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

Here, `h_(1) 3cm , u = -14 cm, f = - 21 cm, v = ?`
As `(1)/(v) - (1)/(u) = (1)/(f)`
`(1)/(v) = (1)/(f) + (1)/(u) = (1)/(-21) + (1)/(-14) = (-2 - 3)/(42) = (-5)/(42)`
`v = (-42)/(5) = -8.4 cm`
`:.` Image is erect, virtual and at `8.4 cm` from the lens on the same side as the object.
As `(h_(2))/(h_(1)) = (v)/(u) :. (h_(2))/(3) = (-8.4)/(-14)`
`h_(2) = 0.6 xx 3 = 1.8 cm`
As the object is moved away from the lens, virtual image moves towards focus of lens (but never beyond focus). The size of image goes on decreasing.