Question

An object of height 3.6 cm is placed at 24 cm in front of a diverging lens of focal length, f = -18 cm. Behind the diverging lens, there is a converging lens of focal length, f = 18 cm. The distance between the lenses is 5 cm. In the next few steps, you will find the location and size of the final image. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is d(il) = . (Use the sign to indicate which side the image is on; positive sign means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) . (Use the sign to Where is the final image formed by the second converging lens? Image distance from second lens is d(i2) = indicate which side the image is on.) (Use the sign to How large is the intermediate image formed by the first diverging lens? Intermediate image height is h(il) = indicate whether the image is upright (positive) or inverted (negative).) (Use the sign to indicate whether How large is the final image formed by the second converging lens? Final image height is h(i2) the image is upright or inverted.)

Answer 1

1) using lens formula,

1/f = 1/v - 1/u

1/v = - 1/18 - 1/24

v = - 10.28 cm

2) height of image = 3.6*10.28/24 =

Hi = 1.54 cm

3) now,

1/v = 1/18 - 1/15.28

v = - 101.11 cm

4) height of image = 1.54*101.11/15.28

Hi = 9.84 cm