Question

A converging lens, which has a focal length equal to 8.4 cm, is separated by 30.5 cm from a second lens. The second lens is a diverging lens that has a focal length equal to -13.7 cm. An object is 16.8 cm to the left of the first lens.

(a) Find the position of the final image using both a ray diagram and the thin-lens equation. __________cm to the right of the object

(b) Is the final image real or virtual?                 Is the final image upright or inverted?

(c) What is the overall lateral magnification?

Answer 1

For the first lens, apply 1/f = 1/p + 1/q

1/8.4 = 1/16.8 + 1/q

q = 16.8 cm behind the first lens

Since the lenses are 30.8 cm apart, that initial image distance is 30.5 - 16.8 = 13.7 cm in front of the diverging lens, so apply the lens equation again...

1/-13.7 = 1/13.7 + 1/q

q = -6.85 cm

For the distance from the object, that is 16.8 + 30.5 - 6.85 = 40.45​ cm to the right of the object.

Since the final value of q is negative, the image is virtual

We can also find the magnification...

M = -q/p times -q/p

M = -16.2/16.2 times -(-6.85)/13.7

M = -0.5

Since the magnification is negative, the final image is inverted