Question

A thin lens has focal length f, and its aperture has diameter D. It forms an image of intensity I. If the central part of the aperture, of diameter D/2, is blocked by an opaque paper, the focal length of the lens and the intensity of image will become(a) $\frac{f}{2},\frac{I}{2}$(b) $f,\frac{I}{4}$(c) $\frac{3f}{4},\frac{I}{2}$(d) $f,\frac{3I}{4}$

Answer 1

[B] is the correct option

Answer 2

Answer:

D) 3I/4

Explanation:

Focal length = f

Diameter = D

Blocked diameter = D/2

The focal length will remain the same.  The intensity is directly proportional to the area of the aperture of the lens receiving the light.  

Thus, the total area of the aperture is,

A = π×d²/4

Remaining area after covering -

A' = π(d²/4  - d²/16) = 3×π× d²/16

I/I' = A/A' = 4/3

I' = 3 I/4

Therefore, the focal length of the lens and the intensity of image will become 3I/4