Physics Solution Manual for 1100 and 2101

7 chapter 26 the refraction of light lenses and

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Unformatted text preview: tance. 1296 THE REFLECTION OF LIGHT: MIRRORS SOLUTION According to the mirror equation, we have 1 1 1 + = d o1 d i1 f 14 244 4 3 and First position of object 1 1 1 + = d o2 d i2 f 14 244 4 3 Second position of object Since the focal length is the same in both cases, it follows that 1 1 1 1 + = + d o1 d i1 d o2 d i2 1 1 1 1 1 1 1 = + − = + − = −0.071 cm −1 d i2 d o1 d i1 d o2 25 cm −17 cm 19 cm b g d i2 = −14 cm The negative value for di2 indicates that the image is located 14 cm behind the mirror. 21. REASONING a. We are dealing with a concave mirror whose radius of curvature is 56.0 cm. Thus, the focal length of the mirror is f = 1 R = 28.0 cm (Equation 25.1) The object distance is 2 do = 31.0 cm. With known values for f and do, we can use the mirror equation to (Equation 25.3) find the image distance. b. To determine the image height hi, recall that it is related to the object height ho by the magnification m; hi = mho. The magnification is related to the image and object distances by the magnification equation, m = −di / d o (Equation 25.4). Since we know values for ho, di, and do, we can find the image height hi. SOLUTION a. The image distance is given by the mirror equation as follows: 1 1 1 1 1 = − = − di f do 28.0 cm 31.0 cm so di = 290 cm (25.3) b. Using the magnification equation, we find that the image height is 290 cm d hi = mho = − i ho = − ( 0.95 cm ) = 31.0 cm do −8.9 cm (25.4) Chapter 25 Problems 1297 c. Since hi is negative, the image is inverted relative to the object. Thus, to make the picture on the wall appear normal, the slide must be oriented upside down in the projector. 22. REASONING For an image that is in front of a mirror, the image distance is positive. Since the image is inverted, the image height is negative. Given the image distance, the mirror equation can be used to determine the focal length, but to do so a value for the object distance is also needed. The object and image heights, together with the knowledge that the image is inverted, allows us to calculate the magnification m. The magnification m is given by m = –di/do (Equation 25.4), where di and do are the image and object distances, respectively. SOLUTION According to Equation 25.4, the magnification is m= hi d =− i ho do or do = − d iho hi Substituting this result into the mirror equation, we obtain FI GJ HK − F 1 IL b1.5 cm g 1P 0.11 cm =G − = JMb cm g+ O H cm K 3.5 13 N Q h h 1 1 1 1 1 = + =− i + = − i +1 f do di d i ho d i d i ho −1 or f = 9 .1 cm 23. REASONING Since the image is behind the mirror, the image is virtual, and the image distance is negative, so that di = −34.0 cm . The object distance is given as do = 7.50 cm . The mirror equation relates these distances to the focal length f of the mirror. If the focal length is positive, the mirror is concave. If the focal length is negative, the mirror is convex. SOLUTION According to the mirror equation (Equation 25.3), we have 1 11 = + f d o di or f= 1 1 = = 9.62 cm 11 1 1 + + do di 7.50 cm ( −34.0 cm ) Since the focal length is positive, the mirror is concave . 1298 THE REFLECTION OF LIGHT: MIRRORS 24. SSM REASONING The object distance do is the distance between the object and the 111 mirror. It is found from += (Equation 25.3), where di is the distance between the d o di f mirror and the image, and f is the focal length of the mirror. We are told that the image appears in front of the mirror, so, according to the sign conventions for spherical mirrors, the image distance must be positive: di = +97 cm. SOLUTION Solving 111 + = (Equation 25.3) for do yields d o di f 1 11 =− d o f di do = or 1 11 − f di = 1 1 1 − ( +42 cm ) ( +97 cm ) = 74 cm 25. REASONING Since the focal length f and the object distance do are known, we will use the mirror equation to determine the image distance di. Then, knowing the image distance as well as the object distance, we will use the magnification equation to find the magnification m. SOLUTION a. According to the mirror equation (Equation 25.3), we have 111 =− di f do or di = 1 1 = = −4.3 m 11 1 1 − − f d o ( −7.0 m ) 11 m The image distance is negative because the image is a virtual image behind the mirror. b. According to the magnification equation (Equation 25.4), the magnification is m=− di do =− ( −4.3 m ) = 11 m 0.39 26. REASONING The magnification m is given by m = –di/do (Equation 25.4), where di and do are the image and object distances, respectively. The object distance is known, and we can obtain the image distance from the mirror equation: 1 11 += do di f (25.3) Chapter 25 Problems 1299 SOLUTION Solving the mirror equation (Equation 25.3) for the image distance di gives 1 1 1 + = do di f 1 1 1 do − f =− = di f do fd o or di = or fd o do − f Substituting this result into the magnification equation (Equation 25.4) gives m=− di do =− c fd o / d o − f do h = f f − do Using this result with the given values for the focal length and object distances, we find Smaller object distance m= f −27 .0 cm = = 0.750 f − do −27 .0 cm − 9 .0 cm Greater object distance m= f −27 .0 cm = = 0...
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