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Two converging lenses with focal lengths of 40 cm and 20 cm
are 16 cm apart. A 4.0 cm -tall object is 12 cm in front of the
40 cm -focal-length lens.
Calculate the image position.
Calculate the image height.
Solution
Using lens law for the first lens
1/f = 1/p + 1/q
1/40 = 1/12 + 1/q
q = -17.14 cm
Since the lenses are 16 cm apart, that image is 33.14 cm from
the second lens, and we will use that as the object
distance for the second lens
1/20 = 1/33.14 + 1/q
q = 50.4 cm
This means the final image is 50.4 cm in back of the second
lens
The distance between the object and the image is adding the
distance from the object to the first lens + the lens
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separation + the distance the final image is away from the
second lens.
D = (12 + 16 + 50.4) = 78.4 cm
The maginifcation for the first lens is -q/p = -(-17.14)/(12) =
1.43
The maginfication for the second lens is -q/p = -50.4/33.14 =
-1.52
Total magnification is (1.43)(-1.52) = -2.17
So, image height is (-2.17)(4) = -8.70 cm image is inverted..

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