An object of size `3.0 cm` is placed `14 cm` in front of a concave lens of focal length `21 cm`. Describe the image produced by the lens. What happens if the object is moved further from the lens ?

Explanation

Using the relation,`1/f=1/v-1/u,` we get
`1/v=1/f+1/u=1/(-21)+1/((-14))`
`therefore v=-8.4 cm`
`therefore` The image is virtual, erect and located at 8.4 cm from the lens on the same side as the object.
Also, we know that
`m=I/O=v/u`
`therefore I=v/uxxO=(-8.4)/(-14)xx3=1.8 cm`
i.e., the image is of diminshed size.
If the object is moved away from the lens, the virtual image moves towards the focus of the lens (but never beyond focus).