Question
Download Solution PDFA 100 W light bulb is able to convert 10% of its power to visible radiation. The average intensity of the visible radiation at a distance of 1 m from the bulb is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Power of the source P=100W
The distance from the source of radiation r=1m
Concept:
The intensity of visible radiation is the quantity of visible light emitted per unit time, per unit area
- \(I=\frac{P}{4\pi r^2}\)
where P is the power of the source, Pi is the power of visible radiation, and r is the distance from the source of radiation.
Explanation:
10% of P=Pi
\(P_i=\frac{10}{100}\times100\)
\(\implies P_i=10\ W\)
Now, \(I=\frac{P}{4\pi r^2}\)
Put all values in above formula, we get,
\(I=\frac{10}{4\times3.14\times(1)^2}\)
\(I=0.796\ W/m^2\)
\(I=0.8\ W/m^2\)
Hence, the correct answer is Option-2-\(I=0.8\ W/m^2\).
Last updated on Jun 19, 2025
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