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 up to 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. $\frac{3 f}{4}$ and $\frac{I}{2}$
• D. $f$ and $\frac{3 I}{4}$

Answer 1

Answer: (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.