Intensity of a 3 kW light bulb

Problem

An industrial light bulb isotropically emits a power $P = 3000\;\mathrm{W}$. Calculate the light intensity at $r = 20\;\mathrm{m}$ and compare it with midday sunlight ($\sim 1000\;\mathrm{W/m^2}$).

Solution

Setup. Isotropic emission: the power is distributed over a sphere of radius $r$, surface area $4\pi r^2$.

$I = \frac{P}{4\pi r^2}$

Substitution.

$I = \frac{3000}{4\pi\,(20)^2} = \frac{3000}{4\pi\cdot 400} = \frac{3000}{5026{,}5}\;\mathrm{W/m^2}$

$\ev{I \approx 0{,}60\;\mathrm{W/m^2}}$

Comparison. The intensity is about $1000/0{,}60 \approx 1700$ times smaller than midday sunlight: even a $3\;\mathrm{kW}$ lamp, at $20\;\mathrm{m}$, illuminates far less than the Sun. The inverse-square law causes intensity to drop off rapidly with distance.

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Topics: Onde Concepts: Intensità di un’onda Skills: Analisi dimensionale