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

Answer

`11cm`

Explanation

`m=(h_(2))/(h_(1))=(v)/(u)=(f)/(f+u)`
For the real image m is negative `-m=(f)/(f-16)............(1)`
For virtual image m is positive `m=(f)/(f-6)....(2)`
From(1)&(2)
`-(f)/(f-6)=(f)/(f-16)rArr16f-f^(2)=-6f+f^(2)rArr2f^(2)=22f`
`f=11cm`