Q. The distance between a point source of light and a screen which is $60\ cm$ is increased to $180\ cm$. The intensity on the screen as compared with the original intensity will be
A) $(1/9)$ times
B) $(1/3)$ times
C) 3 times
D) 9 times
Explanation
(a) $I \propto \frac{1}{r^{2}} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{60^{2}}{180^{2}} = \frac{1}{9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.