Three-lens systems. In the figure, stick figure O (the object) stands on the com
ID: 2258515 • Letter: T
Question
Three-lens systems. In the figure, stick figure O (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to O, which is at object distance p1. Lens 2 is mounted within the middle boxed region, at distance d12 from lens 1. Lens 3 is mounted in the farther boxed region, at distance d23 from lens 2. For this problem, p
1 = 7.4 cm, lens 1 is diverging, d12 = 8.9 cm, lens 2 is diverging, d23 = 6.9 cm, and lens 3 is converging. The distance between the lens and either focal point is 7.8 cm for lens 1, 17 cm for lens 2, and 8.7 cm for lens 3. (You need to provide the proper sign).
Find (a) the image distance i3 for the (final) image produced by lens 3 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real (enter 1) or virtual (enter 0), (d) inverted (enter 1) from object O or noninverted (enter 0), and (e) on the same side of lens 3 as object O (enter 1) or on the opposite side (enter 0).
Please show work, label answers and type out final answers
Explanation / Answer
Given
Obejct distance for first lens is P1=7.4 cm
Focal length of diverging lens f1= -7.8cm
focal length of converging lens f2=+17 cm
distance between the two lenses d= 8.9 cm
Solve:
a)
The image distance for the first lens is
1/f1=1/p1+1/i1
or siplyifyin for i1 we get
i1=f1*p1/p1-f1
= (-7.8cm)*(7.4cm)/7.4cm+7.8cm
= -3.79736842105 cm
Here the image i1 is 3.79736842105 cm to the left of first lens and as the two lenses are separated by 8.9cm.
so the oject distance for the second lens is 8.9cm+3.79736842105 cm= 12.6973684211 cm = p2
The image distance for the second lens is
i2=f2*p2/p2-f2
= (17cm)(12.69cm)/(12.697cm-17cm)
= -50.1347896816 cm
b)
The overall magnification is
M=m1*m2
= (-i1/p1)*(-i2/p2)
= (-3.79736842105cm/7.4cm)*(-50.1345cm/12.69cm)
=2.02714093349
c)
As the final image distance is positive thus the final image is real
d)
As the overall magnification is +ve the final image is real
e)
As the final image is real,the final image is on the side opposite from the objectrelative to lens 2.
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