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

1. Focal length depends on the radius of curvature and the refractive index of the lens, not the aperture. Thus, $f$ remains the same.
2. Intensity $I$ is proportional to the area of the aperture available for light to pass: $I\propto A\propto d^2$.Original Area $A_1=\pi (d/2{)}^2$.Area of the blocked central part $A_2=\pi (d/4{)}^2=A_1/4$.Available Area $A^{\prime }=A_1-A_2=A_1-\frac{A_1}{4}=\frac{3A_1}{4}$.New intensity $I^{\prime }=\frac{3I}{4}$.