1v+1u=1f
By differentiating both sides
⇒dvv2=−duu2
du: size of object =L; dv:size of image;
u: object distance
v: image distance
dv=v2u2du (1v=1f−1u⇒v=fuu−f)
=−f2u2(u−f)2u2du=−f2(b−f)2L
Negative sing implies that object is lying between u and u+ and the image will lie between v and v−dv