A diverging lens with magnitude of focal length `25 cm` is placed at a distance of `15 cm` from a converging lens of magnitude of focal length `20 cm`. A beam of parallel light falls on the diverging lens. The final image formed is.

Options

A. real and it distance of `40 cm` from the divergent lens
B. real and at a distance of `6 cm` from the convergent lens
C. real and at a distance of `40 cm` from convergent lens
D. virtual and at a distance of `40 cm` from convergent lens

Answer

C

Explanation

Here, `f_(1) = - 25 cm, f_(2) + 20 cm, d = 15 cm`
as parallel beam of light falls an diverging lens, it would form a virtual image at `25 cm` from diverging lens. The distance of virtual image from convex lens `= u = -(25 + 15) cm = - 40 cm`. If `v` is distance of final image from convex lens then from `(1)/(v) - (1)/(u) = (1)/(f)`

`(1)/(v) - (1)/(-u) = (1)/(20)`
`(1)/(v) = (1)/(20) - (1)/(40) = (1)/(40), v = 40 cm`
Hence a real image is formed at a distance of `40 cm` from convergent lens as shown in figure here.