Physics Solution Manual for 1100 and 2101

Reasoning and solution using equation 263 we find n

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Unformatted text preview: ated to the focal length f and the image distance di by the thin-lens equation: 1 11 1 1 =−= − or d o = 28.0 cm (26.6) d o f di 35.1 cm −138 cm ______________________________________________________________________________ 78. REASONING When reading the newspaper with her glasses on, the woman focuses on the virtual image appearing at her near point, which is a distance N from her eyes. The image is a distance di from the glasses, and because the image is virtual, the image distance di is negative. Therefore, we can express the near point distance N as a positive quantity in the following fashion: 1400 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS N = x − di (1) where x = 2.00 cm is the distance between her eyes and the glasses. [For example, if the image distance is di = −55 cm, then her near point is N = 2.00 cm − (−55 cm) = 57 cm from 111 += her eyes.] We will determine the image distance di from the thin-lens equation d o di f (Equation 26.6), where do is the distance between the newspaper and her glasses. The focal length f of the glasses (in meters) is the reciprocal of the refractive power (in diopters), according to Equation 26.8: f= 1 = 0.6024 m = 60.24 cm 1.660 diopters (2) We are given the distance D = 42.00 cm between her eyes and the newspaper, so that the distance do between the glasses and the newspaper is given by do = D − x SOLUTION Solving (3) 111 + = (Equation 26.6) for di yields d o di f di = 1 11 − f do (4) Substituting Equation (3) into Equation (4), we obtain di = 1 1 = 11 1 1 − − f do f D−x Substituting Equation (5) into Equation (1), we find that N = x − di = x − 1 1 = 2.00 cm − = 121 cm 1 1 1 1 − − f D−x 60.24 cm 42.00 cm − 2.00 cm (5) Chapter 26 Problems 79. 1401 SSM REASONING AND SOLUTION An optometrist prescribes contact lenses that have a focal length of 55.0 cm. a. The focal length is positive (+55.0 cm); therefore, we can conclude that the lenses are converging . b. As discussed in the text (see Section 26.10), farsightedness is corrected by converging lenses. Therefore, the person who wears these lens is farsighted . c. If the lenses are designed so that objects no closer than 35.0 cm can be seen clearly, we have do = 35.0 cm . The thin-lens equation (Equation 26.6) gives the image distance: 1 11 1 1 =− = − di f do 55.0 cm 35.0 cm di = –96.3 cm or Thus, the near point is located 96.3 cm from the eyes. ______________________________________________________________________________ 80. REASONING The required refractive power of the contact lenses (in diopters) is the reciprocal of the focal length fC (in meters) of the contact lenses, according to Equation 26.8: ( Refractive power )C = 1 fC (1) The focal length fC of the contact lenses must be such that, when a book is a distance of 0.280 m from his eyes, the image is formed at the same distance di, as it is when the man is 111 += wearing his glasses. From the thin-lens equation (Equation 26.6), then, we do di f have that 1 1 1 += doC di fC and 1 doG + 1 1 = di f G (2) where fG is the focal length of the glasses and doG is the distance between the book and the glasses. Because the glasses are a distance s = 0.025 m from his eyes, the object distance doG is shorter than the object distance doC = 0.280 m by 0.025 m: doG = 0.280 m − 0.025 m = 0.255 m. The focal length fG is related to the refractive power of the glasses (4.20 diopters) by Equation 26.8: 1402 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 1 fG ( Refractive power )G = SOLUTION According to Equation (1), (3) 1 is the refractive power of the contact lenses, fC so the first of Equations (2) becomes ( Refractive power )C = 1 1 1 = + f C d oC d i Solving the second of Equations (2) for 1/di yields (4) 1 1 1 = − . Substituting this result di f G doG into Equation (4) gives ( Refractive power )C = 1 1 1 1 1 += + − d oC d i d oC f G d oG (5) Substituting Equation (3) into Equation (5), we find that ( Refractive power )C = 1 1 1 1 1 + − = + ( Refractive power )G − d oC f G d oG d oC d oG Therefore, the refractive power of the contact lenses should be 1 1 + 3.80 diopters − = 3.45 diopters 0.280 m 0.255 m ______________________________________________________________________________ ( Refractive power )C = 81. REASONING AND SOLUTION a. The far point of 6.0 m tells us that the focal length of the lens is f = –6.0 m. The image distance can be found using 111 1 1 =− = − di f do −6.0 m 18.0 m or di = −4.5 m b. The image size as obtained from the magnification is d −4.5 m hi = ho − i = (2.0 m) − = 0.50 m do 18.0 m ______________________________________________________________________________ Chapter 26 Problems 1403 82. REASONING AND SOLUTION We need to determine the focal lengths for Bill's glasses and for Anne's glasses. Using the thin-lens equation we have [Bill] 1 11 1 1 = += + f B do di 23.0 cm −123 cm [Anne] 1 11 1 1 = += + f A do di 23.0 cm −73.0 cm or fB = 28.3 cm or fA = 33.6 cm ′ Now find do for Bill and Anne when they switch glasses. a. Anne: 1 1 1 1 1 = ...
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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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