Answer:

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