InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 |
|
Answer» On blocking the central part of the lens, its focal length does not change. It remains `f` only. Intensity of image is directly proportional to the area of the lens through which light passes. Now, initial area `A_1 = pi (d//2)^2 = pi d^2//4` On blocking the central part of the aperture upto diameter `d//2`, the area left out is `A_2 = pi (d//2) - pi (d//4)^2 = pi (d^2)/(4) - pi (d^2)/(16) = (3 pi d^2)/(16)` As `(I_2)/(I_1) = (A_2)/(A_1) = (3 pi d^2 4)/(16 pi d^2) = (12)/(16) = (3)/(4) :. I_2 = (3)/(4) I_1`. |
|