A convex mirror has a radius of curvature of 60 cm. If a 2.2 m tall object is placed 4 m in front of the
mirror. Where is the image located and how tall is it?
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Answer:
Consider the diagram:
enter image source here
Where:
o
=
0.8
m
=
object distance;
i
=
image distance;
f
=
r
2
=
1.5
2
=
0.75
m
=
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
=
−
0.4
m
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
´
=
0.125
m
=
12.5
c
m
positive because it has the same orientation of the object.
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