A thin lens focal length `f_(1)` and its aperture has diameter `d`. It forms an image of intensity `I`. Now the central part of the aperture up to diameter`(d)/(2)` is blocked by an opaque paper. The focal length and image intensity will change to

Explanation

On blocking the central part of the lens, its focal length does not change. It remains `f` only. Intensity of image is directly proportional to the area of the lens through which light passes. Now, initial area `A_1 = pi (d//2)^2 = pi d^2//4`
On blocking the central part of the aperture upto diameter `d//2`, the area left out is
`A_2 = pi (d//2) - pi (d//4)^2 = pi (d^2)/(4) - pi (d^2)/(16) = (3 pi d^2)/(16)`
As `(I_2)/(I_1) = (A_2)/(A_1) = (3 pi d^2 4)/(16 pi d^2) = (12)/(16) = (3)/(4) :. I_2 = (3)/(4) I_1`.