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)

$f$ and $\frac{3I}{4}$

Answer 2

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, $A_1 = \pi \bigg( \frac{d}{2} \bigg)^2 = \frac{ \pi d^2}{ 4}$
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 $\, \, \, I_2 =\frac{3}{4}I_1 =\frac{3}{4} I$ $( \because I_1=I)$
Hence, focal length of a lens = f, intensity of the
image = $\frac{ 3I}{ 4}$