A lens having focal length f and aperture of diameter d forms an image of intensity $I$ . Aperture of diameter $\frac{d}{2}$  in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

$\begin{array}{l}Focal length of the lens remains same. Intensity of image formed by lens is proportional \\ to area exposed to incident light from object.\\ \therefore \frac{I_2}{I_1}=\frac{A_2}{A_1}\\ Initially, A_1 = \pi {\left(\frac{d}{2}\right)}^2 = \frac{\pi d^2}{4}\\ Finally, after blocking exposed area is \\ A_2 = \frac{\pi d^2}{4} - \pi {\left(\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}^2 = \frac{3\pi d^2}{16}\\ \therefore \frac{I_2}{I_1}=\frac{A_2}{A_1} = \frac{ \frac{3\pi d^2}{16}}{\frac{\pi d^2}{4}} = \frac{3}{4}\\ \Rightarrow I_2 = \frac{3}{4}{I}_1 = \frac{3}{4}I\\ Hence, focal length of a lens = f \\ and Intensity of the image = 3I/4\end{array}$