# Homework Help: Reflection off mirror problem

1. Oct 21, 2005

### positron

If you place an object a distance so from a spherical mirror, how do you calculate the image location given that the image is right side up and magnified by two times? I know the equation but I always get confused about what signs to use for lenses vs. mirrors, etc.

I know the magnification is si/so = 2. I don't understand how to determine whether the image should be virtual or real, from the equation 1/so+1/si = 1/f. Maybe if someone can show me how to use this equation in the context of a reflecting mirror I will understand how to do the problem.
Thanks

Last edited: Oct 21, 2005
2. Oct 21, 2005

### Chi Meson

With a spherical mirror that is concave (the magnifying kind) any image that is "real" will be inverted. Any image that is upright will be virtual.

In this case the magnification is 2.0, and image distance will be a negative value according to convention.

3. Oct 21, 2005

### positron

How do you know that any image that is real is inverted and that any image that is virtual is upright, based on the equation? I know I can just memorize this fact, but is there another way to think about it?

4. Oct 21, 2005

### Chi Meson

It is a fact. The equation is modified to fit the fact. Unfortunately, the equation changes depending on the text. In many cases the magnification equation is: m=si/so=-hi/ho where the negative is thrown in to have a "positive magnification" indicate a real image.

Sometimes the negative is left out in order to have a negative magnification indicate an inverted image.

These are only conventions (something people agreed to do). But to repeat: it starts as a fact, due to the simple geometric nature of basic *single* lens or *single* mirror optics, virtual images are always upright, real images are always inverted.