A 5.95-mm-high firefly sits on the axis of, and 12.9 cm in front of, the thin le
ID: 1389852 • Letter: A
Question
A 5.95-mm-high firefly sits on the axis of, and 12.9 cm in front of, the thin lens A, whose focal length is 6.21 cm. Behind lens A there is another thin lens, lens B, with local length 29.5 cm. The two lenses share a common axis and are 57.1 cm apart. Is the image of the firefly that lens B forms real or virtual? How far from lens B is this image located (expressed as a positive number)? What is the height of this image (as a positive number)? Is this image upright or inverted with respect to the firefly?Explanation / Answer
Note that
1/f = 1/do + 1/di
where
f = focal length = 6.21 cm
do = object distance = 1.29E+01 cm
di = image distance
Thus, solving for di,
di = 11.97443946 cm
Thus, the magnification, m = -di/do,
m = -0.928251121
Thus, as
hi = 5.95E+00 mm
And hf = m * hi, then
hf = -5.52309417 mm
Thus, for the second interface,
Note that
1/f = 1/do + 1/di
where
f = focal length = 29.5 cm
do = object distance = 4.51E+01 cm
di = image distance
Thus, solving for di,
di = 85.19386136 cm [IT IS REAL, PART A and B]
Thus, the magnification, m = -di/do,
m = -1.887927504
Thus, as
hi = -1.12E+01 mm
And hf = m * hi, then
hf = 21.20746723 mm [ANSWER, PART C]
IT IS UPRIGHT AS hf > 0. [ANSWER, PART D]
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