Physics Solution Manual for 1100 and 2101

# In doing so we must be careful to take into account

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Unformatted text preview: do2 (2) Substituting Equations (2) into Equation (1) yields x = d i1 − d i2 = 1 1 1 1 − = − = 0.011 m 1 1 1 1 1 1 1 1 − − − − f d o1 f d o2 0.2000 m 3.5 m 0.2000 m 50.0 m 55. REASONING AND SOLUTION a. According to the thin-lens equation, we have 111 1 1 =− = − di f do −25 cm 38 cm or di = −15 cm b. The image is virtual since the image distance is negative. ______________________________________________________________________________ Chapter 26 Problems 1383 56. REASONING A converging lens must be used, because a diverging lens cannot produce a real image. Since the image is one-half the size of the object and inverted relative to it, the image height hi is related to the object height ho by hi = − 1 ho , where the minus sign indicates that the 2 image is inverted. According to the magnification equation, Equation 26.7, the image distance di is related to the object distance do by di / d o = − hi / ho . But we know that hi / ho = − 1 , so 2 () di / d o = − − 1 = 1 . 2 2 SOLUTION a. Let d be the distance between the object and image, so that d = do + di. However, we 3 know from the REASONING that d i = 1 d o , so d = d o + 1 d o = 2 d o . The object distance 2 2 is, therefore, d o = = 2 d = 2 ( 90.0 cm ) = 60.0 cm 3 3 b. The thin-lens equation, Equation 26.6, can be used to find the focal length f of the lens: 1 1 1 1 1 3 = += + = f d o di d o 1 d o d o 2 do 60.0 cm = = 20.0 cm 3 3 ______________________________________________________________________________ f= 57. REASONING AND SOLUTION a. The sun is so far from the lens that the incident rays are nearly parallel to the principal axis, so the image distance di is nearly equal to the focal length of the lens. The magnification of the lens is m=− di 10.0 × 10 −2 m =− = −6.67 × 10−13 do 1.50 × 1011 m The image height hi is hi = mho = ( −6.67 × 10−13 )(1.39 × 109 m) = −9.27 × 10−4 m The diameter of the sun’s image on the paper is the magnitude of hi, or 9.27 × 10–4 m. The area A of the image is A = 1 π (9.27 × 10 −4 m) 2 = 6.74 × 10−7 m 2 4 1384 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS b. The intensity I of the light wave is the power P that strikes the paper perpendicularly divided by the illuminated area A (see Equation 16.8) 0.530 W P = = 7.86 × 105 W/m 2 A 6.74 × 10−7 m 2 ______________________________________________________________________________ I= 58. REASONING A diverging lens always produces a virtual image, so that the image distance di is negative. Moreover, the object distance do is positive. Therefore, the distance between the object and the image is do + di = 49.0 cm, rather than do – di = 49.0 cm. The equation do + di = 49.0 cm and the thin-lens equation constitute two equations in two unknowns, and we will solve them simultaneously to obtain values for di and do. SOLUTION a. Solving the equation do + di = 49.0 cm for do, substituting the result into the thin-lens equation, and suppressing the units give 111 += do di f 1 1 1 += 49.0 – di di –233.0 or (1) Grouping the terms on the left of Equation (1) over a common denominator, we have di + 49.0 – d i di ( 49.0 – di ) = 49.0 1 = di ( 49.0 – di ) –233.0 (2) Cross-multiplying and rearranging in Equation (2) gives di ( 49.0 – di ) = –11 417 or di2 – 49.0di –11 417 = 0 (3) Using the quadratic formula to solve Equation (3), we obtain di = – ( –49.0 ) ± ( –49.0 )2 – 4 (1.00)( –11 417 ) = 2 (1.00 ) –85.1 cm We have discarded the positive root, because we know that di must be negative for the virtual image. b. Using the fact that do + di = 49.0 cm, we find that the object distance is do = 49.0 cm – di = ( 49.0 cm ) – ( –85.1 cm ) = 134.1 cm ______________________________________________________________________________ Chapter 26 Problems 59. 1385 SSM REASONING AND SOLUTION a. A real image must be projected on the drum; therefore, the lens in the copier must be a converging lens . b. If the document and its copy have the same size, but are inverted with respect to one another, the magnification equation (Equation 26.7) indicates that m = – d i / d o = –1 . Therefore, d i / d o = 1 or d i = d o . Then, the thin-lens equation (Equation 26.6) gives 1 1 1 2 + == di do f do do = di = 2 f or Therefore the document is located at a distance 2 f from the lens. c. Furthermore, the image is located at a distance of 2 f from the lens. ______________________________________________________________________________ 60. REASONING AND SOLUTION The focal length of the lens can be obtained from the thin-lens equation as follows: 1 1 1 = + f 4.00 m 0.210 m or f = 0.200 m The same equation applied to the projector gives 1 1 1 = − or do = 0.333 m do 0.200 m 0.500 m ______________________________________________________________________________ 61. SSM REASONING The magnification equation (Equation 26.7) relates the object and image distances do and di , respectively, to the relative size of the of the image and object: m = –( d i / d o ) ....
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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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