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

`(1)/(v) - (1)/(u) = (1)/(f) " " f = -21 ` cm , u = - 14 cm , `h_(0)` = 3 cm
`(1)/(v) = (1)/(f) + (1)/(u) = (1)/(-21) + (1)/(-14) " " therefore v = (-42)/(5) = - 8.4 ` cm
`(h_(i))/(h_(0)) = (v)/(u) , h_(i) = h_(0) ((v)/(u)) = 3xx (-8.4)/(-14) = 1.8` cm
image is virtual, erect diminished, formed on the same side of the object if the object is moved away from the lens, image moves towards the focus and decreases in size.