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

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 magniture of magnification.
In the first case, when the image is real:
`muv=-16`
`v=+16n`
So, applying `(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/(16)+(1)/(16)=(1)/(f)`
or `1+(1)/(n)=(16)/(f)`
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)`
Adding (i) and (ii), we get
`2=(22)/(f) or f=11cm`