# Concave Spherical Mirrors

1. Aug 11, 2011

### roam

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

A concave spherical mirror has a radius of curvature of magnitude 27.1 cm. Determine the object position for which the resulting image is inverted and larger than the object by a factor of 4.00.

2. Relevant equations

Mirror equation in terms of focal length: 1/p + 1/q = 1/f

3. The attempt at a solution

I'm not getting the correct answer for this question. Here's what I've done so far:

First we find the focal length

$$f=\frac{R}{2} \implies \frac{27.1}{2}=13.55 \ cm$$

$$M = \frac{-q}{p}=4 \implies q=-4p$$

Substituting in

$$\frac{1}{p} - \frac{1}{4p} = \frac{1}{13.55}$$

$$\frac{3}{4p} = \frac{1}{13.55} \implies p =10.162$$

I don't know why the answer is wrong. I know that image is inverted when Magnification is negative. Here it is not -ve because the questions says "larger" so I took it as positive...

Last edited: Aug 11, 2011
2. Aug 11, 2011

### PeterO

What are you trying to find??? You haven't said.

3. Aug 11, 2011

### roam

Oops. The question I was trying to ask was:

Determine the object position for which the resulting image is inverted and larger than the object by a factor of 4.00.

4. Aug 11, 2011

### PeterO

You are having trouble with the minus signs.

There are two conventions out there one says magnification is -p/q the other says it is p/q

one of them has 1/p + 1/q = 1/r , the other has 1/p - 1/q = 1/r

make everything positive, and work on a positive magnification meaning inverted.

5. Aug 12, 2011

### roam

So, are you saying I should have used M=q/p so that q=4p? Because I am using the +ve equation 1/p+1/q=1/f.

6. Aug 12, 2011

### PeterO

Try it and see.

7. Aug 12, 2011

### roam

YES! It worked. Thanks for clarifying this, I REALLY appreciate all your help.