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

Given, object.s height, `h_(0)=3.0cm`
Object distance, `u=-14cm`,
Focal length, `f=-21cm`
Using lens formula, we get
`(1)/(v)-(1)/(u)=(1)/(f)`
`rArr" "(1)/(v)+(1)/(14)=-(1)/(21)`
`rArr" "(1)/(v)=-(1)/(14)-(1)/(21)`
`=(-3-2)/(42)=-(5)/(42)`
`rArr" "v=-(42)/(5)=-8.4cm`
The image is erect, virtual and is located 8.4 cm from the lens on the same side as the object. Using the relation,
`(h_(i))/(h_(o))=(v)/(u)`
`rArr" "h_(i)=(8.4)/(14)xx3=1.8cm" (diminished)"`
As the object is moved away from the lens, the virtual image moves towards the focus of the lens and progressively diminishes in size.