Question

A thin lens has a focal length $f$ and its aperture has a 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 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

Answer: (D)

The focal length remains $f$ , as any small portion of the lens forms a complete image of the object and has the same focal length.
While the intensity of the image depends on the area as,
$Intensity\propto {\left(Area\right)}^2$
$\frac{I_2}{I_1}=\frac{A_2}{A_1}$
$\frac{I_2}{I}=\frac{\frac{\pi d^2}{4}-\frac{\pi {\left(\frac{d}{2}\right)}^2}{4}}{\frac{\pi d^2}{4}}$
$\frac{I_2}{I}=\frac{3}{4}$
$I_2=\frac{3}{4}I$