# Mirrors and images

1. May 25, 2007

### -_-'

A virtual image is formed by a convex mirror that has a 30cm focal length. Where must the object be placed if the image is to be half the size of the object?

I know that since the mirror is convex F is negative (F is the focal length). Useful equations are 1/Do + 1/Di = 1/F, where Do is the objects distance and Di is the images distance. I/O = Di/Do, where I is the height of the image and O is the height of the object. And O/I = F/F-Di.

I'm not really sure on where to start could some one give me some pointers.

I = Ox(1/2) or 2I = O

F= -30cm

2. May 25, 2007

### variation

Hello,

The amplification I/O also equls to $$\frac{|D_i|}{|D_o|}$$ when we talk about near-axis light.
Be careful and recall that the image is a virtual one and therefore $$D_i$$ is a negative value in Gauss' formula.

Best regards

3. May 25, 2007

### -_-'

Thanks for the tip :D