Two-lens systems. In figure 34-44, stick figure O (the object) stands on the com
ID: 2051039 • Letter: T
Question
Two-lens systems. In figure 34-44, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p1. Lens 2 is mounted within the farther boxed region, at distance d. For this problem, p1 = 9.0 cm, lens 1 is diverging, d = 13 cm, and lens 2 is converging. The distance between the lens and either focal point is 5.5 cm for lens 1 and 5.7 cm for lens 2. (You need to provide the proper sign).Find (a) the image distance i2 for the image produced by lens 2 (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 2 as object O (enter 0) or on the opposite side (enter 1).
(a) unit cm
(b) no units
(c) 1 - real
(d) 1 - inverted
(e) 1 - opposite side
just need help on (a) and (b)
Explanation / Answer
Given data:Obejct distance for first lens, p1 = 9.0 cm
Focal length of diverging lens, f1 = -5.5 cm Focal length of converging lens, f2 = +5.7 cm Distance between the two lenses, d = 13 cm -------------------------------------------------------------------------------------------- Solution: a) The image distance for the first lens can be calculated as 1/f1 = 1/p1+1/i1 i1 = f1p1/p1-f1 = (-5.5 cm)(9.0 cm)/(9.0 cm+5.5 cm) = -3.4138 cm Here the image i1 is 3.4138 cm to the left of first lens and as the two lenses are separated by 13 cm. So, the oject distance for the second lens can be calculated as p2 = 13 cm + 3.4138 cm = 16.4138 cm The image distance for the second lens can be calculated as i2 = f2p2/p2-f2 = (5.7 cm)(16.4138 cm)/(16.4138 cm - 5.7 cm) = 8.73 cm ----------------------------------------------------------------------------------------------- = (5.7 cm)(16.4138 cm)/(16.4138 cm - 5.7 cm) = 8.73 cm ----------------------------------------------------------------------------------------------- b)
The overall magnification can be calculated as M = m1m2 = (-i1/p1)(-i2/p2) = (3.4138 cm / 9.0 cm)(-8.73 cm / 16.4138 cm) = -0.2018 ----------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- c)
As the final image distance is positive thus the final image is real. ----------------------------------------------------------------------------------------------- d)
As the overall magnification is negative thus the final image is inverted. ----------------------------------------------------------------------------------------------- e) As the final image is real,the final image is on the side opposite from the object relative to lens 2.
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