posted by Jordan on .
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?
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
di/do = (36-do)/do = (36/do) -1
will tell you the magnification in each case.