optics.....................................................................................

first there will be formation of virtual image due to the lense at v1 given by 1/v1-1/u=1/f
now this virtual image will act as an object for the concave lens.let now suppose image formed at a distance v2 is given by1/v2-1/v1=1/f
=>1/v2=1/f-1/v1=1/f-1/f/-1/u=-1/u
now again it will refract again through the lens and will form the final image at v3
1/v3-1/v2=1/f
=>1/v3+1/u=1/f
so effective f will same as tht of f