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 $=\frac{3 I}{4}$