Physics Solution Manual for 1100 and 2101

# When the plane mirror replaces the convex mirror the

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Unformatted text preview: these values into Equation (1), we obtain vIY = (–0.90 m/s) – (+0.90 m/s) = –1.80 m/s 5. SSM REASONING The geometry is shown below. According to the law of reflection, the incident ray, the reflected ray, and the normal to the surface all lie in the same plane, and the angle of reflection θ r equals the angle of incidence θ i . We can use the law of reflection and the properties of triangles to determine the angle θ at which the ray leaves M2. 1286 THE REFLECTION OF LIGHT: MIRRORS θ γ β M2 65° φ α 120° M1 SOLUTION From the law of reflection, we know that φ = 65° . We see from the figure that φ + α = 90° , or α = 90° – φ = 90° – 65° = 25° . From the figure and the fact that the sum of the interior angles in any triangle is 180°, we have α + β + 120° = 180° . Solving for β , we find that β = 180° – (120° + 25°) = 35° . Therefore, since β + γ = 90° , we find that the angle γ is given by γ = 90° – β = 90° – 35° = 55° . Since γ is the angle of incidence of the ray on mirror M2, we know from the law of reflection that θ = 55° . 6. REASONING AND SOLUTION a. The height of the shortest mirror would be one-half the height of the person. Therefore, h = H/2 = (1.70 m + 0.12 m)/2 = 0.91 m b. The bottom edge of the mirror should be above the floor by h' = (1.70 m)/2 = 0.85 m 7. REASONING AND SOLUTION The two arrows, A and B are located in front of a plane mirror, and a person at point P is viewing the image of each arrow. As discussed in Conceptual Example 1, light emanating from the arrow is reflected from the mirror and is reflected toward the observer at P. In order for the observer to see the arrow in its entirety, both rays, the one from the top of the arrow and the one from the bottom of the arrow, must pass through the point P. Chapter 25 Problems Plane mirror P 1287 Plane mirror P A B According to the law of reflection, all rays will be reflected so that the angle of reflection is equal to the angle of incidence. The ray from the top of arrow A strikes the mirror and reflects so that it passes through point P. Likewise, the ray from the bottom of the arrow is reflected such that it too passes through point P. Therefore, the observer at P sees the arrow at A in its entirety . Similar reasoning shows that the ray from the top of arrow B passes through point P. However, as the drawing shows, the ray from the bottom of the arrow does not pass through P. This conclusion is true no matter where the bottom ray strikes the mirror. The observer does not see the arrow at B in its entirety . 8. REASONING The drawing shows the ray of light reflecting from the mirror and striking the floor. The angle of reflection θ is the same as the angle of incidence. The angle θ is also the angle that the light ray makes with the floor (see the drawing). Therefore, we can use the inverse tangent function to find θ as a function of y and x; y θ = tan −1 . Since the mirror is facing x east and the sun is rising, the angle of incidence θ becomes larger (see the drawing). As the angle θ becomes larger, the distance x becomes smaller. θ θ y θ x SOLUTION When the horizontal distance is x1 = 3.86 m, the angle of incidence θ1 is y −1 1.80 m = tan 3.86 m = 25.0° x1 θ1 = tan −1 When the horizontal distance is x2 = 1.26 m, the angle of incidence θ2 is 1288 THE REFLECTION OF LIGHT: MIRRORS y −1 1.80 m = tan 1.26 m = 55.0° x2 θ 2 = tan −1 As the sun rises, the change in the angle of incidence is 55.0° − 25.0° = 30.0°. Since the earth rotates at a rate of 15.0° per hour, the elapsed time between the two observations is 1h Elapsed time = ( 30.0° ) = 2.00 h 15.0° 9. REASONING AND SOLUTION a. After the mirror has been rotated, the new angle of incidence is θi = 45° + 15° = 60°. The angle of reflection, then, is also equal to 60°. The reflected ray which was originally 90° (45° + 45°) from the original angle of incidence, is now 120° (60° + 60°) from the incident ray's direction. Therefore, the reflected ray has been rotated through β = 120° − 90° = 30° b. The angle through which the reflected ray is rotated depends only on the angle through which the mirror is rotated, and is independent of the angle of incidence. Therefore, β ′ = 30° . 10. REASONING a. The incident ray from the laser travels a distance of L = 50.0 km due north to the mirror (see the drawing, which is not to scale). The angle θ we seek, that between the normal to the surface of the mirror and due south, is the angle of incidence. The angle of incidence must equal the angle of reflection, so the angle α between the incident and reflected rays is bisected by the normal to the surface of the mirror. Therefore, we have that Mirror North θ L θ East α South Laser Detector d θ = 1α 2 (1) Because the laser, the mirror, and the detector sit at the corners of a right triangle, we will use α = tan −1 ( d L ) (Equation 1.6) to determine the angle α, where d = 117 m is the distance between...
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