# Find the image distance - optical instruments

1. Dec 15, 2008

### ally1h

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?

2. Dec 15, 2008

### mgb_phys

Slight calculator typo
(1/-10) - (1/30) = - (1/10 + 1/30) = -0.133

3. Dec 15, 2008

### ally1h

Well now... don't I feel silly?

Thank you.

4. Dec 15, 2008

### mgb_phys

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