Question

A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

• A. $f$ and $\frac{I}{4}$
• B. $\frac{3f}{4}$ and $\frac{I}{2}$
• C. $f$ and $\frac{3I}{4}$
• D. $\frac{f}{2}$ and $\frac{I}{2}$

Answer 1

Answer: (C)

Focal length of the lens remains same.
Intensity of image formed by lens is proportional to area exposed to incident light from object. i.e. Intensity $\propto$ area
or $\frac{I_2}{I_1}=\frac{A_2}{A_1}$
Initial area, $A_1=\pi {\left(\frac{d}{2}\right)}^2=\frac{\pi d^2}{4}$
After blocking, exposed area,
$A_2=\frac{\pi d^2}{4}-\frac{\pi (d/2{)}^2}{4}=\frac{\pi d^2}{4}-\frac{\pi d^2}{16}=\frac{3\pi d^2}{16}$
$\frac{I_2}{I_1}=\frac{A_2}{A_2}=\frac{16}{\frac{\pi d^2}{4}}=\frac{3}{4}$
or $I_2=\frac{3}{4}{I}_1=\frac{3}{4}I\left(\because I_1=I\right)$
Hence, focal length of a lens $=f$, intensity of the image $=\frac{3I}{4}$