Answer:
The intensity will be 1/9 as much.
Explanation:
The intensity of the light or any source is inversely related to the square of the distance.
[tex]I\alpha \frac{1}{r^{2} }[/tex]
Now according to the question the distance is increased by three times than,
[tex]\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }[/tex]
Therefore,
[tex]\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }[/tex]
Therefore the intensity will become 1/9 times to the initial intensity.