# Focal Distance

1. Aug 10, 2008

### Gear2d

1. The problem statement, all variables and given/known data
A converging lens (convex lens) creates an image that is 2 times the size of the object. The object is placed:

2. Relevant equations

-di/do = hi/ho
1/f = 1/di + 1/do

3. The attempt at a solution

I thought the answer was that the object is place two focal lengths in front of the lens. But the answer says its half a focal length in front of the lens.

I mean: -di/do = hi/ho = 2
-di = 2do
1/f = 1/di + 1/do
= -1/2do + 1/do = 1/2do
f = 2do

Isn't f=2do saying two focal lengths?

Last edited: Aug 10, 2008
2. Aug 10, 2008

### alphysicist

I'm not sure what you mean. If $f=10$ for example, then $d_o=5$, so $d_o$ is half the focal length.

3. Aug 10, 2008

### Gear2d

So simple, thanks. I guess I just missed an easy answer by thinking too much (or thinking to less).

Thanks again

4. Aug 10, 2008

### Redbelly98

Staff Emeritus
No.

f=2do is saying do=f/2, or half a focal length.

(do=2f would be saying two focal lengths.)