When an object is placed at the proper distance ot the left of a converging lens, the image is focused in a screen 30 cm to the right of lens. A dieverging lens is now placed 15 cm to the right of the converging lens and it is found that the screen must be moved `19.2cm` farther to the right to obtain a sharp image. Find the focal length of the diverging lens.

Explanation

Image formed due to converging lens acts as ovject for diverging lens
`(1)/(v)-(1)/(u) =-(1)/(f)=(1)/(34.2)-(1)/(15)=(15-34.2)/(34.2xx15)=(-19.2)/(34.2xx15)`
`f=-26.7cm`