Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a × 10⁻² W / m². The value of 'a' will be _________.

Solution

Physics sir · 4 steps

Only 11 W is visible light, and it spreads over a sphere, so intensity falls as 1/r².

1. Visible power. 10% of 110 W is P = 11 W. Only this part counts, because the question asks about visible radiation.
2. Spread over sphere. The bulb is a point source, so the power spreads evenly over a sphere of area 4π r². The intensity at distance r is I = P/(4π r²).

   Δ I = P/4π(1/1²-1/5²) = 11/4π×24/25

3. Put in numbers. 11/4π = 0.875 W m⁻² at 1 m and 11/100π = 0.035 W m⁻² at 5 m. Their difference is 264/100π = 0.840 W m⁻².
4. So. Δ I = 84.0 × 10⁻² W m⁻², so a = 84.

The common slip: 840. Using the full 110 W instead of the visible 11 W gives 8.40 W m⁻², so a = 840.

The idea to keep. For a point source, intensity is power divided by 4π r². First pick out the part of the power the question is about, then apply the inverse-square law.

Answer from the NTA final answer key. Solution by PrepWiser.