Question

An object 25 cm tall is placed 80 cm in front of a convex mirror that has a radius of curvature of 1.5 m. What is the image distance? What is the image height?

Answer 1

I tried this:

Explanation:

Consider the diagram:

Where:
`o=0.8m=`object distance;
`i=`image distance;
`f=r/2=1.5/2=0.75m=`focal length. BUT, this focal length is "inside" the mirror in the virtual region so it will be NEGATIVE (this is a convention for the virtual side).
We can use the relationship:
`1/o +1/f=1/f`
or:
`1/0.8+1/i=-1/0.75`
rearranging:
`i=-0.4m`
This wil be a VIRTUAL image placed at a negative distance and produced by EXTENSIONS of real rays.

For the magnification `m` we get:
`m=(h')/h=-i/o`
where:
`h´=` image height;
`h=` object height:
we get:
`(h´)/0.25=-(-0.4)/(0.8)`
so:
`h´=0.125m=12.5cm` positive because it has the same orientation of the object.