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?
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
You didn't do anything wrong. However, when dealing with magnifications, it's important to consider the sign of the magnification. In this case, the negative sign indicates an inverted image.
When using the formula M = -di/do, you found two possible values for the magnification based on the two possible values of di and do:
For do = -33.12 cm, di = 75.12 cm:
M = -75.12 / -33.12 = 2.27 (rounded to two decimal places)
For do = -8.88 cm, di = 50.88 cm:
M = -50.88 / -8.88 = 5.73 (rounded to two decimal places)
However, since the magnification can be positive or negative based on the orientation of the image, the two additional possible magnifications are the negative values of the ones you found:
For do = -33.12 cm, di = 75.12 cm:
M = -2.27
For do = -8.88 cm, di = 50.88 cm:
M = -5.73
So, in total, there are two possible magnifications: 2.27 and 5.73, with their negative counterparts -2.27 and -5.73.