# A lens of focal $30\; cm$ and diameter 5 cm forms an image of intensity I. If central part of the lens is covered by black circular paper of diameter $2.5 \;cm$ What will be the new focal length and new intensity of image.

$(a)\;15\;cm,I \\ (b)\;30\;cm,\frac{I}{4} \\ (c)\;15\;cm, \frac{3I}{4} \\ (d)\;30\;cm, \frac{3I}{4}$

The focal length of the lens does not change.
Intensity of light is proportional to area.
$\large\frac{I}{I_1}=\frac{A}{A_1}$
Where $A$= initial area
$A_1$ =final area
$I_1= \large\frac{I A_1}{A}$
$A= \pi \bigg( \large\frac{d}{2}\bigg)^2$
$\qquad= \large\frac{\pi d^2}{4}$
$A_1 =\pi \large\frac{ d^2}{4} -\frac{\pi (d/2)^2}{4}$
$\qquad= \large\frac{ 3 \pi d^2}{16}$
$I_1= I \large\frac{\Large \frac{3 \pi d^2}{16}}{\Large\frac{\pi d^2 }{4}}$
$\qquad=\large\frac{3}{4}$$I$
Hence d is the correct answer.