A) 9 : 4
B) 2 : 3
C) 3 : 2
D) 4 : 9
Correct Answer: A
Solution :
The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity\[ie\], \[I=\frac{P}{4\pi {{r}^{2}}}\] or \[I\propto \frac{1}{{{r}^{2}}}\] or \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\] Here, \[{{r}_{1}}=2\,\,m,\,\,{{r}_{2}}=3\,\,m\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\]
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