Question

A diverging lens with a focal length of -11.3 cm and a converging lens with a focal length of 14.6 cm have a common central axis. Their separation is 30.9 cm. An object of height 1.0 cm is 33.1 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?

Answer 1

given diverging lens, f1 = -11.3 cm
converging lens, f2 = 14.6 cm
common central axis
seperation d = 30.9 cm

ho = 1 cm
u = -33.1 cm, in front of the diverging lens
hence image distance for diverging lens = v'
1/v' - 1/u = 1/f1
1/v' + 1/33.1 = -1/11.3
v' = -8.424099099 cm
hence
object distance for converging lens = u'
u' = v' - d = -39.324099099099 cm
image distance = v
1/v - 1/u' = 1/f2
1/v - 1/u' = 1/14.6
v = 23.22154771122751 cm

a. location of final image wwrt converging lens = v = 23.2215477112275 cm
b. image height = ho*(v'/u)(v/u') =-0.150289227964 cm
-ve sign shows that the final imge is inverted