A thin convex lens of focal length f, has its aperture of diameter d. It forms a real iomage of intensity I. The central part of the aperture is blocked by an opaque paper of diameter `(d)/(2)`. What is the focal length of the lens and the intensity of image formed by the lens ?

Options

A. `(f)/(2), (I)/(2)`
B. `f, (3I)/(4)`
C. `(3f)/(4), (1)/(2)`
D. `f, (I)/(4)`

Answer

B

Explanation

The focal length of the lens will not change and n, `R_(1)` and `R_(2)` are not changed.
`[ because (1)/(f)=(n-1)((1)/(R_(1))-(1)/(R_(2)))]`
However, the intensity of the image is decreased.
Intensity is proportional to the effective transparent area of the lens.
In the first case, transparent are `=pi r^(2) therefore I_(1) prop pi r^(2)`
In th second case, the area of the opaque paper
`=pi((r)/(2))^(2)=(pi r^(2))/(4)`
`therefore ` Effective transparent area `=pi r^(2)-(pi^(2))/( 4)=(3)/(4) pi r^(2)`
`therefore (I_(2))/(I_(1)) =(3)/(4) therefore I_(2) =(3)/(4)I_(1) or I_(2)=(3)/(4)I`