A 2 cm object is placed 36 cm from a screen. Where should a converging lens of focal length 8 cm be placed between them to form a clear image on the screen? What two possible magnifications could the image have?

Answer 1

Let do be the distance of the object from the lens, and di be the distance of the image (on the other side) from the lens.
do + di = 36

1/do + 1/(36 - do) = 1/f = 1/8
Turn that into a quadratic equation for do. There will be two roots.
[(36-do) + do]/[do*(36-do)] = 1/8
36 = (1/8)*[36 do-do^2)]
288 = 36 do - do^2
do^2 - 36 do + 288 = 0
(do -12)(do - 24) = 0
do = 12 or 24 cm

The ratio
di/do = (36-do)/do = (36/do) -1
will tell you the magnification in each case.