Two-lens systems. In the figure, stick figure O (the object) stands on the commo
ID: 1483020 • Letter: T
Question
Two-lens systems. In the figure, 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 = 8.6 cm, lens 1 is diverging, d = 16 cm, and lens 2 is converging. The distance between the lens and either focal point is 5.4 cm for lens 1 and 6.7 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 or virtual, (d) inverted from object O or noninverted, and (e) on the same side of lens 2 as object O or on the opposite side.
Explanation / Answer
Given,
f1 = -5.4 (as its a diverging lens) ; p1 = +8.6 ; d = 16 ; f2 = 6.7
(a)we need to determine the final image distance i2.
we know from lens equation that:
i/f1 = i/p1 + 1/i1
i1 = p1 x f1 / (p1 - f1 ) = 8.6 x (-5.4) / (8.6 + 5.4) = 46.44/14 = -3.32
p2 = d - i1 = 16 - (-3.32) = 19.32
again using the same formula:
i2 = p2 x f2 / (p2 - f2) = 19.32 x 6.7 / (19.32 - 6.7) = 129.44/12.62 = 10.26
Hence, i2 = 10.26
(b)overall magnification M can be calculated as follows:
M = m1 x m2
m1 = -i1/p1 = -(-3.32)/8.6 = 0.39
m2 = -i2/p2 = -i2/p2 = -10.26/19.32 = -0.53
M = m1 x m2 = 0.39 x (-0.53) = -207
Hence, M = -207.
c)Since the final image distance is a positive value, the image is real.
d)Since the magnification is negative the final image is inverted.
e)On the opposite side of the lens.
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