# Concave Mirror question?

1. May 18, 2014

### ichilouch

1. The problem statement, all variables and given/known data
Where do you put an object in front of a concave mirror of focal length 10cm to produce an image that is inverted and 2.5 times greater than the object?

2. Relevant equations

$\frac{1}{f}$=$\frac{1}{P}$+$\frac{1}{Q}$
Where:
Q=Image distance from mirror
P=Object's distance from mirror
f = Focal length

m=$\frac{-Q}{P}$
m=Magnification

3. The attempt at a solution

m=-2.5
Q=-mP ------ (1)

$\frac{1}{f}$=$\frac{1}{P}$+$\frac{1}{Q}$
$\frac{1}{f}$=$\frac{1}{P}$+$\frac{1}{-mP}$

P=$\frac{f}{\frac{m}{m-1}}$=$\frac{0.1}{\frac{-2.5}{-2.5-1}}$
P=0.011 meter

2. May 18, 2014

Focal length of a concave mirror is to be taken negative.

3. May 18, 2014

### dauto

No it isn't. The most common sign convetion - the Gaussian sign convention - which is used by most elementary optic texts has concave mirrors with positive focal distance and convex mirrors with negative focal distance.

4. May 18, 2014

### ehild

$\frac{0.1}{\frac{-2.5}{-2.5-1}}$ is not 0.11 m.

ehild

5. May 19, 2014

### ichilouch

Sorry, Its 0.14m

6. May 19, 2014

### ehild

That is correct.

ehild

7. May 19, 2014

Oh sorry! I didn't notice that the Poster took object distance as positive. Here in India, we have a different sign convention. According my textbooks, sign of the distance with respect to the pole of the mirror depends on whether the distance is parallel or antiparallel to the incident light ray. If the 'vector' from pole to focus is parallel to the light ray, it is taken positive and negative otherwise.

8. May 19, 2014

### dauto

First time a hear of this convention. Could you elaborate?

9. May 19, 2014

I have posted an image of the sign convention.

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10. May 19, 2014

### dauto

That seems to be the Cartesian sign convention which is the second most common sign convention as far as I know. I thought you were talking about some other convention.