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If nearly 10% of the power of a 110W light bulb is converted to visible radiation. Find the change in average intensities of visible radiation,at a distance of 1m from the bulb to a distance of 5m

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$84 \times 10^{- 2} w / m^{2}$

$84 \times 10^{- 1} w / m^{2}$

$84 \times 10^{- 4} w / m^{2}$

$84 \times 10^{- 3} w / m^{2}$

answer is A.

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Detailed Solution

$I_{1} - I_{2} = \frac{P^{'}}{4 π {r_{1}}^{2}} - \frac{p'}{4 π {r_{2}}^{2}} p^{'} = \frac{10}{100} \times 110 = 11 W$

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