Q. A thin lens has 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
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 $\frac{3I}{4}$
Explanation
(d) $I \propto A^{2} \Rightarrow \frac{I_{2}}{I_{1}} = \left( \frac{A_{2}}{A_{1}} \right)^{2} = \frac{\pi r^{2} - \frac{\pi r^{2}}{4}}{\pi r^{2}} = \frac{3}{4}$
$\Rightarrow I_{2} = \frac{3}{4}I_{1}$ and focal length remains unchanged
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.