A photosensitive material is at 9,m to the le | vedclass

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(B) At $t = 0$,the position of the photosensitive material is $x_m(0) = -9\,m$ and the position of the source is $x_s(0) = +7\,m$. The distance betwee

Vedclass · Accepted Answer

$1 : 16$

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(B) At $t = 0$,the position of the photosensitive material is $x_m(0) = -9\,m$ and the position of the source is $x_s(0) = +7\,m$. The distance between them is $r_1 = |7 - (-9)| = 16\,m$.At $t = 3\,s$,the positions are given by $x = x_0 + vt$. For the material: $x_m(3) = -9 + (8 	imes 3) = -9 + 24 = 15\,m$.For the source: $x_s(3) = 7 + (4 	imes 3) = 7 + 12 = 19\,m$.The distance between them is $r_2 = |19 - 15| = 4\,m$.The intensity $I$ of light from a point source follows the inverse square law,$I \propto 1/r^2$.Therefore,the ratio of intensities is $\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} = \left(\frac{4}{16}ight)^2 = \left(\frac{1}{4}ight)^2 = \frac{1}{16}$.

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