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Physics Solution Manual for 1100 and 2101

Solving equation 253 for 1 and then taking the

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Unformatted text preview: the laser and the detector. Chapter 25 Problems b. Turning the surface normal too far from due south increases both the angle of incidence and the angle of reflection, in this case by 0.004° each. Thus, the angle α between the incident and reflected rays increases by twice that amount. We will use the new angle α, with tan α = D L (Equation 1.3), to determine the distance D between the laser and the reflected beam (see the drawing). The distance x by which the reflected beam misses the detector is the distance D minus the distance d between the laser and the detector: 1289 Mirror θ θ α L Detector Laser d x D x= D−d (2) We note that the distance d is given in meters, while the distance L is given in kilometers, where 1 km = 103 m. SOLUTION a. Substituting α = tan −1 ( d L ) (Equation 1.6) into Equation (1) yields d 117 m o = 0.0670 3 50.0 × 10 m θ = 1 α = 1 tan −1 = 1 tan −1 2 2 2 L b. When the mirror is misaligned, the angle of incidence is larger than the result found in part (a) by 0.004°. From Equation (1), then, the angle α between the incident and reflected rays becomes α = 2θ = 2 0.0670o + 0.004o = 0.142o ( ) Solving tan α = D L (Equation 1.3) for the distance D between the laser and the reflected beam, we obtain D = L tan α (3) Substituting Equation (3) into Equation (2), and using α = 0.142o , we find that ( ) x = D − d = L tan α − d = 50.0 ×103 m tan 0.142o − 117 m = 7 m 1290 THE REFLECTION OF LIGHT: MIRRORS 11. REASONING According to the discussion about relative velocity in Section 3.4, the velocity vIY of your image relative to you is the vector sum of the velocity vIM of your image relative to the mirror and the velocity vMY of the mirror relative to you: vIY = vIM + vMY As you walk toward the stationary mirror, you perceive the mirror moving toward you in the opposite direction. Thus, vMY = –vYM. The velocity vYM has the components vYM,x and vYM,y, while the velocity vIM has the components vIM,x and vIM,y. The x direction is perpendicular to the mirror, and the two x components have the same magnitude. This is because the image in a plane mirror is always just as far behind the mirror as the object is in front of it. For instance, if an object moves 1 meter perpendicularly toward the mirror in 1 second, the magnitude of its velocity relative to the mirror is 1 m/s. But the image also moves 1 meter toward the mirror in the same time interval, so that the magnitude of its velocity relative to the mirror is also 1 m/s. The two x components, however, have opposite directions. The two y components have the same magnitude and the same direction. This is because an object moving parallel to a plane mirror has an image that remains at the same distance behind the mirror as the object is in front of it and moves in the same direction as the object. SOLUTION According to the discussion in the REASONING, vIY = vIM + vMY = vIM – vYM This vector equation is equivalent to two equations, one for the x components and one for the y components. For the x direction, we note that vYM,x = –vIM,x. vIY,x = vIM,x − vYM,x = − ( 0.90 m/s ) cos 50.0° − ( 0.90 m/s ) cos 50.0° = −1.2 m/s For the y direction, we note that vYM,y = vIM,y. vIY,y = vIM,y − vYM,y = ( 0.90 m/s ) sin 50.0° − ( 0.90 m/s ) sin 50.0° = 0 m/s Since the y component of the velocity vIY is zero, the velocity of your image relative to you points in the –x direction and has a magnitude of 1.2 m/s . Chapter 25 Problems 12. REASONING The time proportional to the total each path. Therefore, distances identified in the have t reflected t direct = of travel is distance for using the drawing, we 1291 lamp, L 2s mirror, M d LM d LM + d MP d LP (1) d LP 30.0° 30.0° We know that the angle of incidence is equal to the angle of reflection. We also know that the lamp L is a distance 2s from the mirror, while the person P is a distance s from the mirror. d MP s person, P Therefore, it follows that if d MP = d , then d LM = 2 d . We may use the law of cosines (see Appendix E) to express the distance d LP as 2 bd g+ d d LP = 2 2 b gg b g b – 2 2 d d cos 2 30.0 ° SOLUTION Substituting the expressions for d MP , d LM , and d LP into Equation (1), we find that the ratio of the travel times is t reflected t direct = 2d + d 2 bd g+ d 2 2 = bg b g bg – 2 2 d d cos 2 30.0 ° 3 5 – 4 cos 60.0 ° = 1.73 13. REASONING When an object is located very far away from a spherical mirror (concave or convex), the image is located at the mirror’s focal point. Here, the image of the distant object is located 18 cm behind the convex mirror, so that the focal length f of the mirror is f = −18 cm and is negative since the mirror is convex. SOLUTION In constructing a ray diagram, we will need to know the radius R of the mirror. 1 2 The focal length of a convex mirror is related to the radius by f = − R (Equation 25.2). We can use this expression to determine the radius: 1 2 f =− R or R = −2 f = −2 ( −18 cm ) = 36 cm In the ray diagram that follows, we denote the focal point by F and the center of curvature by C. Note t...
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