Object O1 is 15.2 cm to the left of a converging lens with a 10.5-cm focal lengt

Question

Object O1 is 15.2 cm to the left of a converging lens with a 10.5-cm focal length. A second lens is positioned 10.0 cm to the right of the first lens and is observed to form a virtual image at the position of the original object O1. (a) What is the focal length of the second lens? cm (b) What is the overall magnification of this system? 34.5 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. (c) What is the nature (i.e., real or virtual, upright or inverted) of the final image? virtual and inverted real and inverted real and upright virtual and upright

Explanation / Answer

(a) analyzing for the first lens,
1/s + 1/s' = 1/f
1/15.2 + 1/s' = 1/10.5
s' = 33.96 cm

analyzing for the second one,
s = d - s'
s = 10 - 33.96
s = -23.96 cm
1/s + 1/s' = 1/f
1/(-23.96) + 1/(-15.2 - 10) = 1/f
f = - 12.28 cm
it's diverging lens.

(b) M = (s'/s)(s'/s)
M = (33.96/15.2)(-25.2/-23.96)
M = 2.35 times

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