Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch34

Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch34

(Parte 1 de 2)

1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm.

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 = 5.0 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

3. The intensity of light from a point source varies as the inverse of the square of the distance from the source. Before the mirror is in place, the intensity at the center of the screen is given by IP = A/d 2 , where A is a constant of proportionality. After the mirror is in place, the light that goes directly to the screen contributes intensity IP, as before. Reflected light also reaches the screen. This light appears to come from the image of the source, a distance d behind the mirror and a distance 3d from the screen. Its contribution to the intensity at the center of the screen is

P r

The total intensity at the center of the screen is

P Pr P P

The ratio of the new intensity to the original intensity is I/IP = 10/9 = 1.1.

4. 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

5. We apply the law of refraction, assuming all angles are in radians:

sin

sin nw air which in our case reduces to θ' ≈ θ/nw (since both θ and θ ' are small, and nair ≈ 1). We refer to our figure below.

The object O is a vertical distance d1 above the water, and the water surface is a vertical distance d2 above the mirror. We are looking for a distance d (treated as a positive number) below the mirror where the image I of the object is formed. In the triangle O AB

1||tan,ABddθθ=≈ and in the triangle CBD d BC d d

Finally, in the triangle ACI, we have |AI| = d + d2. Therefore, tan d dAC AB BC dA I d d d d d d nθ θθ θ θ

6. We note from Fig. 34-32 that m = 1

2 when p = 5 cm. Thus Eq. 34-7 (the magnification equation) gives us i = −10 cm in that case. Then, by Eq. 34-9 (which applies to mirrors and thin-lenses) we find the focal length of the mirror is f = 10 cm. Next, the problem asks us to consider p = 14 cm. With the focal length value already determined, then Eq. 34-9 yields i = 35 cm for this new value of object distance. Then, using Eq. 34-7 again, we find m = i/p = −2.5.

7. We use Eqs. 34-3 and 34-4, and note that m = –i/p. Thus, p m f r −==.

We solve for p:

p r

m =− FHG IKJ =− FHG IKJ =2

. cm

8. The graph in Fig. 34-3 implies that f = 20 cm, which we can plug into Eq. 34-9 (with p = 70 cm) to obtain i = +28 cm.

(b) Eq. 34-9 yields i = pf /( p−f ) = +36 cm
(c)Then, by Eq. 34-7, the lateral magnification is m = −i/p = −2.0
(e) The magnification computation produced a negative value, so it is inverted (I)

(d) Since the image distance computation produced a positive value, the image is real (R).

(f) For a mirror, the side where a real image forms is the same as the side where the object is.

10. A concave mirror has a positive value of focal length. (a) Then (with f = +10 cm and p = +15 cm), the radius of curvature is 220 cmrf==+.

(b) Eq. 34-9 yields i = pf /( p−f ) = +30 cm
(c)Then, by Eq. 34-7, m = −i/p = –2.0
(e) The magnification computation produced a negative value, so it is inverted (I)

(d) Since the image distance computation produced a positive value, the image is real (R).

(f) For a mirror, the side where a real image forms is the same as the side where the object is.

1. 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)
inverted] (NI)

(e) The magnification computation produced a positive value, so it is upright [not

(f) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

12. A concave mirror has a positive value of focal length. (a) Then (with f = +36 cm and p = +24 cm), the radius of curvature is r = 2f = + 72 cm.

(b) Eq. 34-9 yields i = pf /( p−f ) = –72 cm
(c) Then, by Eq. 34-7, m = −i/p = +3.0
(d) Since the image distance is negative, the image is virtual (V)
inverted] (NI)

(e) The magnification computation produced a positive value, so it is upright [not

(f) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

13. A convex mirror has a negative value of focal length. (a) Then (with f = –10 cm and p = +8 cm), the radius of curvature is r = 2f = –20 cm.

(b) Eq. 34-9 yields i = pf /( p−f ) = – 4.4 cm
(c) Then, by Eq. 34-7, m = −i/p = +0.56
(d) Since the image distance is negative, the image is virtual (V)
inverted] (NI)

(e) The magnification computation produced a positive value, so it is upright [not

(f) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

14. A convex mirror has a negative value of focal length. (a) Then (with f = –35 cm and p = +2 cm), the radius of curvature is r = 2f = –70 cm.

(b) Eq. 34-9 yields i = pf /( p−f ) = –14 cm
(c) Then, by Eq. 34-7, m = −i/p = +0.61
(d) Since the image distance is negative, the image is virtual (V)
inverted] (NI)

(e) The magnification computation produced a positive value, so it is upright [not

(f) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

(a) Then (with f = –8 cm and p = +10 cm), the radius of curvature is r = 2f = –16 cm
(b) Eq. 34-9 yields i = pf /( p−f ) = –4.4 cm
(c) Then, by Eq. 34-7, m = −i/p = +0.4
(d) Since the image distance is negative, the image is virtual (V)

15. A convex mirror has a negative value of focal length.

inverted] (NI)

(e) The magnification computation produced a positive value, so it is upright [not

(f) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

16. A convex mirror has a negative value of focal length. (a) Then (with f = –14 cm and p = +17 cm), the radius of curvature is r = 2f = –28 cm.

(b) Eq. 34-9 yields i = pf /( p−f ) = –7.7 cm
(c) Then, by Eq. 34-7, m = −i/p = +0.45
(d) Since the image distance is negative, the image is virtual (V)
inverted] (NI)

(e) The magnification computation produced a positive value, so it is upright [not

(f) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

17. (a) From Eqs. 34-3 and 34-4, we obtain i = pf/(p – f ) = pr/(2p – r).

Differentiating both sides with respect to time and using vO = –dp/dt, we find

didt ddt prpr rv p r v prpr r pr b g .

(b) If p = 30 cm, we obtain

2 cm

(c) If p = 8.0 cm, we obtain

(d) If p = 1.0 cm, we obtain

15cm 5.0cm/s 6.7cm/s.

is a convex mirrorWe examine the point where m = 0.50 and p = 10 cm. Combining
our expectation that the mirror is convex)Now, for p = 21 cm, we find m = –f/(p – f)

18. We note that there is “singularity” in this graph (Fig. 34-34) like there was in Fig. 34- 3), which tells us that there is no point where p = f (which causes Eq. 34-9 to “blow up”). Since p > 0, as usual, then this means that the focal length is not positive. We know it is not a flat mirror since the curve shown does decrease with p, so we conclude it Eq. 34-7 and Eq. 34-9 we obtain m = – i /p = –f/(p – f). This yields f = –10 cm (verifying = +0.32.

19. (a) The mirror is concave. (b) f = +20 cm (positive, because the mirror is concave). (c) r = 2f = 2(+20 cm) = +40 cm. (d) The object distance p = +10 cm, as given in the Table.

(e) The image distance is i = (1/f – 1/p)–1 = (1/20 cm – 1/10 cm)–1 = –20 cm.

(f) m = –i/p = –(–20 cm/10 cm) = +2.0. (g) The image is virtual (V). (h) The image is upright or not inverted (NI).

(i) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

20. (a) The fact that the magnification is 1 means that the mirror is flat (plane).

(b) Flat mirrors (and flat “lenses” such as a window pane) have f = ∞ (or f = –∞ since the sign does not matter in this extreme case).

(c) The radius of curvature is r = 2f = ∞ (or r = –∞) by Eq. 34-3. (d) p = + 10 cm, as given in the Table.

(e) Eq. 34-4 readily yields i = pf /( p−f ) = –10 cm. (f) The magnification is m = –i/p = +1.0. (g) The image is virtual since i < 0. (h) The image is upright, or not inverted (NI).

(i) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

21. (a) Since f > 0, the mirror is concave. (b) f = + 20 cm, as given in the Table. (c) Using Eq. 34-3, we obtain r = 2f = +40 cm. (d) p = + 10 cm, as given in the Table.

(e) Eq. 34-4 readily yields i = pf /( p−f ) = +60 cm. (f) Eq. 34-6 gives m = –i/p = –2.0. (g) Since i > 0, the image is real (R). (h) Since m < 0, the image is inverted (I).

(i) For a mirror, the side where a real image forms is the same as the side where the object is.

2. (a) Since m = − 1/2 < 0, the image is inverted. With that in mind, we examine the various possibilities in Figs. 34-7, 34-9 and 34-10, and note that an inverted image (for reflections from a single mirror) can only occur if the mirror is concave (and if p > f ).

(b) Next, we find i from Eq. 34-6 (which yields i = mp = 30 cm) and then use this value (and Eq. 34-4) to compute the focal length; we obtain f = +20 cm.

(c) Then, Eq. 34-3 gives r = 2f = +40 cm. (d) p = 60 cm, as given in the Table. (e) As already noted, i = +30 cm.

(f) m = − 1/2, as given. (g) Since i > 0, the image is real (R). (h) As already noted, the image is inverted (I).

(i) For a mirror, the side where a real image forms is the same as the side where the object is.

23. (a) Since r < 0 then (by Eq. 34-3) f < 0, which means the mirror is convex. (b) The focal length is f = r/2 = –20 cm. (c) r = – 40 cm, as given in the Table. (d) Eq. 34-4 leads to p = +20 cm. (e) i = –10 cm, as given in the Table. (f) Eq. 34-6 gives m = +0.50. (g) The image is virtual (V). (h) The image is upright, or not inverted (NI).

(i) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

24. (a) Since 0 < m < 1, the image is upright but smaller than the object. With that in mind, we examine the various possibilities in Figs. 34-7, 34-9 and 34-10, and note that such an image (for reflections from a single mirror) can only occur if the mirror is convex.

(b) Thus, we must put a minus sign in front of the “20” value given for f, i.e., f = – 20 cm. (c) Eq. 34-3 then gives r = 2f = –40 cm.

(d) To solve for i and p we must set up Eq. 34-4 and Eq. 34-6 as a simultaneous set and solve for the two unknowns. The results are p = +180 cm = +1.8 m, and

(e) i = –18 cm. (f) m = 0.10, as given in the Table. (g) The image is virtual (V) since i < 0. (h) The image is upright, or not inverted (NI), as already noted.

(i) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

25. (a) The mirror is convex, as given.

(b) Knowing the mirror is convex means we must put a minus sign in front of the “40” value given for r. Then, Eq. 34-3 yields f = r/2 = –20 cm.

(c) r = – 40 cm.

(d) The fact that the mirror is convex also means that we need to insert a minus sign in front of the “4.0” value given for i, since the image in this case must be virtual (see Figs. 34-7, 34-9 and 34-10). Eq. 34-4 leads to p = +5.0 cm.

(e) As noted above, i = – 4.0. (f) Eq. 34-6 gives m = +0.8. (g) The image is virtual (V) since i < 0. (h) The image is upright, or not inverted (NI).

(i) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

26. (a) Since the image is inverted, we can scan Figs. 34-7, 34-9 and 34-10 in the textbook and find that the mirror must be concave.

(b) This also implies that we must put a minus sign in front of the “0.50” value given for m. To solve for f, we first find i = –pm = +12 cm from Eq. 34-6 and plug into Eq. 34-4; the result is f = +8 cm.

(c) Thus, r = 2f = +16 cm. (d) p = +24 cm, as given in the Table. (e) As shown above, i = –pm = +12 cm. (f) m = –0.50, with a minus sign. (g) The image is real (R) since i > 0. (h) The image is inverted (I), as noted above.

(i) For a mirror, the side where a real image forms is the same as the side where the object is.

convex
(f) From Eq. 34-7, we get m = +1/2 = 0.50

27. (a) The fact that the focal length is given as a negative value means the mirror is (b) f = –30 cm, as given in the Table. (c) The radius of curvature is r = 2f = –60 cm. (d) Eq. 34-9 gives p = if /(i – f) = +30 cm. (e) i = –15, as given in the Table.

(g) The image distance is given as a negative value (as it would have to be, since the mirror is convex), which means the image is virtual (V).

(h) Since m > 0, the image is upright (not inverted: NI). (i) The image is on the side of the mirror opposite to the object.

28. (a) We are told that the image is on the same side as the object; this means the image is real (R) and further implies that the mirror is concave.

(b) The focal distance is f = +20 cm

(c) The radius of curvature is r = 2f = +40 cm. (d) p = +60 cm, as given in the Table. (e) Eq. 34-9 gives i = pf/(p – f) = +30 cm.

(f) Eq. 34-7 gives m = −i/p = −0.50. (g) As noted above, the image is real (R). (h) The image is inverted (I) since m < 0.

(i) For a mirror, the side where a real image forms is the same as the side where the object is.

29. (a) As stated in the problem, the image is inverted (I) which implies that it is real (R). It also (more directly) tells us that the magnification is equal to a negative value: m =

−0.40. By Eq. 34-7, the image distance is consequently found to be i = +12 cm. Real images don’t arise (under normal circumstances) from convex mirrors, so we conclude that this mirror is concave.

(b) The focal length is f = +8.6 cm, using Eq. 34-9 f = +8.6 cm.

(c) The radius of curvature is r = 2f = +17.2 cm ≈ 17 cm. (d) p = +30 cm, as given in the Table. (e) As noted above, i = +12 cm.

(f) Similarly, m = −0.40, with a minus sign. (g) The image is real (R). (h) The image is inverted (I).

(i) For a mirror, the side where a real image forms is the same as the side where the object is.

30. (a) From Eq. 34-7, we get i = −mp = +28 cm, which implies the image is real (R) and on the same side as the object. Since m < 0, we know it was inverted (I). From Eq. 34-9, we obtain f = ip/(i + p) = +16 cm, which tells us (among other things) that the mirror is concave.

(b) f = ip/(i + p) = +16 cm. (c) r = 2f = +32 cm. (d) p = +40 cm, as given in the Table.

(e) i = −mp = +28 cm.

(f) m = −0.70, as given in the Table. (g) The image is real (R). (h) The image is inverted (I).

(i) For a mirror, the side where a real image forms is the same as the side where the object is.

31. (a) The fact that the magnification is equal to a positive value means that the image is upright (not inverted: NI), and further implies (by Eq. 34-7) that the image distance (i) is

than unity is only possible for convex mirrors

equal to a negative value the image is virtual (V). Looking at the discussion of mirrors in sections 34-3 and 34-4, we see that a positive magnification of magnitude less

(b) For 0< m < 1 this will only give a positive value for p = f /(1 – 1/m) if f < 0. Thus, with a minus sign, we have f = −30 cm.

(c) r = 2f = –60 cm. (d) p = f /(1 – 1/m) = + 120 cm = 1.2 m. (e) i = –mp = –24 cm. (f) m = +0.20, as given in the Table. (g) The image is virtual (V). (h) The image is upright, or not inverted (NI).

(i) For a mirror, the side where a virtual image forms is opposite from the side where the object is.

32. (a) We use Eq. 34-8 and note that n1 = nair = 1.0, n2 = n, p = ∞, and i = 2r:

−nr n r

We solve for the unknown index: n = 2.0. (b) Now i = r so Eq. 34-8 becomes nr n r which is not valid unless n→∞ or .r→∞ It is impossible to focus at the center of the sphere.

3. We use Eq. 34-8 (and Fig. 34-10(d) is useful), with n1 = 1.6 and n2 = 1 (using the rounded-off value for air).

Using the sign convention for r stated in the paragraph following Eq. 34-8 (so that 5.0 cmr=−), we obtain i = –2.4 cm for objects at p = 3.0 cm. Returning to Fig. 34-36

(and noting the location of the observer), we conclude that the tabletop seems 7.4 cm away.

34. 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

cm cm cm.

(f) The object and its image are in the same side. The ray diagram would be similar to Fig. 34-1(c) in the textbook.

35. In addition to n1 =1.0, we are given (a) n2=1.5, (b) p = +10 and (d) 13i=−. (c) Eq. 34-8 yields n rn n

(Parte 1 de 2)

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