A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length
ID: 1380646 • Letter: A
Question
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. Use ray tracing to draw image 1 and image 2 (the final image), and determine di 2 and hi2. Use the thing lens equation to verify your results and calculate the total magnification mt. Is image 1 real or virtual, upright or inverted, reduced or enlarged. Classify image 2 as well, relative to both image 1 and to the original object.
Explanation / Answer
Note that
1/f = 1/do + 1/di
where
f = focal length = -6 cm
do = object distance = 5 cm
di = image distance
Thus, solving for di,
di = -2.727272727 cm
Thus, the magnification, m = -di/do,
m = 0.545454545
Which is 10.7272 cm to the left of the converging lens.
Thus, it is VIRTUAL, ERECT, REDUCED. [ANSWER]
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Thus, for he second lens,
Note that
1/f = 1/do + 1/di
where
f = focal length = 6 cm
do = object distance = 10.7272 cm
di = image distance
Thus, solving for di,
di = 13.61550178 cm
Thus, the magnification, m = -di/do,
m = -1.269250296
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Thus, for the final image,
mtot = m1 m2 = 0.5454 * -1.269
mtot = -0.6923 [ANSWER]
Also,
di2 = 13.62 cm
hi2 = h mtot = -2.769 cm
which is REAL, INVERTED, REDUCED. [FINAL IMAGE]
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