Posted by Jordan on Tuesday, March 10, 2009 at 11:30pm.
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?

HELP Physics  drwls, Wednesday, March 11, 2009 at 3:18am
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.
[(36do) + do]/[do*(36do)] = 1/8
36 = (1/8)*[36 dodo^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 = (36do)/do = (36/do) 1
will tell you the magnification in each case.
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