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

Options

A. f and `I/4`
B. `(3f)/(4)` and ` I/2`
C. f and `(3I)/(4)`
D. `(f)/(2)` and ` I/2`

Answer

C

Explanation

Focal length of the lens remains same.
Intensity of image formed by the lens is propotional to area exposed to incident light from object.
i.e., `"Intensity" prop "area"` or `(I_(2))/(I_(1)) = (A_(2))/(A_(1))`
Initial area, `A_(1) = pi(d/2)^(2) = (pid^(2))/(4)`
After blocking, exposed area,
`A_(2) = (pid^(2))/(4) - (pi(d//2)^(2))/(4) = (pid^(2))/(4) - (pid^(2))/(16) = (3rd^(2))/(16)`
`:. (I_(2))/(I_(1)) = (A_(2))/(A_(1)) = ((3rd^(2))/(16))/((pid^(2))/(4)) = 3/4` or `I_(2) = 3/4I_(1) = 3/4I , ( :' I_(1) = I)`
Hence, focal length of lens `= f`,
Intensity of the image ` = (3I)/(4)`