Question

A diverging lens with a focal length of -19.7 cm and a converging lens with a focal length of 11.0 cm have a common central axis. Their separation is 39.5 cm. An object of height 1.1 cm is 26.4 cm in front of the diverging lens, on the common central axis. Find the location of the final image produced by the combination of the two lenses.
Where is the image located as measured from the converging lens?

What is the height of that image?
With respect to the original object, is the image is real or virtual, upright or inverted? Check all statements that are true.
Select all that are True.
The image is real
The image is upright
The image is inverted
The image is virtual

Answer 1

v = image distance , u = object distance , m = magnificatin

lens formula

1/v -1/u=1/f

1/v-1/-26.4 = 1/-19.7

from here v≈-11.2816 cm virtual upright

m1 = v/u = -11.2816/ -26.4 =0.427333333

in left same side of the object now this image act as object for 2nd lens so its distance from conveging lense = 11.2816 +39.5 = 50.7816 cm

apply again

1/v -1/u=1/f

1/v -1/-50.7816 = 1/11

from here 2nd image v= 14.0416 cm in right answer  

m1 = v/u = 14.0416/ -50.7816 = -0.276509602

so total m = m1*m2 =0.427333333*-0.276509602 = - 0.1181

height of image = 1.1* 0.1181 = 0.1299 cm answer

image is virtual and inverted