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

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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.

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