An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Explanation

`h_(c )= 3cm, u= -14cm, f= -21cm`
This lens formula: `(1)/(v)- (1)/(u)= (1)/(f)`
`v= (uf)/(u+ f) = (-14 xx -21)/(-14-21)= -8.4cm`
Position of image: 8.4cm from lens on same side as object
Nature of image: Virtual and erect
Size of image `= (v)/(u)= (h_(i))/(h_(0)) OR h_(i)= h_(0) xx (v)/(u)= (3 xx (-8.4))/(-14)= 1.8cm`
Image is dimished