A lens haivng focal length 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. `(3 f)/(4) and (I)/(2)`
C. `f and (3 I)/(4)`
D. `(f)/(2) and (I)/(2)`

Answer

C

Explanation

By convering the aperture of diameter `d//2`, focal length of lens is not affected. Area reduces by `1//4 th`. So does the intensity.
New focal length `= f` and
New intensity `= I - (I)/(4) = (3 I)/(4)`.