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A point source is kept at a distance of \(1000 \mathrm{~m}\) has an illumination \(I\). To change the illumination to \(16 I\) the new distance should become
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Verified Answer
The correct answer is:
\(250 \mathrm{~m}\)
For a point source, illuminance at a distance \(r\) is given by
Illuminance \(I=\frac{\text { luminous flux } \times \phi}{4 \pi r^2}\)
\(\begin{aligned}
& \therefore \quad \frac{I}{I^{\prime}}=\left(\frac{r^{\prime}}{r}\right)^2 \\
& \frac{I}{16 I}=\frac{r^{\prime 2}}{(1000 \mathrm{~m})^2} \\
& \text { or } r^{\prime 2}=\frac{(1000 \mathrm{~m})^2}{16} \text { or } r^{\prime}=\frac{(1000 \mathrm{~m})}{4} \\
& r^{\prime}=250 \mathrm{~m} \\
&
\end{aligned}\)
Illuminance \(I=\frac{\text { luminous flux } \times \phi}{4 \pi r^2}\)
\(\begin{aligned}
& \therefore \quad \frac{I}{I^{\prime}}=\left(\frac{r^{\prime}}{r}\right)^2 \\
& \frac{I}{16 I}=\frac{r^{\prime 2}}{(1000 \mathrm{~m})^2} \\
& \text { or } r^{\prime 2}=\frac{(1000 \mathrm{~m})^2}{16} \text { or } r^{\prime}=\frac{(1000 \mathrm{~m})}{4} \\
& r^{\prime}=250 \mathrm{~m} \\
&
\end{aligned}\)
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