Posted by **Bob** on Sunday, March 22, 2009 at 9:58pm.

A 2 cm object is placed 42 cm from a screen. Where should a converging lens of focal length 7 cm be placed between them to form a clear image on the screen? What two possible magnifications could the image have?

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1/do + 1/(42-do) = 1/7

do = -33.12 or -8.88

di = 42-do = 75.12 or 50.88

M = -di/do

= -75.12/-33.12 or -75.12/-8.88

= 2.27 or 8.46

= -50.88/-33.12 or -50.88/-8.88

= 1.54 or 5.73

I got 4 possible magnifications, but it said there was only 2. Did I do something wrong?

- Physics -
**drwls**, Sunday, March 22, 2009 at 11:47pm
When you compute di/do, you have to use the do, di pairs that go together. You should not be getting negative values for do.

1/do + 1/(42-do) = 1/7

(42 - do + do)/[do(42-do)] = 1/7

do(42-do)= 7*42 = 294

do^2 -42do +294 = 0

do = (1/2)[42 +/- sqrt588] = 33.1 or 8.87

di = 8.9 goes with do = 33.1 and vice versa

The magnification is either 3.7 or 1/3.7

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