Physics Solution Manual for 1100 and 2101

00m it follows that l1 400 m l2 400 m 325 m 075

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Unformatted text preview: distance between the eye and the near point, and fo is the focal length of the objective. We can calculate fe from this equation, provided that we have a value for M, since values for all of the other variables are given in the problem statement. Although M is not given directly, we do have a value for the angular size of the image θ ′ = −8.8 ×10−3 rad and ( ) the reference angular size seen by the naked eye when the object is located at the near point (θ = 5.2 × 10−5 rad). From these two angular sizes we can determine the angular magnification using the definition in Equation 26.10: M= θ′ θ (26.10) SOLUTION Substituting Equation 26.10 into Equation 26.11 and solving for fe shows that M= ( L − fe ) N θ′ ≈− fo fe θ fe ≈ or θ ′ f o f e ≈ −θ ( L − f e ) N ( ) or θ ′ f o f e − θ f e N ≈ −θ LN − 5.2 × 10−5 rad (16 cm )( 25 cm ) −θ LN = = 0.86 cm θ ′ fo − θ N −8.8 × 10−3 rad ( 2.6 cm ) − 5.2 × 10−5 rad ( 25 cm ) ( ) ( ) 107. SSM REASONING AND SOLUTION a. The index of refraction n2 of the liquid must match that of the glass, or n2 = 1.50 . b. When none of the light is transmitted into the liquid, the angle of incidence must be equal to or greater than the critical angle. According to Equation 26.4, the critical angle θc is given by sin θc = n2/n1, where n2 is the index of refraction of the liquid and n1 is that of the glass. Therefore, n2 = n1 sin θc = (1.50) sin 58.0° = 1.27 If n2 were larger than 1.27, the critical angle would also be larger, and light would be transmitted from the glass into the liquid. Thus, n2 = 1.27 represents the largest index of refraction of the liquid such that none of the light is transmitted into the liquid. ______________________________________________________________________________ Chapter 26 Problems 1417 108. REASONING AND SOLUTION The actual height d of the diving board above the water can be obtained by using Equation 26.3. As usual, n1 is the index of refraction of the medium (air) associated with the incident ray, and n2 is that of the medium (water) associated with the refracted ray. Taking the refractive index of water from Table 26.1, we find n 1.00 = 3.0 m d = d ′ 1 = (4.0 m) 1.33 n 2 ______________________________________________________________________________ 109. SSM REASONING The far point is 5.0 m from the right eye, and 6.5 m from the left eye. For an object infinitely far away (do = ∞), the image distances for the corrective lenses are then –5.0 m for the right eye and –6.5 m for the left eye, the negative sign indicating that the images are virtual images formed to the left of the lenses. The thin-lens equation [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ] can be used to find the focal length. Then, Equation 26.8 can be used to find the refractive power for the lens for each eye. SOLUTION Since the object distance do is essentially infinite, 1 / do = 1 / ∞ = 0 , and the thin-lens equation becomes 1 / di = 1 / f , or di = f . Therefore, for the right eye, f = –5.0 m , and the refractive power is (see Equation 26.8) [Right eye] Refractive power 1 1 == = –0.20 diopters (in diopters) f (–5.0 m) Similarly, for the left eye, f = –6.5 m , and the refractive power is Refractive power 1 1 == = –0.15 diopters (in diopters) f (–6.5 m) ______________________________________________________________________________ [Left eye ] 110. REASONING AND SOLUTION Using Equation 26.3, we find n 1.546 d = 1 d ' = 2.5 cm = 3.9 cm n2 1.000 ______________________________________________________________________________ 111. SSM REASONING AND SOLUTION Equation 26.6 gives the thin-lens equation which relates the object and image distances do and di , respectively, to the focal length f of the lens: (1 / do ) + (1 / di ) = (1 / f ) . 1418 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS The optical arrangement is similar to that in Figure 26.26. The problem statement gives values for the focal length ( f = 50.0 mm ) and the maximum lens-to-image-sensor distance ( di = 275 mm ). Therefore, the maximum distance that the object can be located in front of the lens at a distance that is as small as do, where do is found as follows: 1 11 1 1 =– = – or do = 61.1 mm do f di 50.0 mm 275 mm ______________________________________________________________________________ 112. REASONING Since we are given the focal length and the object distance, we can use the thin-lens equation to calculate the image distance. From the algebraic sign of the image distance we can tell if the image is real (image distance is positive) or virtual (image distance is negative). Knowing the image distance and the object distance will enable us to use the magnification equation to determine the height of the image. SOLUTION a. Using the thin-lens equation to obtain the image distance di from the focal length f and the object distance do, we find 111 1 1 =− = − = 0.00491 cm −1 di f do 88.00 cm 155.0 cm or di = 204 cm b. The fact that the image distance is...
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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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