How Do You Determine Object Location and Magnification in a Concave Mirror?

[KingNothing]

Well, here is the problem:

A girl has her 2-cm high eye 12 cm directly in front of a concave mascara mirror or focal length 18 cm. Where is the object located, and what is the magnification?

I'm really lost on this one.

EDIT: Sorry, I just realized this would be better suited for the homework help k12 forum.

 

[Doc Al]

mirror equation

I presume that her eye is the object and that you are trying to find where the image is.

You'll need the lens/mirror equation:
\frac{1}{f} = \frac{1}{o} + \frac{1}{i}

And you'll also need the linear magnification:
M = -\frac{i}{o}

To understand what these mean, and how to use the sign conventions, read your text. Give it a try.

 

[Chen]

Just in case your text uses the same variable names as the one we used to study this, Doc Al' o signifies the distance of the object from the mirror and i signifies the distance of the image from the mirror. Sometimes they are replaced by u and v repsectively, so your formula might look like this:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

 

[KingNothing]

Okay, thanks chen and doc al.
I didn't have those equations, and didn't have access to the text (end of the year, books had been turned in already).

So I did use the right equation...good.

Thanks Chen for the variables thing, honestly, thanks.

 

[Chen]

No problem. Since you already returned your books, you might want to a take a look at HyperPhysics:
http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html
It provides a clear and concise explanation of many basic subjects in physics (and math).

 

[killerinstinct]

Don't forget, when doing concave mirrors/converging lens, focal length is always positive.
When doing convex mirrors/diverging lens, focal length is always negative.

*Just in case you were doing similar problems involving mirrosrs