Explanation:
Height of object =5cm
Position of object, u=β25cm
Focal length of the lens, f=10cm
Position of image, v=?
We know that,
v
1
β
u
1
=
f
1
v
1
+
25
1
=
10
1
v
1
=
10
1
β
25
1
So,
v
1
=
50
(5β2)
That is,
1
=
50
3
So,
v=
3
50
=16.66cm
Thus, distance of image is 16.66cm on the opposite side of lens.
Now, magnification =
u
v
That is,
m=
β25
16.66
=β0.66
Also,
m=
heightofobject
heightofimage
or
β0.66=
5cm
heightofimage
Therefore, Height of image =3.3cm