Mar 11, 2018

$A . \frac{4}{3}$

Explanation:

Given, $L \propto \frac{1}{d} ^ 2$

so, $L = \frac{k}{d} ^ 2$ where, $k$ is a constant

Given, $L = 12$ for , $d = 3$

so,we can write,

$12 = \frac{k}{3} ^ 2$

or, $k = 12 \cdot 9$

Now,if intensity is $L '$ when the distnace $d '$ is $9 m$,we can write,

$L ' = \frac{k}{9} ^ 2 = \frac{12 \cdot 9}{9 \cdot 9} = \frac{4}{3}$ (as, $k = 12 \cdot 9$)