Question

5. An object 10 cm tall is placed at x = 0 on the optical axis. A converging lens with focal length of magnitude 10 cm is placed at x = 15 cm, and a diverging lens with focal length of magnitude 20 cm is placed at x = 50 cm. What is the height of the final image?

Answer 1

here,

the height of object , ho = 10 cm

for converging lens

the focal length , f1 = 10 cm

the object distance , do1 = -15 cm

let the image distance be di1

using the lens formula

1/f1 = 1/di1 - 1/do1

1/10 = 1/di1 - 1/(-15)

solving for di1

di1 = 30 cm

for diverging lens

the focal length , f2 = - 20 cm

the object distance , do2 = - (50 - 15 - di1) cm = - 5 cm

let the image distance be di2

using the lens formula

1/f2 = 1/di2 - 1/do2

1/(-20) = 1/di2 - 1/(-5)

solving for di2

di2 = - 4 cm

the magnification , m = (di1/do1) * (do2/di2)

m = (-4/(-5)) * (30/(-15))

m =- 1.6

the height of image , hi = ho * m

hi = 10 cm * (-1.6) = - 16 cm

the height of image is (-16 cm)