Physics
posted by Bob on .
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?

1/do + 1/(42do) = 1/7
do = 33.12 or 8.88
di = 42do = 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?

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