Unformatted text preview: lies inside the focal point of the
converging (#1) lens. According to Figure 26.28, such an object produces a virtual image
that lies to the left of the lens. This image act as the object for the diverging (#2) lens.
Since a diverging lens always produces a virtual image that lies to the left of the lens, the
final image lies to the left of the diverging lens. In part (b), the diverging (#1) lens produces
a virtual image that lies to the left of the lens. This image act as the object for the
converging (#2) lens. Since the object lies outside the focal point of the converging lens,
the converging lens produces a real image that lies to the right of the lens (see Figure 26.26).
Thus, the final image lies to the right of the converging lens.
SOLUTION
a. The focal length of lens #1 is f1 = 15.00 cm, and the object distance is do1 = 10.0 cm.
The image distance di1 produced by the first lens can be obtained from the thinlens
equation, Equation 26.6: 1392 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 1
1
1
1
1
=−
=
−
= − 3.33 × 10−2 cm−1 or di1 = − 30.0 cm
di1 f1 do1 15.00 cm 10.00 cm
This image is located to the left of lens #1 and serves as the object for lens #2. Thus, the
object distance for lens #2 is di2 = 30.0 cm + 50.0 cm = 80.0 cm. The image distance
produced by lens #2 is 1
1
1
1
1
−2
−1
=
−
=
−
= − 6.25 × 10 cm or di2 = −16.0 cm
di2 f 2 do2 −20.00 cm 80.0 cm
The negative value for di2 indicates that, as expected, the final image is to the left of lens #2.
b. The focal length of the lens #1 is f1 = –20.0 cm, and the object distance is
do1 = 10.00 cm. The image distance di1 produced by the first lens can be obtained from the
thinlens equation, Equation 26.6.
1
1
1
1
1
−1
−1
=−
=
−
= − 1.50 × 10 cm or di1 = − 6.67 cm
di1 f1 do1 −20.0 cm 10.00 cm
This image is located to the left of lens #1 and serves as the object for lens #2. Thus the
object distance for lens #2 is di2 = 6.67 cm + 50.0 cm = 56.7 cm. The image distance
produced by lens #2 is
1
1
1
1
1
−2
−1
=
−
=
−
= 4.90 × 10 cm or di2 = 20.4 cm
di2 f 2 do2 15.00 cm 56.7 cm
The positive value for di2 indicates that, as expected, the final image is to the right of lens #2.
______________________________________________________________________________
69. SSM REASONING The problem can be solved using the thinlens equation
[Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ] twice in succession. We begin by using the thin
lensequation to find the location of the image produced by the converging lens; this image
becomes the object for the diverging lens.
SOLUTION
a. The image distance for the converging lens is determined as follows: 1
1
1
1
1
=–
=
–
d i1 f d o1 12.0 cm 36.0 cm or d i1 = 18.0 cm This image acts as the object for the diverging lens. Therefore, Chapter 26 Problems 1
1
1
1
1
=–
=
–
d i2
f d o2 –6.00 cm (30.0 cm –18.0 cm ) or 1393 d i2 = –4.00 cm Thus, the final image is located 4.00 cm to the left of the diverging lens .
b. The magnification equation (Equation 26.7: hi / ho = – di / do ) gives d i1
18.0 cm
mc = –
=–
= –0.500
d o1
36.0 cm
144444 2444444
4
3 d i2
–4.00 cm
md = –
=–
= 0.333
d
12.0 cm
1444o2 4 2444444
44
3 Converging lens Diverging lens Therefore, the overall magnification is given by the product m c md = –0.167 .
c. Since the final image distance is negative, we can conclude that the image is virtual .
d. Since the overall magnification of the image is negative, the image is inverted .
e. The magnitude of the overall magnification is less than one; therefore, the final image is
smaller .
______________________________________________________________________________
70. REASONING The following drawing shows the arrangement of the lenses and the location
of the object. The thinlens equation can be used to locate the final image produced by the
converging lens. We know the focal length of this lens, but to determine the final image
distance from the thinlens equation, we also need to know the object distance, which is not
given. To obtain this distance, we recall that the image produced by one lens (the diverging
lens) is the object for the next lens (the converging lens). We can use the thinlens equation
to locate the image produced by the diverging lens, since the focal length and the object
distance for this lens are also given. The location of this first image relative to the converging
lens will tell us the object distance that we need. To find the height of the final image, we
will use the magnification equation twice, once for the diverging lens and once for the
converging lens.
20.0 cm 10.0 cm
30.0 cm 1394 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS SOLUTION
a. Using the thinlens equation to obtain the distance di of the first image from the diverging
lens, which has a focal length f and an object distance do, we find 111
1
1
=−
=
−
= −0.200 cm 1
di f do −10.0 cm 10.0 cm or di = −5.00 cm The minus sign indicates that this first image...
View
Full
Document
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

Click to edit the document details