Physics Solution Manual for 1100 and 2101

Reasoning and solution from the drawing in the text

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Unformatted text preview: to the distance L between the lenses minus the focal length fe of the eyepiece, or di1 ≈ L – fe. Thus, the object distance is given by 1 1 1 1 1 = − ≈ − do1 f o di1 f o L − fe The magnification due to the objective is given by Equation 26.7 as mobjective = –di1/do1. Since both di1 and do1 are now known, the magnification can be evaluated. SOLUTION a. According to Equation 26.11, the angular magnification of the compound microscope is 1410 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS M ≈− ( L − fe ) N = − ( 26.0 cm − 6.50 cm ) ( 35.0 cm ) = ( 3.50 cm )( 6.50 cm ) fo fe −30.0 b. Using the thin-lens equation, we can determine the object distance from the objective as follows: 1 1 1 1 1 = − = − = 0.234 cm −1 do1 f o L − f e 3.50 cm 26.0 cm − 6.50 cm or d o1 = 4.27 cm . c. The magnification m of the objective is given by Equation 26.7 as di1 26.0 cm − 6.50 cm =− = −4.57 do1 4.27 cm ______________________________________________________________________________ mobjective = − 96. REASONING The angular magnification M of an astronomical telescope is given by f M ≈ − o (Equation 26.12), where fo is the focal length of the objective lens, and fe is the fe focal length of the eyepiece. The length of the barrel must be adjusted so that the image formed by the objective appears very close to the focal point of the eyepiece. The result is that the length L of the barrel is approximately equal to the sum of the focal lengths of the telescope’s two lenses: L ≈ fo + fe . Therefore, if the barrel can only be shortened by 0.50 cm, then the replacement eyepiece can have a focal length fe2 that is only 0.50 cm shorter than the focal length fe1 of the current eyepiece. SOLUTION The focal length of the replacement eyepiece is f e 2 = f e1 − 0.50 cm = 1.20 cm − 0.50 cm = 0.70 cm Therefore, from Equation 26.12, the angular magnification with the second eyepiece in place is fo 180 cm =− = −260 fe 0.70 cm ______________________________________________________________________________ M ≈− Chapter 26 Problems 97. 1411 SSM REASONING Knowing the angles subtended at the unaided eye and with the telescope will allow us to determine the angular magnification of the telescope. Then, since the angular magnification is related to the focal lengths of the eyepiece and the objective, we will use the known focal length of the eyepiece to determine the focal length of the objective. SOLUTION From Equation 26.12, we have M= f θ′ =– o θ fe where θ is the angle subtended by the unaided eye and θ′ is the angle subtended when the telescope is used. We note that θ′ is negative, since the telescope produces an inverted image. Thus, using Equation 26.12, we find fo = – fe θ ′ = 1.1 m 8.0 × 10–5 rad ______________________________________________________________________________ θ =– ( 0.032 m ) ( –2.8 × 10–3 rad ) 98. REASONING The refractive power of the eyepiece (in diopters) is the reciprocal of its focal length fe (in meters), according to the definition in Equation 26.8: ( Refractive power )e = 1 fe (1) To obtain the focal length of the eyepiece, we consider the angular magnification of the telescope. According to Equation 26.12, the angular magnification M is M ≈− fo (26.12) fe To use this expression to determine fe, however, we need a value for fo, which is the focal length of the objective. Although a value for fo is not given directly, the value of the refractive power of the objective is given, and it can be used in Equation 26.8 to obtain fo: ( Refractive power )o = 1 fo (2) SOLUTION Substituting fe and fo from Equations (1) and (2) into Equation 26.12, we find 1412 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS M ≈− fo fe =− 1/ ( Refractive power )o 1/ ( Refractive power )e =− ( Refractive power )e ( Refractive power )o Solving for the refractive power of the eyepiece gives ( Refractive power )e ≈ − M ( Refractive power )o = − ( −132 ) (1.50 diopters ) = 198 diopters 99. SSM REASONING AND SOLUTION The angular magnification of an astronomical telescope, is given by Equation 26.12 as M ≈ – f o / f e . Solving for the focal length of the eyepiece, we find fo 48.0 cm = 0.261 cm M ( –184 ) ______________________________________________________________________________ fe ≈ – =– 100. REASONING AND SOLUTION a. The lens with the largest focal length should be used for the objective of the telescope. Since the refractive power is the reciprocal of the focal length (in meters), the lens with the smallest refractive power is chosen as the objective, namely, the 1.3 - diopter lens . b. According to Equation 26.8, the refractive power is related to the focal length f by Refractive power (in diopters) =1/[f (in meters)] . Since we know the refractive powers of the two lenses, we can solve Equation 26.8 for the focal lengths of the objective and the eyepiece. We find that fo = 1/(1.3 diopters) = 0.77 m . Similarly, for the eyepiece, fe =...
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