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About $5\%$ of the power of a low light bulb is converted to visible radiation . What is the average intensity of visible radiation , at a distance of 1m from the bulb ?

$\begin{array}{1 1} 0.4\;wm^{-2} \\ 0.4\;wm^{2} \\ 4.0\;wm^{-2} \\ 4.0\;wm^{2} \end{array}$

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A)
Solution :
$I= \large\frac{\text{Power of visible light}}{\text{area}}$
$\quad= \large\frac{100 \times (5/100)}{4 \pi (1)^2}$
$\quad=0.4\;wm^{-2}$