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. $\frac{1}{9}$ times
• B. $\frac{1}{3}$ time
• C. $3$ times
• D. $9$ times

Answer 1

Answer: (A)

$\frac{1}{9}$ times

As we know,

$I\propto \frac{1}{r^2}\Rightarrow \frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}=\frac{{60}^2}{{180}^2}=\frac{1}{9}$