An object 1.80 cm tall is placed 100 cm in front of a diverging lens having a focal length of magnitude 25 cm.A converging lens with a focal length of magnitude 33.33 cm is placed 30 cm past the first lens.What is the lateral magnification of this system of lenses?
A) 2.5
B) -2.5
C) -0.40
D) 1.0
E) 0.40

Question

Quizplus · Accepted Answer

First, we can use the thin lens equation to find the image distance for the diverging lens:
1/f = 1/di - 1/do
1/(-25) = 1/di - 1/100
di = -50 cm
(Note that the negative sign indicates that the image is virtual and upright.)

Next, we can use the thin lens equation again to find the final image distance for the whole system:
1/f = 1/di - 1/do
1/33.33 = 1/-50 - 1/30
di = -20 cm
(Note that the negative sign again indicates a virtual and upright image.)

Finally, we can use the magnification equation to find the lateral magnification:
m = -di / do
m = -(-20) / 100
m = 0.2

But we need to be careful with the sign of the magnification. Since the object is upright, a positive magnification would mean that the image is also upright (but smaller), while a negative magnification would mean that the image is inverted. In this case, the magnification is less than 1 in magnitude, so the image is smaller than the object.
Therefore, the correct answer is (C) -0.40, which indicates a virtual, upright, and smaller image.

An Object 1