Question

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

Answer 1

On blocking the central part of the lens, its focal length does not change. It remains `f` only. Intensity of image is directly proportinal 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) = (3pid^(2))/(16)`
As `(I_(2))/(I_(1)) = (A_(2))/(A_(1)) = (3pid^(2)4)/(16pid^(2)) = (12)/(16) = (3)/(4) :. I_(2) = (3)/(4)I_(1)`