Question

A diverging lens with f = -22.5 cm is placed 10.0 cm behind a converging lens with f = 17.5 cm . Part A Where will an object at infinity be focused? Express your answer using two significant figures.

d = ........ cm beyond second lens.

Answer 1

he image of the converging lens will be the object of the diverging lens

so 1/f = 1/s + 1/s' since s = infinity s' = f = 17.50cm

this is 10 - 17.5 = -7.50cm for the image distance of the diverging lens

therefore 1/f = 1/s + 1/s'

or 1/s' = 1/f - 1/s
so s' = f*s/(s - f) = -22.5*(-7.50)/(-7.50 - -22.5) = 11.25cm (behind the diverging lens)