Physics 107 HOMEWORK ASSIGNMENT #20
Cutnell & Johnson, 7th edition
Chapter 26: Problems 52, 72, 74, 104, 114
*52 Concept Simulation 26.4 provides the option of exploring the ray diagram that applies to
this problem. The distance between an object and its image formed by a diverging lens is 49.0
cm. The focal length of the lens is −233.0 cm. Find (a) the image distance and (b) the object
distance.
72 A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive
power is 1.660 diopters. Since these eyeglasses do not completely correct her vision, she must
hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm
from her eyes. How far is her near point from her eyes?
**74 The contacts worn by a farsighted person allow her to see objects clearly that are as close
as 25.0 cm, even though her uncorrected near point is 79.0 cm from her eyes. When she is
looking at a poster, the contacts form an image of the poster at a distance of 217 cm from her
eyes. (a) How far away is the poster actually located? (b) If the poster is 0.350 m tall, how tall is
the image formed by the contacts?
104 A slide projector has a converging lens whose focal length is 105.00 mm. (a) How far (in
meters) from the lens must the screen be located if a slide is placed 108.00 mm from the lens? (b)
If the slide measures
, what are the dimensions (in mm) of its image?
*114 An object is in front of a converging lens (f = 0.30 m). The magnification of the lens is m
= 4.0. (a) Relative to the lens, in what direction should the object be moved so that the
magnification changes to m = −4.0? (b) Through what distance should the object be moved?
52. REASONING A diverging lens always produces a virtual image, so that the image distance
di is negative. Moreover, the object distance do is positive. Therefore, the distance between
the object and the image is do + di = 49.0 cm, rather than do – di = 49.0 cm. The equation
do + di = 49.0 cm and the thin-lens equation constitute two equations in two unknowns, and
we will solve them simultaneously to obtain values for di and do.
SOLUTION
a. Solving the equation do + di = 49.0 cm for do, substituting the result into the thin-lens
equation, and suppressing the units give
1 1 1
 
d o di f
1
1
1
 
49.0 – di di –233.0
or
(1)
Grouping the terms on the left of Equation (1) over a common denominator, we have
di  49.0 – di
di  49.0 – di 

49.0
1

di  49.0 – di  –233.0
(2)
Cross-multiplying and rearranging in Equation (2) gives
di  49.0 – di   –11 417
or
di2 – 49.0di –11 417  0
(3)
Using the quadratic formula to solve Equation (3), we obtain
di 
–  –49.0  
 –49.0 2 – 4 1.00  –11 417 

2 1.00 
–85.1 cm
We have discarded the positive root, because we know that di must be negative for the
virtual image.
b. Using the fact that do + di = 49.0 cm, we find that the object distance is
do  49.0 cm – di   49.0 cm  –  –85.1 cm   134.1 cm
72. REASONING The eyeglasses form a virtual image of the newspaper page at her near point.
The image distance is negative, and we can calculate it by using the thin-lens equation. In
doing so, we need to keep two things in mind. First, we must take into account the fact that
she wears her eyeglasses 2.00 cm from her eyes. This is because the distances that appear in
the thin-lens equation are measured with respect to the lens of the eyeglasses, not with respect
to her eyes. Thus, the object distance (the location of the newspaper page) is
do = 42.00 cm  2.00 cm. Second, in using the thin-lens equation we must have a value for
the focal length f, and this is not directly given. However, we do know the refractive power
of the lens, which is the reciprocal of the focal length, according to Equation 26.8. After
calculating the image distance, we can obtain her near point with respect to the eyeglasses by
taking the magnitude of this negative number. Finally, we will account for the fact that the
eyeglasses are 2.00 cm from her eyes.
SOLUTION According to the thin-lens equation, the reciprocal of the image distance di is
1 1 1
 
di f d o
To use this expression, we need a value for the focal length, and we use Equation 26.8 for the
refractive power to obtain it. This equation indicates that the refractive power in diopters is
the reciprocal of the focal length in meters, so we have
Refractive power =
1
f
or
f 
1
1

 0.6024 m  60.24 cm 
Refractive power 1.660 diopters
Remembering that the eyeglasses are worn 2.00 cm from the eyes, we can now apply the thinlens equation:
1 1 1
1
1
 


 0.00840 cm-1
di f do 60.24 cm 42.00 cm  2.00 cm
or
di  119 cm
The magnitude of this value for di is 119 cm and gives the location of the woman’s near point
with respect to the eyeglasses. Since the eyeglasses are worn 2.00 cm from the eyes, the near
point is located at a distance of 121 cm from the eyes .
74. REASONING AND SOLUTION If the near point is 79.0 cm, then di = – 79.0 cm, and
do = 25.0 cm. Using the thin-lens equation, we find that the focal length of the correcting
lens is
d d
(25.0 cm)(79.0 cm)
f  o i 
 36.6 cm
do  di 25.0 cm   79.0 cm 
a. The distance d o to the poster can be obtained as follows:
1 1 1
1
1
  

do
f di 36.6 cm 217 cm
b. The image size is
 d 
 217 cm 
hi  ho   i   (0.350 m)  
  2.43 m
 31.3 cm 
 do 
or
d o = 31.3 cm
104. REASONING The distance from the lens to the screen, the image distance, can be
obtained directly from the thin-lens equation, Equation 26.6, since the object distance and
focal length are known. The width and height of the image on the screen can be determined
by using Equation 26.7, the magnification equation.
SOLUTION
a. The distance di to the screen is
1 1 1
1
1
 


 2.646  104 mm1
di f do 105.00 mm 108.00 mm
so that di  3.78  103 mm = 3.78 m .
b. According to the magnification equation, the width and height of the image on the screen are
Width
 d 
 3.78  103 mm 
2
hi  ho   i    24.0 mm   
   8.40  10 mm
 d 
108 mm 

 o
The width is 8.40  102 mm .
Height
 d 
 3.78  103 mm 
3
hi  ho   i    36.0 mm   
   1.26  10 mm
 d 
108 mm 

 o
The height is 1.26  103 mm .
114. REASONING To find the distance through which the object must be moved, we must
obtain the object distances for the two situations described in the problem. To do this, we
combine the thin-lens equation and the magnification equation, since data for the
magnification is given.
SOLUTION
a. Since the magnification is positive, the image is upright, and the object must be located
within the focal point of the lens, as in Figure 26.28. When the magnification is negative
and has a magnitude greater than one, the object must be located between the focal point and
the point that is at a distance of twice the focal length from the lens, as in Figure 26.27.
Therefore, the object should be moved away from the lens .
b. According to the thin-lens equation, we have
1
1
1


di do
f
or
di 
do f
do – f
(1)
According to the magnification equation, with di expressed as in Equation (2), we have
m–
di
do
–
1  d o f 
f

 
d o d o – f  f – d o
or
do 
f  m – 1
m
(2)
Applying Equation (2) to the two cases described in the problem, we have
 d o + m 
f  m –1 f  4.0 –1 3.0 f


m
4.0
4.0
(3)
 do  – m 
f  m –1 f  –4.0 –1 5.0 f


m
–4.0
4.0
(4)
Subtracting Equation (3) from Equation (4), we find that the object must be moved away
from the lens by an additional distance of
f
 do – m –  do +m  5.0
4.0
–
3.0 f 2.0 f 0.30 m


 0.15 m
4.0
4.0
2.0