When an small object is placed at a distance `x_(1)` and `x_(2)` from a lens on its principal axis, then real image and a virtual image are formed respectively having same magnitude of transverse magnification. Then the focal length of the lens is :

Options

A. `x_(1) -x_(2)`
B. `(x_(1)-x_(2))/(2)`
C. `(x_(1) +x_(2))/(2)`
D. `x_(1) +x_(2)`

Answer

C

Explanation

`m=(f)/(f+u)m_(1)=(f)/(f-x_(1))`
`m_(2)=(f)/(f-x_(2))m_(1)=-m_(2)`
`2f =(x_(1)+x_(2))f=(x_(1)+x_(2))/(2)`.