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

Answer

D

Explanation

`IpropA^(2)implies(I_(2))/(I_(1))=((A_(2))/(A_(1)))^(2)=(pir^(2)-(pir^(2))/4)/(pi^(2))=(3)/(4)`
`impliesI_(2)=(3)/(4)I_(1)` and focal length remains unchanged .