**Física I, 7ªED Halliday Fundamentos da física mecânica, cap.22**

Física I, 7ªED Halliday Fundamentos da física mecânica, cap.22

(Parte **1** de 4)

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

(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

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

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

(Parte **1** de 4)