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 ?

Options

A. Virtual image moves towards the focus
B. Erect, image moves towards the centre
C. Both (a) and (b)
D. None of the above

Answer

A

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`
So, 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)/(u)xx O=(-8.4)/(-14)xx3=1.8 cm`
i.e., the image is of diminished size.
If the object is moved away from the less, the virtual image moves towards the focus of the lens (but never beyond focus).