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Unformatted text preview: on (5) into Equation (2) yields f2 d o2 − f 2 (5) Chapter 26 Problems m = m1m2 = − m1 f 2 do2 − f 2 1397 (6) Substituting Equation (1) into Equation (6), we obtain the overall magnification:
m=− m1 f 2
m1 f 2
( +0.600 ) ( +4.00 m ) = +2.00
=−
=−
d o2 − f 2
1.60 m + −1.20 m − 4.00 m
L + di1 − f 2 73. REASONING AND SOLUTION Let d be the separation of the lenses. The first lens forms
its image at
111
1
1
=−
=
−
or
di = 80.0 cm
di f do 16.0 cm 20.0 cm This first image is inverted and enlarged relative to the coin [see Figure 26.27(a)]. Moreover
this image serves as the object for the second lens, which is placed in such a position that it
produces a final image that has the same orientation and size as the coin. Therefore the
second lens must invert the first image and reduce its size. In other words, the second lens
must be located to the right of the first image [see Figure 26.26(a)], with the result that the
object distance for the second lens is d – 80.0 cm. According to the thinlens equation, the
reciprocal of the image distance for the second lens is
11
1
=−
di′ f d − 80.0 cm (1) The magnification of the first lens is
m=− 80.0 cm
= −4.00
20.0 cm Since the overall magnification of the combination must be +1.000, the magnification of the
second lens must be –0.250. Therefore, applying the magnification equation to the second
lens, we have
di' = 0.250 (d – 80.0 cm) (2) Substituting Equation (2) into Equation (1) and rearranging yields d = 160.0 cm .
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74. REASONING The thinlens equation, Equation 26.6, can be used to find the distance from
the blackboard to her eyes (the object distance). The distance from her eye lens to the retina
is the image distance, and the focal length of her lens is the reciprocal of the refractive
power (see Equation 26.8). The magnification equation, Equation 26.7, can be used to find
the height of the image on her retina. 1398 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION
a. The thinlens equation can be used to find the object distance do. However, we note
–1
from Equation 26.8 that 1/f = 57.50 m and di = 0.01750 m, so that 1
11
1
= − = 57.50 m −1 −
= 0.36 m −1 or di = 2.8 m
d o f di
0.01750 m
b. The magnification equation can be used to find the height hi of the image on the retina d 0.01750 m −2
hi = ho − i = ( 5.00 cm ) −
(26.7) = −3.1 × 10 cm
d
2.8 m o
______________________________________________________________________________ 75. REASONING We will apply the thinlens equation to solve this problem. In doing so, we
must be careful to take into account the fact that the lenses of the glasses are worn at a
distance of 2.2 or 3.3 cm from her eyes.
SOLUTION
a. The object distance is 25.0 cm – 2.2 cm, since it is measured relative to the lenses, which
are worn 2.2 cm from the eyes. As discussed in the text, the lenses form a virtual image
located at the near point. The image distance must be negative for a virtual image, but the
value is not –67.0 cm, because the glasses are worn 2.2 cm from the eyes. Instead, the
image distance is –67.0 cm + 2.2 cm. Using the thinlens equation, we can find the focal
length as follows: 1
11
1
1
=
+=
+
f do di 25.0 cm − 2.2 cm −67.0 cm + 2.2 cm or f = 35.2 cm b. Similarly, we find
1
11
1
1
=
+=
+
or
f = 32.9 cm
f do di 25.0 cm − 3.3 cm −67.0 cm + 3.3 cm
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76. REASONING The glasses form an image of a distant object at her far point, a distance
L = 0.690 m from her eyes. This distance is equal to the magnitude di of the image
distance, which is measured from the glasses to the image, plus the distance s between her
eyes and the glasses:
L = di + s (1) Chapter 26 Problems 1399 Because distant objects may be considered infinitely distant, the object distance in the
111
thinlens equation
+=
(Equation 26.6) is do = ∞ , and we see that the image
do di f
distance is equal to the focal length f of the eyeglasses: 111
11
= + = 0+ =
f ∞ di
di di or di = f (2) The focal length f (in meters) is equal to the reciprocal of the refractive power (in diopters)
1
of the glasses, according to f =
(Equation 26.8).
Refractive power
SOLUTION Solving Equation (1) for s yields s = L − d i . Substituting Equation (2) into
this result, we obtain
s = L − di = L − f Substituting f = (3) 1
(Equation 26.8) into Equation (3), we find that
Refractive power 1
1
= 0.690 m −
= 0.023 m
Refractive power
− 1.50 diopters
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s = L− f = L− 77. SSM REASONING The closest she can read the magazine is when the image formed by
the contact lens is at the near point of her eye, or di = −138 cm. The image distance is
negative because the image is a virtual image (see Section 26.10). Since the focal length is
also known, the object distance can be found from the thinlens equation. SOLUTION The object distance do is rel...
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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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