# Really simple question about mirros

1. Nov 12, 2007

### Vidatu

1. The problem statement, all variables and given/known data

Where must you place an object in front of a concave mirror with radius R so that the image is erect and 2.5 times the size of the object? Where is the image?

2. Relevant equations

$$1/s + 1/s' = 2/R$$
$$m = h'/h = -s'/s$$

3. The attempt at a solution

I know that the object will be within the focal length of the mirror, and the image will be virtual. However, I'm having problems with the signs of my values. I know that the object distance will be positive, the radius of curvature will be positive, and the image distance will be negative. What's the right way to put this into the equations?

From $$m = h'/h = -s'/s$$ I get $$s = -s'/2.5$$, and from there, I substitute s into $$1/s + 1/s' = 2/R$$ to get $$1/s - 1/(s'/2.5) = 2/R$$. This gives me $$s' = -3/4 R$$. I don't think this is right, as I get a negative object distance and positive image distance, which shouldn't happen. Can anyone give me some help with this?

Last edited: Nov 12, 2007
2. Nov 12, 2007

### Vidatu

Anyone able to help with this?

3. Nov 13, 2007

### Vidatu

No help at all?

4. Nov 14, 2007

### Shooting Star

My earlier post deleted. Possibly wrong calculation.

5. Nov 14, 2007

### Vidatu

Hmm. I think I may have been really, really stupid with this. Anyone who wants a good laugh at me, look at my calculations again. Yeah, I really thought, after writing that out and typing it in that that negative was a positive. And that two negatives make another negative. Wow, go me.

In any case, this is done. How exactly do I mark it as solved?

6. Nov 14, 2007

### Shooting Star

Reposting it: quite correct.

Source is at (3/5)f.

1/do + 1/di = i/f and -di/do = 2.5.

You can find di now.

7. Nov 14, 2007

### Shooting Star

Oh, we must have been typing at the same time (almost).