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?

1 Answer
Apr 2, 2017

I tried this:

Explanation:

Consider the diagram:
enter image source here
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.