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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.
Solution
Power rating of bulb, $P=100 W$
It is given that about $5 \%$ of its power is converted into visible radiation.
$\therefore$ Power of visible radiation, $P^{\prime}=\frac{5}{100} \times 100=5 W$
Hence, the power of visible radiation is $5 W$.
$(a)$ Distance of a point from the bulb, $d =1 m$ Hence, intensity of radiation at that point is given as:
$I=\frac{P^{\prime}}{4 \pi d^{2}}$
$=\frac{5}{4 x(1)^{2}}=0.398 W / m^{2}$
$(b)$ Distance of a point from the bulb, $d _{1}=10 m$ Hence, intensity of radiation at that point is given as:
$I=\frac{P^{\prime}}{4 \pi\left(d_{1}\right)^{2}}$
$=\frac{5}{4 x(10)^{2}}=0.00398 W / m ^{2}$
