Question

A 2.05-cm-tall object is placed 30.0 cm to the left of a converging lens with a focal length f1 = 20.5 cm . A diverging lens, with a focal length f2 = -42.5 cm , is placed 30.0 cm to the right of the first lens.

Q:How tall is the final image of the object?

Answer 1

let

h = 2.05 cm
u1 = 30 cm
f1 = 20.5 cm
let v1 is the image distance for the first lens.

use, 1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = 1/20.5 - 1/30

v1 = 64.7 cm

magnification, m1 = -v1/u1 = -64.7/30 = -2.157

for second lens

object distance, u2 = -64.7 + 30 = -34.7 cm
f2 = -42.5 cm

let v2 is the image distance fro the second lens

use, 1/u2 + 1/v2 = 1/f2

1/v2 = 1/f2 - 1/u2

1/v2 = 1/(-42.5) - 1/(-34.7)

v2 = 189 cm

magnification, m2 = -v2/u2 = -189/(-34.7) = 5.45

overall magnifcation, M = m1*m2 = -2.157*5.45 = -11.7

image height = |M|*object height

= 11.7*2.05

= 24.0 cm <<<<<<<<<-------------------Answer