An object is placed at a distance of `f//2` from a convex lens. The image will be

Options

A. at f, real and inverted
B. at 2f, virtual and erect
C. at `(3f)/2`, real and inverted
D. at one of the foci, virtual and double its size

Answer

D

Explanation

We have,
`1/v=1/f+1/u=1/f+1/(-f/2)=1/f-2/f=-1/f`
`thereforev=-f`
So, `m=v/u=(-f)/(-f/2)=2`
We get a virtual image of double the size at one of the foci of the lens.