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 away from the lens?

Explanation

0' = 3.0 cm
`u=-14cm, f=-21cm`
`(1)/(upsilon)+(1)/(u)=(1)/(f) rArr (1)/(upsilon)+(1)/(14)=(-1)/(21)`
`(1)/(upsilon)=(-1)/(21)-(1)/(14)`
`upsilon=(-42)/(5)=-8.4cm`
Image is erect, virutal and located 8.4 cm from the lens on the same side as the object. Using the relation.
`(I)/(O)=(upsilon)/(u)`
`upsilon=(8.4)/(15)xx5=1.8cm`
As the object is moved away from the lens, the virutal image moves towards the focus of the lens and progressively diminishes in size. (When u = 21 cm, `v=-10.5cm` and when `u=oo, v=-21cm`)