A lens having focal length f and aperture of diameter d forms an image of intensity I . Aperture of diameter d2  in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Focal length of the lens remains same. Intensity of image formed by lens is proportional to area exposed to incident light from object. ∴I2I1=A2A1Initially,   A1 = πd22  = πd24Finally, after blocking exposed area is A2 = πd24 − π(d4)2 = 3πd216∴I2I1=A2A1 =  3πd216πd24 = 34 ⇒I2 =  34I1 = 34IHence, focal length of a lens = f    and     Intensity of the image = 3I/4