Physics Solution Manual for 1100 and 2101

56 sin 398 from equation 261 n c v we find that the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ex mirror of radius R, so its focal length is given by f =−1 R (25.2) 2 b. Both the spacecraft and its image would have the same angular speed ω about the center of the moon because both have the same orbital period. The linear speed v of an orbiting body is related to its angular speed ω by v = rω (Equation 8.9), where r is the radius of the orbit. Therefore, the speeds of the spacecraft and its image are, respectively, vo = roω vi = riω and (1) We have used vo and ro in Equation (1) to denote the orbital speed and orbital radius of the spacecraft, because it is the object. The symbols vi and ri denote orbital speed and orbital radius of the spacecraft’s image. The orbital radius ro of the spacecraft is the sum of the radius R of the moon and the spacecraft’s altitude do = 1.20×105 m above the moon’s surface: ro = R + do (2) The “orbital radius” ri of the image is its distance from the center of the moon. The image is below the surface, and the image distance di is negative. Therefore, the distance between the center of the moon and the image is found from ri = R + di (3) SOLUTION a. Solving Equation 25.3 for 1 and then taking the reciprocal of both sides, we obtain di 1 1 1 do − f =− = di f d o do f or di = Substituting Equation 25.2 into Equation (4), we find that do f do − f (4) Chapter 25 Problems di = do f do − f = ( do − 1 R 2 do − ( −1 2 ) R ) = − 1 do R 2 do + R 1 2 = ( 1307 )( ) = −1.07 ×105 m 6 m + 1 (1.74 ×10 m ) 2 − 1 1.22 × 105 m 1.74 × 106 m 2 1.22 × 10 5 Therefore, the image of the spacecraft would appear 1.07 × 105 m below the surface of the moon. b. Solving the first of Equations (1) for ω yields ω = vo ro . Substituting this result into the second of Equations (2), we find that v r vi = riω = ri o = i r r o o vo (5) Substituting Equations (2) and (3) into Equation (5) yields ( 1.74 × 106 m + −1.07 × 105 m ri R + di vi = vo = vo = 1.74 × 106 m + 1.22 × 105 m ro R + do 36. REASONING AND SOLUTION The ray diagram is shown in the figure (Note: f = 5.0 cm and do = 15.0 cm). a. The ray diagram indicates that the image distance is 7.5 cm in front of the mirror. b. The image height is 1.0 cm , and the image ) (1620 m/s ) = 1420 m/s Object F C Real Image is inverted relative to the object. f di do 1308 THE REFLECTION OF LIGHT: MIRRORS 37. SSM REASONING We have seen that a convex mirror always forms a virtual image as shown in Figure 25.21a of the text, where the image is upright and smaller than the object. These characteristics should bear out in the results of our calculations. SOLUTION The radius of curvature of the convex mirror is 68 cm. Therefore, the focal length is, from Equation 25.2, f = –(1 / 2 ) R = –34 cm . Since the image is virtual, we know that d i = –22 cm . a. With d i = –22 cm and f = –34 cm , the mirror equation gives 1 11 1 1 =– = – do f d i –34 cm –22 cm or d o = +62 cm b. According to the magnification equation, the magnification is m=– di do =– –22 cm = +0.35 62 cm c. Since the magnification m is positive, the image is upright . d. Since the magnification m is less than one, the image is smaller than the object. 38. REASONING a. We are given that the focal length of the mirror is f = 45 cm and that the image distance is one-third the object distance, or di = 1 d o . These two pieces of information, along with the 3 mirror equation, will allow us to find the object distance. b. Once the object distance has been determined, the image distance is one-third that value, since we are given that di = 1 d o . 3 SOLUTION a. The mirror equation indicates that 1 1 1 += d o di f Substituting di = 1 d o and solving for do gives 3 (25.3) Chapter 25 Problems 1 1 1 +1 = do 3 d o f or 4 1 = do f 1309 do = 4 f = 4 ( 45 cm ) = 180 cm or b. The image distance is di = 1 do = 1 (180 cm ) = 6.0 × 101 cm 3 3 39. REASONING The drawing at the right shows the laser beam after reflecting from the plane mirror. The angle of reflection is α, and it is equal to the angle of incidence, which is 33.0°. Note that the angles labeled β in the drawing are also 33.0°, since they are angles formed by a line that intersects two parallel lines. Knowing these angles, we can use trigonometry to determine the distance dDC, which locates the spot where the beam strikes the floor. L β 1.10 m 33.0° α A B 1.80 m β D Floor C SOLUTION Applying trigonometry to triangle DBC, we see that tan β = d BC or d DC d DC = d BC tan β (1) The distance dBC can be determined from d BC = 1.80 m − d AB , which can be substituted into Equation (1) to show that d DC = d BC tan β = 1.80 m − d AB tan β (2) The distance dAB can be found by applying trigonometry to triangle LAB, which shows that d tan β = AB or d AB = (1.10 m ) tan β 1.10 m Substituting this result into Equation (2), gives d DC = 1.80 m − d AB 1.80 m − (1.10 m ) tan β 1.80 m − (1.10 m ) tan 33.0° = = = 1.67 m tan β tan β tan 33.0° 1310 THE REFLECTION OF LIGHT: MIRRORS 40. REASONING The magnification of the mirror is related to th...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online