Question

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

Answer 1

Answer: (A)

(1 / 9) times

Answer 2

$I \propto \frac{1}{r^{2}} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{60^{2}}{180^{2}} = \frac{1}{9}$