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Unformatted text preview: 1/(11 diopters) = 0.091 m . Therefore, the distance between the lenses should be L ≈ f o + f e = 0.77 m + 0.091 m= 0.86 m
c. The angular magnification of the telescope is given by Equation 26.12 as
fo 0.77 m
=–
= –8.5
fe
0.091 m
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M≈– 101. REASONING From the discussion of the telescope in Section 26.13, we know that the
length L of the barrel is approximately equal to the focal length fo of the objective plus the
focal length fe of the eyepiece; L ≈ f o + f e . In addition, the angular magnification M of a
telescope is given by M ≈ − f o / f e (Equation 26.12). These two relations will permit us to
determine the focal lengths of the objective and the eyepiece. Chapter 26 Problems 1413 SOLUTION
a. Since L ≈ f o + f e , the focal length of the objective can be written as fo ≈ L − fe (1) Solving the expression for the angular magnification (Equation 26.12) for fe gives
f e ≈ − f o / M . Substituting this result into Equation (1) gives f
fo ≈ L − − o M or fo ≈ L
1
1−
M = 1.500 m
= 1.482 m
1
1−
−83.00 b. Using Equation (1), we have
f e ≈ L − f o = (1.500 m ) − (1.482 m ) = 0.018 m
______________________________________________________________________________ 102. REASONING The angular magnification M of a telescope is given by Equation 26.12: M ≈− fo (26.12) fe To achieve large values for M, the focal length fo of the objective needs to be greater than the
focal length fe of the eyepiece.
In an astronomical telescope one of the focal points of the objective falls virtually at the same
place as one of the focal points of the eyepiece (see Section 26.13). Therefore, the length L
of the telescope is approximately equal to the sum of the focal lengths: L ≈ fo + fe (1) Solving Equation (1) for fo and substituting the result into Equation 26.12 gives M ≈− fo
fe ≈− L − fe
fe SOLUTION Applying Equation (2) to each telescope, we find (2) 1414 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS MA ≈ − MB ≈ − MC ≈ − LA − f e
fe
LB − f e
fe
LC − f e =− ( 455 mm − 3.00 mm ) = −151 =− ( 615 mm − 3.00 mm ) = −204 =− ( 824 mm − 3.00 mm ) = 3.00 mm 3.00 mm −274
fe
3.00 mm
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103. REASONING AND SOLUTION
a. According to Equation 26.12, the magnification is M =− fo
fe =− 19.4 m
= –194
0.100 m b. The angular size of the crater is θ= ho
do = 1500 m
= 4.0 ×10−6 rad
3.77 ×108 m The angular magnification is, M = θ ' /θ ( Equation 26.12), so that θ ' = Mθ = (–194)(4.0 × 10–6 rad) = –7.8 × 10–4 rad
Since hi′ = θ ' fe, we have
hi′ = θ ′ f e = (–7.8 × 10–4 rad)(0.100 m) = –7.8 × 10 –5 m c. The apparent distance is shorter by a factor of 194, so
3.77 × 108 m
= 1.94 × 106 m
194
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Apparent distance = 104. REASONING The angular magnification of a refracting telescope is 32 800 times larger
when you look through the correct end of the telescope than when you look through the
wrong end. We wish to find the angular magnification, M = − fo / fe (see Equation 26.12) of
the telescope. Thus, we proceed by finding the ratio of the focal lengths of the objective and
the eyepiece and using Equation 26.12 to find M. Chapter 26 Problems 1415 SOLUTION When you look through the correct end of the telescope, the angular
magnification of the telescope is M c = − f o / f e . If you look through the wrong end, the
roles of the objective and eyepiece lenses are interchanged, so that the angular
magnification would be M w = − f e / f o . Therefore, Mc
Mw = − fo / fe 2 f = o = 32 800 − fe / fo fe or fo
fe = ± 32 800 = ±181 The angular magnification of the telescope is negative, so we choose the positive root and
obtain M = − f o / f e = – ( +181) = −181 .
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105. SSM REASONING The ray diagram is constructed by drawing the paths of two rays
from a point on the object. For convenience, we choose the top of the object. The ray that is
parallel to the principal axis will be refracted by the lens and pass through the focal point on
the right side. The ray that passes through the center of the lens passes through undeflected.
The image is formed at the intersection of these two rays on the right side of the lens. SOLUTION The following ray diagram (to scale) shows that di = 18 cm and reveals a
real, inverted, and enlarged image.
Scale: 3 cm
Ob ject F
F
Image ______________________________________________________________________________
106. REASONING The focal length fe of the eyepiece can be determined by using
Equation 26.11:
( L − fe ) N
M ≈−
(26.11)
fo fe 1416 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS where M is the angular magnification, L is the distance between the objective and the
eyepiece, N is the...
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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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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