Physics Solution Manual for 1100 and 2101

0400 sin 2 red solving this equation for n2 violet

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Unformatted text preview: = − ′ do f B di 28.3 cm −73 cm or ′ do = 20.4 cm Relative to the eyes, this becomes 20.4 cm + 2.00 cm = 22.4 cm . b. Bill: 1 1 1 1 1 = −= − ′ do f A di 33.6 cm −123 cm or ′ do = 26.4 cm Relative to the eyes, this becomes 26.4 cm + 2.00 cm = 28.4 cm . ______________________________________________________________________________ 83. SSM REASONING The angular size of a distant object in radians is approximately equal to the diameter of the object divided by the distance from the eye. We will use this definition to calculate the angular size of the quarter, and then, calculate the angular size of the sun; we can then form the ratio θ quarter / θ sun . SOLUTION The angular sizes are θ quarter ≈ 2.4 cm = 0.034 rad 70.0 cm and θ sun ≈ 1.39 × 109 m 1.50 × 1011 m = 0.0093 rad Therefore, the ratio of the angular sizes is θ quarter 0.034 rad = 3.7 0.0093 rad ______________________________________________________________________________ θ sun = 1404 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 84. REASONING The angular magnification M of a magnifying glass is given by 1 1 M ≈ − f d i N (26.10) where f is the focal length of the lens, di is the image distance, and N is the near point of the eye. The focal length and the image distance are related to the object distance do by the thin-lens equation: 11 1 −= (26.6) f di d o These two relations will allow us to determine the angular magnification. SOLUTION Substituting Equation 26.6 into Equation 26.10 yields 1 1 N 72 cm M ≈ − N = = = 18 f d d o 4.0 cm i ______________________________________________________________________________ 85. REASONING The distance between the work and the magnifying glass is the object distance do. This distance can be calculated by using the thin-lens equation, since the image distance and the focal length are known. The angular magnification of the magnifying glass is given by Equation 26.10. SOLUTION a. The object distance do is 1 11 1 1 =−= − do f di 9.50 cm −25.0 cm or do = 6.88 cm (26.6) Note that di = −25.0 cm, since the image falls to the left of the lens; see Figure 26.39b. b. The angular magnification M of the magnifying glass is 1 1 1 1 ( (26.10) M = − N = − 25.0 cm ) = 3.63 9.50 cm −25.0 cm f di ______________________________________________________________________________ Chapter 26 Problems 1405 86. REASONING a. When the filling is viewed from a distance do without magnifying glasses its angular size h θ (in radians) is given by θ ≈ o , where ho is the object height (the diameter of the filling, do in this case). The greatest angular size occurs when the object distance do is as small as possible. Without magnifying glasses, the smallest object distance is the dentist’s near point N: do = N = 17.0 cm. b. Wearing the magnifying glasses enables the dentist to get closer to the filling and see an enlarged image of the filling. The increased angular size θ ′ of the filling then depends upon θ′ 1 1 the focal length f of the lens and the image distance di, according to ≈ − N θ f di (Equation 26.10), where θ is the angular size of the filling found in part (a). The largest possible angular size θ ′ is obtained when the image appears at the near point N, so we will take di = −N, where the negative sign indicates that the image is virtual. SOLUTION a. The diameter of the filling is ho = 2.4 mm, and the object distance is equal to the dentist’s near point: do = N = 17.0 cm. One centimeter equals ten millimeters, so the object distance h can be expressed as do = 170 mm. From θ ≈ o , the angular size of the filling is do θ≈ b. Solving ho do = 2.4 mm = 0.014 rad 170 mm θ′ 1 1 ≈ − N (Equation 26.10) for θ ′ yields θ f di 1 1 − f di θ′ ≈θ N (1) To obtain the maximum possible angular size, we take di = −N in Equation (1), which gives 1 1 1 N 1 17.0 cm + 1 = 0.054 rad N = θ + N = θ + 1 = ( 0.014 rad ) f (−N ) 6.0 cm f N f ______________________________________________________________________________ θ′ ≈θ − 1406 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 87. REASONING The angular magnification M of a magnifying glass is given by Equation 26.10 as θ′ 1 1 M = ≈ − N θ f di where θ ′ = 0.0380 rad is the angular size of the final image produced by the magnifying glass, θ = 0.0150 rad is the reference angular size of the object seen at the near point without the magnifying glass, and N is the near point of the eye. The largest possible angular magnification occurs when the image is at the near point of the eye, or di = –N, where the minus sign denotes that the image lies on the left side of the lens (the same side as the object). This equation can be solved to find the focal length of the magnifying glass. SOLUTION Letting di = –N, and solving Equation 26.10 for the focal length f gives 21.0 cm = 13.7 cm 0.0380 rad −1 0.0150 rad ______________________________________________________________________________ f=...
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