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find the image distance - optical instruments |
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| Dec15-08, 04:01 PM | #1 |
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find the image distance - optical instruments
1. The problem statement, all variables and given/known data
An object is placed at 30 cm in front of a diverging lens with a focal length of 10 cm. What is the image distance? 2. Relevant equations 1/f = 1/do + 1/di 3. The attempt at a solution I thought this was a straight-forward question: 1/di = 1/-f - 1/do 1/di = (1/-10 cm) - (1/30 cm) 1/di = -0.066 cm di = 1/-0.066 cm = -15 cm BUT- the answer provided to me states that the answer is -7.5cm. So what is it that I'm missing here? |
| Dec15-08, 04:11 PM | #2 |
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Recognitions:
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Slight calculator typo
(1/-10) - (1/30) = - (1/10 + 1/30) = -0.133 |
| Dec15-08, 04:12 PM | #3 |
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Well now... don't I feel silly?
Thank you. |
| Dec15-08, 04:17 PM | #4 |
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Recognitions:
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find the image distance - optical instruments
It's very easy to get those sort of sums wrong,
It's worth rearranging them so that you know if the answer you expect is +ve or -ve And in a form where you can estimate the magnitude, eg (1/10+1/30) is obviously going to be a bit bigger than 1/10 |
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