Physics Solution Manual for 1100 and 2101

# Since the magnification m is negative the image is

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Unformatted text preview: hat the horizontal and vertical distances in this drawing are to scale. This means that the mirror is represented by a circular arc that is also drawn to scale. Note that we have 1292 THE REFLECTION OF LIGHT: MIRRORS used only rays 1 and 3 in constructing this diagram. Only two of the three rays discussed in the text are needed. 1 3 F 18 cm 6.0 cm C From the drawing, we see that the image is located 6.0 cm behind the mirror . 14. REASONING The object distance (do = 11 cm) is shorter than the focal length (f = 18 cm) of the mirror, so we expect the image to be virtual, appearing behind the mirror. Taking Figure 25.18a as our model, we will trace out: three rays from the tip of the object to the surface of the mirror, then three reflected rays, and finally three virtual rays extending behind the mirror and meeting at the tip of the image. The scale of the ray tracing will determine the location and height of the image. The three sets of rays are: 1. An incident ray from the object to the mirror, parallel to the principal axis and then reflected through the focal point F. 2. An incident ray from the object to the mirror, directly away from the focal point F and then reflected parallel to the principal axis. (The incident ray cannot pass from the object through the focal point, as this would take it away from the mirror, and it would not be reflected.) 3. An incident ray from the object to the mirror, directly away from the center of curvature C, then reflected back through C. SOLUTION 2 2 2 3 C 1 3 3 1 7.6 cm tall 1 F 11 cm Image (virtual) 28 cm Object 3.0 cm tall Scale: 6.0 cm Chapter 25 Problems 1293 a. The ray diagram indicates that the image is 28 cm behind the mirror . b. We see from the ray diagram that the image is 7.6 cm tall . 15. SSM REASONING AND SOLUTION The ray diagram is shown below. (Note: f = –50.0 cm and do = 25.0 cm ) Scale Reflected ray 10.0 cm Virtual Image Object 10.0 cm 25.0 cm 6.67 cm 16.7 cm F C a. The ray diagram indicates that the image is 16.7 cm behind the mirror . b. The ray diagram indicates that the image height is 6.67 cm . 16. REASONING We will start by drawing the two situations in which the object is 25 cm and 5 cm from the mirror, making sure that all distances (including the radius of curvature of the mirror) and heights are to scale. For each location of the object, we will draw several rays to locate the image (see the Reasoning Strategy for convex mirrors in Section 25.5). Once the images have been located, we can readily answer the questions regarding their positions and heights. SOLUTION The following two ray diagrams illustrate the situations where the objects are at different distances from the convex mirror. 1294 THE REFLECTION OF LIGHT: MIRRORS Image 1 3 Object F C 25 cm Image 1 3 Object F C 5 cm a. As the object moves closer to the mirror, it can be seen that the magnitude of the image distance becomes smaller . b. As the object moves closer to the mirror, the magnitude of the image height becomes larger . c. By measuring the image heights, we find that the ratio of the image height when the object distance is 5 cm to that when the object distance is 25 cm is 3 . 17. REASONING AND SOLUTION a. A ray diagram, which will look similar, but not identical, to that in Figure 25.21a, reveals that the image distance is 20.0 cm behind the mirror, or di = − 20.0 cm . b. The ray diagram also shows that the image height is 6.0 cm , and the image is upright relative to the object. Chapter 25 Problems 1295 18. REASONING AND SOLUTION A plane mirror faces a concave mirror ( f = 8.0 cm ). The following is a ray diagram of an object placed 10.0 cm in front of the plane mirror. Incident ray Reflected ray Object Image from plane mirror F Scale: 5.0 cm Plane mirror Image from concave mirror Concave mirror The ray diagram shows the light that is first reflected from the plane mirror and then the concave mirror. The scale is shown in the figure. For the reflection from the plane mirror, as discussed in the text, the image is upright, the same size as the object, and is located as far behind the mirror as the object is in front of it. When the reflected ray reaches the concave mirror, the ray that is initially parallel to the principal axis passes through the focal point F after reflection from the concave mirror. The ray that passes directly through the focal point emerges parallel to the principal axis after reflection from the concave surface. The point of intersection of these two rays locates the position of the image. By inspection, we see that the image is located at 10.9 cm from the concave mirror. 19. SSM REASONING This problem can be solved using the mirror equation, Equation 25.3. SOLUTION Using the mirror equation with d i = +26 cm and f = 12 cm , we find 1 11 1 1 = – = – do f di 12 cm 26 cm or d o = +22 cm 20. REASONING The mirror equation relates the object and image distances to the focal length. Thus, we can apply the mirror equation once with the given object and image distances to determine the focal length. Then, we can use the mirror equation again with the focal length and the second object distance to determine the unknown image dis...
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