Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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Unformatted text preview: on (5) into Equation (2) yields f2 d o2 − f 2 (5) Chapter 26 Problems m = m1m2 = − m1 f 2 do2 − f 2 1397 (6) Substituting Equation (1) into Equation (6), we obtain the overall magnification: m=− m1 f 2 m1 f 2 ( +0.600 ) ( +4.00 m ) = +2.00 =− =− d o2 − f 2 1.60 m + −1.20 m − 4.00 m L + di1 − f 2 73. REASONING AND SOLUTION Let d be the separation of the lenses. The first lens forms its image at 111 1 1 =− = − or di = 80.0 cm di f do 16.0 cm 20.0 cm This first image is inverted and enlarged relative to the coin [see Figure 26.27(a)]. Moreover this image serves as the object for the second lens, which is placed in such a position that it produces a final image that has the same orientation and size as the coin. Therefore the second lens must invert the first image and reduce its size. In other words, the second lens must be located to the right of the first image [see Figure 26.26(a)], with the result that the object distance for the second lens is d – 80.0 cm. According to the thin-lens equation, the reciprocal of the image distance for the second lens is 11 1 =− di′ f d − 80.0 cm (1) The magnification of the first lens is m=− 80.0 cm = −4.00 20.0 cm Since the overall magnification of the combination must be +1.000, the magnification of the second lens must be –0.250. Therefore, applying the magnification equation to the second lens, we have di' = 0.250 (d – 80.0 cm) (2) Substituting Equation (2) into Equation (1) and rearranging yields d = 160.0 cm . ______________________________________________________________________________ 74. REASONING The thin-lens equation, Equation 26.6, can be used to find the distance from the blackboard to her eyes (the object distance). The distance from her eye lens to the retina is the image distance, and the focal length of her lens is the reciprocal of the refractive power (see Equation 26.8). The magnification equation, Equation 26.7, can be used to find the height of the image on her retina. 1398 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION a. The thin-lens equation can be used to find the object distance do. However, we note –1 from Equation 26.8 that 1/f = 57.50 m and di = 0.01750 m, so that 1 11 1 = − = 57.50 m −1 − = 0.36 m −1 or di = 2.8 m d o f di 0.01750 m b. The magnification equation can be used to find the height hi of the image on the retina d 0.01750 m −2 hi = ho − i = ( 5.00 cm ) − (26.7) = −3.1 × 10 cm d 2.8 m o ______________________________________________________________________________ 75. REASONING We will apply the thin-lens equation to solve this problem. In doing so, we must be careful to take into account the fact that the lenses of the glasses are worn at a distance of 2.2 or 3.3 cm from her eyes. SOLUTION a. The object distance is 25.0 cm – 2.2 cm, since it is measured relative to the lenses, which are worn 2.2 cm from the eyes. As discussed in the text, the lenses form a virtual image located at the near point. The image distance must be negative for a virtual image, but the value is not –67.0 cm, because the glasses are worn 2.2 cm from the eyes. Instead, the image distance is –67.0 cm + 2.2 cm. Using the thin-lens equation, we can find the focal length as follows: 1 11 1 1 = += + f do di 25.0 cm − 2.2 cm −67.0 cm + 2.2 cm or f = 35.2 cm b. Similarly, we find 1 11 1 1 = += + or f = 32.9 cm f do di 25.0 cm − 3.3 cm −67.0 cm + 3.3 cm ______________________________________________________________________________ 76. REASONING The glasses form an image of a distant object at her far point, a distance L = 0.690 m from her eyes. This distance is equal to the magnitude |di| of the image distance, which is measured from the glasses to the image, plus the distance s between her eyes and the glasses: L = di + s (1) Chapter 26 Problems 1399 Because distant objects may be considered infinitely distant, the object distance in the 111 thin-lens equation += (Equation 26.6) is do = ∞ , and we see that the image do di f distance is equal to the focal length f of the eyeglasses: 111 11 = + = 0+ = f ∞ di di di or di = f (2) The focal length f (in meters) is equal to the reciprocal of the refractive power (in diopters) 1 of the glasses, according to f = (Equation 26.8). Refractive power SOLUTION Solving Equation (1) for s yields s = L − d i . Substituting Equation (2) into this result, we obtain s = L − di = L − f Substituting f = (3) 1 (Equation 26.8) into Equation (3), we find that Refractive power 1 1 = 0.690 m − = 0.023 m Refractive power − 1.50 diopters ______________________________________________________________________________ s = L− f = L− 77. SSM REASONING The closest she can read the magazine is when the image formed by the contact lens is at the near point of her eye, or di = −138 cm. The image distance is negative because the image is a virtual image (see Section 26.10). Since the focal length is also known, the object distance can be found from the thin-lens equation. SOLUTION The object distance do is rel...
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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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