Physics Solution Manual for 1100 and 2101

00 m is the focal length of the diverging lens to

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Unformatted text preview: _______________________________ 50. REASONING a. Given the focal length (f = −0.300 m) of the lens and the image distance di, we will 111 employ the thin-lens equation += (Equation 26.6) to determine how far from the d o di f van the person is actually standing, which is the object distance do. The image and the person are both behind the van, so the image is virtual, and the image distance is negative: di = −0.240 m. b. Once we have determined the object distance do, we will use the magnification equation h d m = i = − i (Equation 26.7) to calculate the true height ho of the person from the height ho do hi = 0.34 m of the image. SOLUTION a. Solving Equation 26.6 for do, we obtain 1 11 =− do f di or do = 1 1 = = 1.2 m 11 1 1 − − f di −0.300 m ( −0.240 m ) b. Taking the reciprocal of both sides of hi d = − i (Equation 26.7) and solving for ho ho do yields ho d =− o hi di or d 1.2 m ho = −hi o = − ( 0.34 m ) = 1.7 m d −0.240 m i 1380 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS ______________________________________________________________________________ 51. REASONING Since the object distance and the focal length of the lens are given, the thin-lens equation (Equation 26.6) can be used to find the image distance. The height of the image can be determined by using the magnification equation (Equation 26.7). SOLUTION a. The object distance do, the image distance di, and the focal length f of the lens are related by the thin-lens equation: 1 11 += (26.6) d o di f Solving for the image distance gives 111 1 1 =− = − di f do 12.0 cm 8.00 cm or di = −24 cm b. The image height hi (the height of the magnified print) is related to the object height ho, the image distance di, and the object distance do by the magnification equation: d −24 cm (26.7) hi = − ho i = − ( 2.00 mm ) = 6.0 mm do 8.00 cm ______________________________________________________________________________ 52. REASONING The height of the mountain’s image is given by the magnification equation as hi = –hodi/do. To use this expression, however, we will need to know the image distance di, which can be determined using the thin-lens equation. Knowing the image distance, we can apply the expression for the image height directly to calculate the desired ratio. SOLUTION According to the thin-lens equation, we have 1 1 1 + = di do f (1) For both pictures, the object distance do is very large compared to the focal length f. Therefore, 1/do is negligible compared to 1/f, and the thin-lens equation indicates that di = f. As a result, the magnification equation indicates that the image height is given by hi = – ho di do =– ho f do (2) Chapter 26 Problems 1381 Applying Equation (2) for the two pictures and noting that in each case the object height ho and the focal length f are the same, we find ( hi )5 km ( hi )14 km ho f – d ( do )14 km 14 km o 5 km = = = = 2.8 ho f ( do )5 km 5.0 km – d o 14 km ______________________________________________________________________________ 53. REASONING The distance from the lens to the screen, the image distance, can be obtained directly from the thin-lens equation, Equation 26.6, since the object distance and focal length are known. The width and height of the image on the screen can be determined by using Equation 26.7, the magnification equation. SOLUTION a. The distance di to the screen is 11 1 1 1 =− = − = 2.646 × 10−4 mm −1 di f do 105.00 mm 108.00 mm so that di = 3.78 × 103 mm = 3.78 m . b. According to the magnification equation, the width and height of the image on the screen are Width d 3.78 × 103 mm 2 hi = ho − i = ( 24.0 mm ) − = − 8.40 × 10 mm d 108 mm o The width is 8.40 × 10 2 mm . Height d hi = ho − i d o 3.78 × 103 mm 3 = ( 36.0 mm ) − = − 1.26 × 10 mm 108 mm The height is 1.26 × 103 mm . ______________________________________________________________________________ 1382 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 54. REASONING In order for a sharply focused image to appear on the image sensor, the distance between the sensor and the lens must be equal to the image distance di, which can 111 be found from the thin-lens equation += (Equation 26.6), where d o di f f = 200.0 mm = 0.2000 m is the focal length of the lens and do is the distance between the object and the lens. Since there are two object distances (do1, do2), there will be two image distances (di1, di2). When the converging lens forms a real image, increasing the object distance decreases the image distance. Therefore, the smaller object distance of do1 = 3.5 m will yield the greater image distance di1, and the greater object distance of do2 = 50.0 m will yield the smaller image distance di2. Thus, the distance x that the lens must move is equal to the difference between the image distances: x = d i1 − d i2 (1) 111 1 += (Equation 26.6) for di yields di = . Using this 11 d o di f − f do result with the first and second object distances, we have that SOLUTION Solving di1 = 1 1 1 − f do1 and di2 = 1 1 1 − f...
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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.

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