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

Options

A. `f//2, l//2`
B. `f, l//4`
C. `3f//4, l//2`
D. `f, 3l//4`

Answer

C

Explanation

lnitial area `=(pid^(2))/(4)`
after blockening, area that allows light `=(pid^(2))/(4)-(pid^(2))/(16)=(3)/(4).(pid^(2))/(4)`
its is `(3)/(4)th` of the total area of the lens would allow the light, hence Intensity is now `(3I)/(4)` . There will be no change in focal lenght