Let the illuminance of the candle and the lamp at a distance 1m be IC and
ILrespectively. According to question
$\text{E}=\frac{\text{I}_\text{L}}{(8.0)^2}=\frac{\text{I}_\text{c}}{(0.2)^2}$
$\Rightarrow\frac{\text{I}_\text{L}}{\text{I}_\text{c}}=16$
Again after the wrapping the paper around
$\frac{49}{100}\frac{\text{I}_\text{L}}{(0.8+\text{x})^2}=\frac{\text{I}_\text{c}}{(0.2)^2}$
Either,
x = -0.24
or,
x = -1.36 this value is not acceptable because it is more than 0.8
so, x = 0.24m (-ve sign denotes distance is to be decreased).
x = 24cm
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