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. (a) f and `I/4`
B. (b) `(3f)/(4)` and `I/2`
C. (c) f and `(3I)/(4)`
D. (d) `f/2` and `I/2`

Answer

C

Explanation

Intensity, `IpropA^2`
`I_2/I_1=[A_2/A_1]^2=(pir^2-(pir^2)/(4))/(pir^2)=3/4`
`impliesI_2=3/4I_1`
and focal length remains unchanged.