Optics Question 578
Question: 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 upto diameter $ \frac{d}{2} $ is blocked by an opaque paper. The focal length and image intensity will change to
[CPMT 1989; MP PET 1997; KCET 1998]
Options:
A) $ \frac{f}{2} $ and $ \frac{I}{2} $
B) $ f $ and $ \frac{I}{4} $
C) $ \frac{3f}{4} $ and $ \frac{I}{2} $
D) $ f $ and $ \frac{3I}{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I\propto A^{2}\Rightarrow \frac{I _{2}}{I _{1}}={{( \frac{A _{2}}{A _{1}} )}^{2}}=\frac{\pi r^{2}-\frac{\pi r^{2}}{4}}{\pi r^{2}}=\frac{3}{4} $
$ \Rightarrow I _{2}=\frac{3}{4}I _{1} $ and focal length remains unchanged.
