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)/(4) and (I)/(2)`

Answer

C

Explanation

`f'=f`& Intensity `prop` Area so `I'=I -(I)/(4)=(3I)/(4)`