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Unformatted text preview: g that follows, we locate the image using the two rays labeled a and b,
which originate at the top of the object. a
b F
Object Image 2 0.0 cm 1422 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS a. Measuring according to the scale used in the drawing, we find that the image is located
10.0 cm to the left of the lens. The lens is a diverging lens and forms a virtual image, so the
image distance is di = –10.0 cm .
b. Measuring the heights of the image and the object in the drawing, we find that the
magnification is m = +0.500 .
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119. SSM WWW REASONING The optical arrangement is similar to that in Figure 26.26.
We begin with the thinlens equation, [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ]. Since the
distance between the moon and the camera is so large, the object distance do is essentially
infinite, and 1 / do = 1 / ∞ = 0 . Therefore the thinlens equation becomes 1 / di = 1 / f or
di = f . The diameter of the moon's imagine on the slide film is equal to the image height
hi, as given by the magnification equation (Equation 26.7: hi / ho = – di / do ). When the slide is projected onto a screen, the situation is similar to that in Figure 26.27. In
this case, the thinlens and magnification equations can be used in their usual forms.
SOLUTION
a. Solving the magnification equation for hi gives h i = – ho 50.0 × 10 –3 m –4
= (–3.48 × 10 6 m) m = –4.52 × 10
do 3.85 × 10 8 m di The diameter of the moon's image on the slide film is, therefore, 4.52 × 10 –4 m .
b. From the magnification equation, hi = – ho ( d i / d o ) . We need to find the ratio di / do .
Beginning with the thinlens equation, we have
1
1
1
+
=
do di
f or 1
11
=–
do
f di which leads to di
do = di
f – di
di = di
f –1 Therefore, ( ) d 15.0 m hi = – ho i –1 = – 4.52 ×10 –4 m –1 = –6.12 ×10 –2 m 110.0 ×10 –3 m f The diameter of the image on the screen is 6.12 × 10 –2 m .
______________________________________________________________________________ Chapter 26 Problems 1423 120. REASONING The angular magnification of a magnifying glass is given by Equation
26.10: M ≈ (1/ f ) – (1/ di ) N , where N is the distance from the eye to the nearpoint. For maximum magnification, the closest to the eye that the image can be is at the near point,
with di = – N (where the minus sign indicates that the image lies to the left of the lens and
is virtual). In this case, Equation 26.10 becomes Mmax ≈ N / f + 1 . At minimum
magnification, the image is as far from the eye as it can be ( di = – ∞ ); this occurs when the
object is placed at the focal point of the lens. In this case, Equation 26.10 simplifies to
Mmin ≈ N / f .
Since the woman observes that for clear vision, the maximum angular magnification is
1.25 times larger than the minimum angular magnification, we have Mmax = 1.25Mmin .
This equation can be written in terms of N and f using the above expressions, and then
solved for f.
SOLUTION We have
N
N
+ 1 = 1.25
f
f
{
123
M
M
min
max Solving for f, we find that
f = 0.25 N = (0.25)(25 cm) = 6.3 cm
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121. SSM REASONING A contact lens is placed directly on the eye. Therefore, the object
distance, which is the distance from the book to the lens, is 25.0 cm. The near point can be
determined from the thinlens equation [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ]. SOLUTION
a. Using the thinlens equation, we have 111
1
1
=−
=
−
di f do 65.0 cm 25.0 cm or di = −40.6 cm In other words, at age 40, the man's near point is 40.6 cm. Similarly, when the man is 45,
we have
111
1
1
=−
=
−
or
di = −52.4 cm
di f do 65.0 cm 29.0 cm and his near point is 52.4 cm.
52.4 cm – 40.6 cm = 11.8 cm . Thus, the man’s near point has changed by 1424 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS b. With do = 25.0 cm and di = –52.4 cm , the focal length of the lens is found as follows: 1
1
1
1
1
=
+
=
+
or
f = 47.8 cm
f d o d i 25.0 cm (–52.4 cm)
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122. REASONING The image distance di2 produced by the 2nd lens is related to the object
distance do2 and the focal length f2 by the thinlens equation (Equation 26.6). The focal
length is known, but the object distance is not. However, the problem states that the object
distance is equal to that (do1) of the 1st lens, so do2 = do1. Since the final image distance di1
and the focal length f1 of the 1st lens are known, we can determine the object distance for
this lens by employing the thinlens equation.
SOLUTION The image distance di2 produced by the 2nd lens is related to the object
distance do2 and the focal length f2 by the thinlens equation: 1
1
1
=
−
di2 f 2 do2 (26.6) Since do2 = do1 (the image distance for the 1st lens), Equation 26.6 can be written as
1
1
1
=
−
di2 f 2 do1 (1) The object distance for 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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