About $5 \%$ of the power of a $100 \; W$ light bulb is converted to visible radiation. What is the average intensity of visible radiation
$(a)$ at a distance of $1 \; m$ from the bulb?
$(b)$ at a distance of $10 \; m$? Assume that the radiation is emitted isotropically and neglect reflection.
$(a)$ at a distance of $1 \; m$ from the bulb?
$(b)$ at a distance of $10 \; m$? Assume that the radiation is emitted isotropically and neglect reflection.
(A) Power rating of the bulb,$P = 100 \; W$.
It is given that about $5 \%$ of its power is converted into visible radiation.
$\therefore$ Power of visible radiation,$P' = \frac{5}{100} \times 100 = 5 \; W$.
$(a)$ At a distance of $d = 1 \; m$ from the bulb,the intensity $I$ is given by the formula $I = \frac{P'}{4 \pi d^2}$.
$I = \frac{5}{4 \times 3.14 \times (1)^2} \approx 0.398 \; W/m^2$.
$(b)$ At a distance of $d = 10 \; m$ from the bulb,the intensity $I$ is given by $I = \frac{P'}{4 \pi d^2}$.
$I = \frac{5}{4 \times 3.14 \times (10)^2} = \frac{5}{1256} \approx 0.00398 \; W/m^2$.
It is given that about $5 \%$ of its power is converted into visible radiation.
$\therefore$ Power of visible radiation,$P' = \frac{5}{100} \times 100 = 5 \; W$.
$(a)$ At a distance of $d = 1 \; m$ from the bulb,the intensity $I$ is given by the formula $I = \frac{P'}{4 \pi d^2}$.
$I = \frac{5}{4 \times 3.14 \times (1)^2} \approx 0.398 \; W/m^2$.
$(b)$ At a distance of $d = 10 \; m$ from the bulb,the intensity $I$ is given by $I = \frac{P'}{4 \pi d^2}$.
$I = \frac{5}{4 \times 3.14 \times (10)^2} = \frac{5}{1256} \approx 0.00398 \; W/m^2$.
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