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Unformatted text preview: ated to the focal length f and the image distance di
by the thinlens equation:
1
11
1
1
=âˆ’=
âˆ’
or d o = 28.0 cm
(26.6)
d o f di 35.1 cm âˆ’138 cm
______________________________________________________________________________
78. REASONING When reading the newspaper with her glasses on, the woman focuses on the
virtual image appearing at her near point, which is a distance N from her eyes. The image is
a distance di from the glasses, and because the image is virtual, the image distance di is
negative. Therefore, we can express the near point distance N as a positive quantity in the
following fashion: 1400 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS N = x âˆ’ di (1) where x = 2.00 cm is the distance between her eyes and the glasses. [For example, if the
image distance is di = âˆ’55 cm, then her near point is N = 2.00 cm âˆ’ (âˆ’55 cm) = 57 cm from
111
+=
her eyes.] We will determine the image distance di from the thinlens equation
d o di f
(Equation 26.6), where do is the distance between the newspaper and her glasses. The focal
length f of the glasses (in meters) is the reciprocal of the refractive power (in diopters),
according to Equation 26.8:
f= 1
= 0.6024 m = 60.24 cm
1.660 diopters (2) We are given the distance D = 42.00 cm between her eyes and the newspaper, so that the
distance do between the glasses and the newspaper is given by
do = D âˆ’ x
SOLUTION Solving (3) 111
+ = (Equation 26.6) for di yields
d o di f
di = 1
11
âˆ’
f do (4) Substituting Equation (3) into Equation (4), we obtain
di = 1
1
=
11
1
1
âˆ’
âˆ’
f do
f Dâˆ’x Substituting Equation (5) into Equation (1), we find that
N = x âˆ’ di = x âˆ’ 1
1
= 2.00 cm âˆ’
= 121 cm
1
1
1
1
âˆ’
âˆ’
f Dâˆ’x
60.24 cm 42.00 cm âˆ’ 2.00 cm (5) Chapter 26 Problems 79. 1401 SSM REASONING AND SOLUTION An optometrist prescribes contact lenses that
have a focal length of 55.0 cm.
a. The focal length is positive (+55.0 cm); therefore, we can conclude that the lenses are
converging .
b. As discussed in the text (see Section 26.10), farsightedness is corrected by converging
lenses. Therefore, the person who wears these lens is farsighted .
c. If the lenses are designed so that objects no closer than 35.0 cm can be seen clearly, we
have do = 35.0 cm . The thinlens equation (Equation 26.6) gives the image distance:
1
11
1
1
=âˆ’
=
âˆ’
di
f do 55.0 cm 35.0 cm di = â€“96.3 cm or Thus, the near point is located 96.3 cm from the eyes.
______________________________________________________________________________
80. REASONING The required refractive power of the contact lenses (in diopters) is the
reciprocal of the focal length fC (in meters) of the contact lenses, according to
Equation 26.8: ( Refractive power )C = 1
fC (1) The focal length fC of the contact lenses must be such that, when a book is a distance of
0.280 m from his eyes, the image is formed at the same distance di, as it is when the man is
111
+=
wearing his glasses. From the thinlens equation
(Equation 26.6), then, we
do di f
have that 1
1
1
+=
doC di fC and 1
doG + 1
1
=
di f G (2) where fG is the focal length of the glasses and doG is the distance between the book and the
glasses. Because the glasses are a distance s = 0.025 m from his eyes, the object distance
doG is shorter than the object distance doC = 0.280 m by 0.025 m:
doG = 0.280 m âˆ’ 0.025 m = 0.255 m. The focal length fG is related to the refractive power of
the glasses (4.20 diopters) by Equation 26.8: 1402 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 1
fG ( Refractive power )G =
SOLUTION According to Equation (1), (3) 1
is the refractive power of the contact lenses,
fC so the first of Equations (2) becomes ( Refractive power )C = 1
1
1
=
+
f C d oC d i Solving the second of Equations (2) for 1/di yields (4) 1
1
1
=
âˆ’
. Substituting this result
di f G doG into Equation (4) gives ( Refractive power )C = 1
1
1
1
1
+=
+
âˆ’
d oC d i d oC f G d oG (5) Substituting Equation (3) into Equation (5), we find that ( Refractive power )C = 1
1
1
1
1
+
âˆ’
=
+ ( Refractive power )G âˆ’
d oC f G d oG d oC
d oG Therefore, the refractive power of the contact lenses should be
1
1
+ 3.80 diopters âˆ’
= 3.45 diopters
0.280 m
0.255 m
______________________________________________________________________________ ( Refractive power )C = 81. REASONING AND SOLUTION
a. The far point of 6.0 m tells us that the focal length of the lens is f = â€“6.0 m. The image
distance can be found using
111
1
1
=âˆ’
=
âˆ’
di f do âˆ’6.0 m 18.0 m or di = âˆ’4.5 m b. The image size as obtained from the magnification is d âˆ’4.5 m hi = ho âˆ’ i = (2.0 m) âˆ’ = 0.50 m
do 18.0 m ______________________________________________________________________________ Chapter 26 Problems 1403 82. REASONING AND SOLUTION We need to determine the focal lengths for Bill's glasses
and for Anne's glasses. Using the thinlens equation we have
[Bill] 1
11
1
1
=
+=
+
f B do di 23.0 cm âˆ’123 cm [Anne] 1
11
1
1
=
+=
+
f A do di 23.0 cm âˆ’73.0 cm or fB = 28.3 cm or fA = 33.6 cm â€²
Now find do for Bill and Anne when they switch glasses. a. Anne:
1
1
1
1
1
=
â...
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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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