Search any question & find its solution
Question:
Answered & Verified by Expert
About $5 \%$ of the power of a $100 \mathrm{~W}$ light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of $1 \mathrm{~m}$ from the bulb?
(b) at a distance of $10 \mathrm{~m}$ ?
Assume that the radiation is emitted isotropically and neglect reflection.
(a) at a distance of $1 \mathrm{~m}$ from the bulb?
(b) at a distance of $10 \mathrm{~m}$ ?
Assume that the radiation is emitted isotropically and neglect reflection.
Solution:
1247 Upvotes
Verified Answer
Here, $\mathrm{P}=100 \mathrm{~W}$, visible radiation $=5 \%$ of $100 \mathrm{~W}$
(a) Average intensity at a distance of $1 \mathrm{~m}$ from the bulb $\Rightarrow \mathrm{r}=1 \mathrm{~m}$
$$
\mathrm{I}=\frac{\text { Power }}{\text { area }}=\frac{0.05 \times 100}{4 \pi \mathrm{r}^2}=\frac{5 \times 7}{4 \times 22 \times 1}=0.4 \mathrm{~W} / \mathrm{m}^2
$$
(b) at a distance of $10 \mathrm{~m}$,
$$
I=\frac{0.05 \times 100}{4 \pi \times(16)^2}=0.004 W / \mathrm{m}^2
$$
(a) Average intensity at a distance of $1 \mathrm{~m}$ from the bulb $\Rightarrow \mathrm{r}=1 \mathrm{~m}$
$$
\mathrm{I}=\frac{\text { Power }}{\text { area }}=\frac{0.05 \times 100}{4 \pi \mathrm{r}^2}=\frac{5 \times 7}{4 \times 22 \times 1}=0.4 \mathrm{~W} / \mathrm{m}^2
$$
(b) at a distance of $10 \mathrm{~m}$,
$$
I=\frac{0.05 \times 100}{4 \pi \times(16)^2}=0.004 W / \mathrm{m}^2
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.