Question

A thin lens of focal length $f$ and its aperture diameter $d$, forms a real image of intensity $I$. Now the central part of the aperture upto diameter $\left(\frac{d}{2}\right)$ is blocked by an opaque paper. The focal length and image intensity would change to

• 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

The correct option is D $f,\frac{3I}{4}$

On blocking the central part of the lens, focal length does not change.
Intensity of the image is directly proportional to the area of the lens.
Initial area= $A_1=\pi (\frac{d}{2}{)}^2=\frac{\pi d^2}{4}$
Final area= $A_2=\pi (\frac{d}{2}{)}^2-\pi (\frac{d}{4}{)}^2=\frac{\pi d^2}{4}-\frac{\pi d^2}{16}=\frac{3\pi d^2}{16}$
Hence
$\frac{I_2}{I_1}=\frac{A_2}{A_1}=\frac{3}{4}$
$I_2=\frac{3}{4}{I}_1=\frac{3}{4}I$