Here, `u =- 20 cm, f= -80 cm, v=?, h_(1)=2 cm, h_(2)=?`
From `(1)/(v) - 1/u= (1)/(f) , (1)/(v) = (1)/(f) + 1/u = (1)/(-80) - (1)/(20) = (-1 -4)/(80)= (-5)/(80)`
`v=(-80)/(5) = -16 cm`.
Thus, image is at 16 cm from the lens on the same side as the object.
As `(h_(2))/(h_(1))=v/u :. h_(2)=(v)/(u) h_(1) =(-16)/(-20) xx 2 cm= 1.6cm`
This is the size of the image.