Curved mirror identification & magnification question

[lynchdemartin]

Homework Statement

I can't seem to figure out whether I'm dealing with a + or - radius value. Anyways here's the question:
An object 5.0 cm tall produces an image that is 7.0 cm behind the mirror. If the radius of the curvature of this mirror is 10.0 cm, what is the magnification of the object? What kind of mirror is used?

Homework Equations

The Attempt at a Solution

I'm guessing it is either a virtual erected concave larger image or virtual erected convex smaller image. How can I solve for M in a question like this to find whether it's larger or smaller :S?

 

[kuruman]

What is the focal length of the mirror?
Is the object closer or farther from the mirror than one focal length?

 

[lynchdemartin]

2 x focal length = radius.
so focal length is 10cm/2 = 5cm
However the confusing part for me is to find out whether the f is on the positive side or the negative side behind the mirror. Because you don't know what type of mirror right?

The only thing I know is that the object is right before it touches the mirror & it gives a virtual image because you have an image on the opposite side of the mirror. So I'm thinking the mirror is either like this http://www.physicsclassroom.com/class/refln/u13l3d7.gif (concave virtual mirror) or it is like this http://www.physicsclassroom.com/class/refln/u13l4a3.gif (convex virtual mirror)

 

[kuruman]

You know that the image is 7.00 cm behind the mirror. That means it is virtual. You also know the focal length. Why don't you try finding the position of the object with (a) positive curvature and (b) negative curvature. If it turns out that one of these is negative, then you know that it can't be right and the other one is.

 

[lynchdemartin]

hmm good point.

 

[kuruman]

.. Or behind the mirror with the image between the mirror & the f point.

Don't be silly. When you look at yourself in a mirror (concave, convex or planar) is your face ever behind the mirror?

 

[lynchdemartin]

anyways I think the answer is 2.4 magnification & it's a concave virtual larger image. Because like you said I shouldn't have a 2 negative d values. Thanks a lot for the help.