The distance between a point source of light | vedclass

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Question

(A) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{r^2}$,where $r$ is the distance from the source.

Vedclass · Accepted Answer

$(1/9)$ times

Vedclass · Answer

(A) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{r^2}$,where $r$ is the distance from the source.Given initial distance $r_1 = 60 \ cm$ and final distance $r_2 = 180 \ cm$.The ratio of the intensities is given by $\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}$.Substituting the values: $\frac{I_2}{I_1} = \left( \frac{60}{180} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9}$.Therefore,the new intensity is $(1/9)$ times the original intensity.

The distance between a point source of light and a screen is $60 \ cm$. If this distance is increased to $180 \ cm$,what will be the intensity on the screen compared to the original intensity?

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