r = +30. (d) Eq. 34‐8 yields F n − n − n IJ i=n G H r pK 2 1 1 2 −1 = 15 . FG 15 − 10 − 10 IJ = −18 cm. . . . H 30 cm 10 cmK (e) The image is virtual (V) and upright since i < 0 . (f) The object and its image are in the same side. The ray diagram would be similar to Fig. 34‐12(c) in the textbook. 45. We solve Eq. 34‐9 for the image distance: ⎛1 1⎞ fp i=⎜ − ⎟ = . p− f ⎝ f p⎠ −1 The height of the image is thus hi = mhp = 49. (a) We use Eq. 34‐10: FG i IJ h H pK p = fhp p− f = (75 mm)(1.80 m) = 5.0 mm. 27 m − 0.075 m L F 1 1 I O L . F 1 1 I OP f = M(n − 1) G − J P = M(15 − 1) G − H ∞ −20 cmJK Q N H r r KQ N −1 1 2 −1 = +40 cm. (b) From Eq. 34‐9, F 1 1 I F 1 − 1 IJ i=G − J =G H f p K H 40 cm 40 cmK −1 −1 = ∞. 58. (a) A convex (converging) lens, since a real image is formed. (b) Since i = d – p and i/p = 1/2, p= (c) The focal length is 2d 2 40.0 cm = = 26.7 cm. 3 3 b g ⎛1 1 ⎞ 1⎞ 2d 2 ( 40.0 cm ) ⎛1 f =⎜ + ⎟ =⎜ + = = 8.89 cm . ⎟= 9 9 ⎝ d / 3 2d / 3 ⎠ ⎝i p⎠ −1 −1 88. The minimum diameter of the eyepiece is given by d ey = d ob 75 mm = = 2.1 mm. 36 mθ 90. We refer to Fig. 34‐20. For the intermediate image p = 10 mm and i = (fob + s + fey) – fey = 300 m – 50 mm = 250 mm, so 1 11 1 1 =+ = + ⇒ f ob = 9.62 mm, f ob i p 250 mm 10 mm and s = (fob + s + fey) – fob – fey = 300 mm – 9.62 mm – 50 mm = 240 mm. Then from Eq. 34‐14, M =− 240 mm s 25 cm =− 9.62 mm f ob f ey FG H IJ FG 150 mmIJ = −125. K H 50 mm K 91. (a) Without the magnifier, θ = h/Pn (see Fig. 34‐19). With the magnifier, letting i = – |i| = – Pn, we obtain 1111111 = − = + = + . pfifi f Pn Consequently, mθ = With f = 10 cm, mθ = 1 + θ ′ h / p 1 / f + 1 / Pn P 25 cm = = = 1+ n = 1+ . θ h / Pn 1 / Pn f f 25cm = 3.5 . 10cm (b) In the case where the image appears at infinity, let i = − | i |→ −∞ , so that 1/ p + 1/ i = 1/ p = 1/ f , we have mθ = With f = 10 cm, θ ′ h / p 1/ f Pn 25 cm . = = == f θ h / Pn 1/ Pn f mθ = 25cm = 2.5. 10cm 98. (a) First, the lens forms a real image of the object located at a distance F 1 1 I F 1 1 IJ i =G − J =G − H f p K H f 2f K −1 1 1 1 1 1 −1 = 2 f1 to the right of the lens, or at p2 = 2(f1 + f2) – 2f1 = 2f2 in front of the mirror. The subsequent image formed by the mirror is located at a distance i2 to the left of the mirror, or at F 1 1 I F 1 1 IJ =G − J =G − H f p K H f 2f K −1 2 2 2 2 −1 = 2 f2 p'1 = 2(f1 + f2) – 2f2 = 2f1 to the right of the lens. The final image formed by the lens is at a distance i'1 to the left of the lens, where F 1 1 I F 1 1 IJ i′ = G − J = G − H f p′ K H f 2 f K −1 1 1 1 1 1 −1 = 2 f1. This turns out to be the same as the location of the original object. (b) The lateral magnification is m= − FG i IJ FG − i IJ FG − i ′ IJ = FG − 2 f IJ FG − 2 f IJ FG − 2 f IJ = −10. H p K H p K H p′ K H 2 f K H 2 f K H 2 f K . 1 2 1 1 1 2 2 1 1 1 2 1 (c) The final image is real (R). (d) It is at a distance i'1 to the left of the lens, (e) and inverted (I), as shown in the figure below. ...
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