Physics Solution Manual for 1100 and 2101

# 0 cm 250 cm or di 406 cm in other words at age 40 the

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Unformatted text preview: N θ′ −1 θ = 88. REASONING AND SOLUTION a. With the image at the near point of the eye, the angular magnification is M = N/f +1 (Equation 26.10, with di = −N). Solving this expression for the focal length f and using M = 6.0, we find that f= N 25 cm = = 5.0 cm = 0.050 m M − 1 6.0 − 1 According to Equation 26.8, the refractive power of this lens is Refractive power = 1 1 = = 2.0 × 101 diopters f 0.050 m b. According to Equation26.10, when the image of the stamp is 45 cm from the eye, 1 1 1 1 M = − N = − (0.25 m) = 5.6 0.050 m −0.45 m f di ______________________________________________________________________________ Chapter 26 Problems 1407 89. SSM REASONING AND SOLUTION The information given allows us to determine the near point for this farsighted person. With f = 45.4 cm and do = 25.0 cm, we find from the thin-lens equation that 111 1 1 =– = – di f do 45.4 cm 25.0 cm or di = –55.6 cm Therefore, this person's near point, N, is 55.6 cm. We now need to find the focal length of the magnifying glass based on the near point for a normal eye, i.e., M = N/f + 1 (Equation 26.10, with di = −N), where N = 25.0 cm. Solving for the focal length gives f= N 25.0 cm = = 3.85 cm M − 1 7.50 − 1 We can now determine the maximum angular magnification for the farsighted person N 55.6 cm +1 = + 1 = 15.4 f 3.85 cm ______________________________________________________________________________ M= 90. REASONING AND SOLUTION According to Equation 26.11, the angular magnification of the microscope with the 100-diopter objective is M1 ≈ − ( L − fe ) N f o1 f e and with the 300-diopter objective is M2 ≈ − ( L − fe ) N f o2 f e Division of the equations results in M 2 fo1 300 diopters = = =3 M1 fo2 100 diopters Since the angular magnification of the microscope with the 300-diopter objective is three times greater than that with the 100-diopter objective, the angle will be 3(3 × 10−3 rad) = 9 × 10 –3 rad . ______________________________________________________________________________ 1408 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 91. SSM REASONING The angular magnification of a compound microscope is given by Equation 26.11: ( L − fe ) N M≈– fo fe where fo is the focal length of the objective, fe is the focal length of the eyepiece, and L is the separation between the two lenses. This expression can be solved for fo , the focal length of the objective. SOLUTION Solving for fo , we find that the focal length of the objective is ( L − fe) N (16.0 cm − 1.4 cm)(25 cm) =– = 0.81 cm feM (1.4 cm)(–320) ______________________________________________________________________________ fo = – 92. REASONING AND SOLUTION The angular magnification of a compound microscope is negative, and Equation 26.9 gives M = θ ′ / θ, so that θ′ = M θ = (–160)(4.0 × 10–3 rad) = –0.64 rad The magnitude of this angular size is 0.64 rad . ______________________________________________________________________________ 93. REASONING AND SOLUTION According to Equation 26.11, the angular magnification of the microscope is M ≈– ( L – fe ) N = – (14.0 cm – 2.5 cm ) ( 25.0 cm ) = –2.3 × 102 ( 0.50 cm )( 2.5 cm ) fo f e Now the new angle is ( )( ) θ ′ = M θ = –2.3 × 102 2.1 × 10 –5 rad = – 4.8 × 10 –3 rad The magnitude of the angle is 4.8 × 10 –3 rad . ______________________________________________________________________________ 94. REASONING Equation 26.11 gives the angular magnification of a compound microscope. We can apply this expression to the microscopes of length L and L′ and then set the two angular magnifications equal to one another. From the resulting equation, we will be able to obtain a value for L′. Chapter 26 Problems 1409 SOLUTION According to Equation 26.11, the angular magnification M of a microscope of length L is ( L – fe ) N M ≈– (26.11) fo fe where N is the near point of the viewer’s eye. For a microscope of length L′, Equation 26.11 also applies, but with L replaced by L′ and the focal lengths fo and fe interchanged. The angular magnification M ′ of this microscope is M′ ≈ – ( L′ – f o ) N (1) fe fo Since M′ = M, we have from Equations 26.11 and (1) that – ( L′ – fo ) N = – ( L – fe ) N fe fo fo fe or L′ = L – f e + f o (2) Using Equation (2), we obtain L′ = L – f e + f o = (12.0 cm ) – ( 2.0 cm ) + ( 0.60 cm ) = 10.6 cm ______________________________________________________________________________ 95. REASONING The angular magnification of a compound microscope is given by Equation 26.11. All the necessary data are given in the statement of the problem, so the angular magnification can be calculated directly. In order to find how far the object is from the objective, examine Figure 26.32a. The object distance do1 for the first lens is related to its focal length f1 and image distance di1 by the thin-lens equation (Equation 26.6). From the drawing we see that the image distance is approximately equal...
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