Question

A converging lens with a focal length of 4.8 cm is located 24.8 cm to the left of a diverging lens having a focal length of -15.0 cm. If an object is located 9.8 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens.
Where is the image located as measured from the diverging lens?
What is the magnification?
Also determine, with respect to the original object whether the image is real or virtual, upright or inverted? Check all statements that are true.
The image is virtual
The image is real
The image is upright
The image is inverted

Answer 1

v= image disance , u = object distance m = magnification

first image formation by converging lens

1/v-1/u = 1/f

1/v -1/-9.8= 1/4.8

m1 = v/u = 9.41/-9.8 = -0.96

from here v= 9.41 cm in right of the lens now this image act as object for diverging lens so its distance from diverging lens = 24.8-9.41 =15.39

2nd image formation by diverging lens

1/v-1/u = 1/f

1/v -1/-15.39 = 1/-15

from here v= -7.596 cm answer

in the left of  diverging lens

m2 = v/u =

-7.596/-15.39 = 0.49356

final magnification = m1*m2 = -0.96*0.49356 =-0.4738176 answer

it is virtual and inverted