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.

[(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.

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