An object is kept at a distance of `16 cm` from a thin lens and the image formed is real. If the object is kept at a distance of `6 cm` from the lens, the image formed is virtual. If the sizes of the images formed are equal, the focal length of the lens will be

Options

A. `15 cm`
B. `17 cm`
C. `21 cm`
D. `11 cm`

Answer

D

Explanation

Only convex lens can form a real well as virtual image. So, the given lens is a convex lens.
Let `f` is the focal length of the lens and `n` the magnitude of magnification.
In the first case when the image is real
`v=-16`
`v=+16n`
So, applying `(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/(16n)+(1)/(16)=(1)/(f)`
or `1+(1)/(n)=(16)/(f)` ...(i)
In the second case when image is virtual
`u=-6`
`v=-6n`
`f=+f`
`:. (1)/(-6n)+(1)/(6)=(1)/(f)`
`:. 1-(1)/(n)=(6)/(f)` ...(ii)
Adding `(i)` and `(ii)` , we get
`2-(22)/(f)` or `f=11cm`