A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is:
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A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is:

G1 Physics · Accepted Answer

Radiated power = 0.05×100 = 5 W. Total intensity at r=5 m is I = 5/[4π(5)²] = 1/(20π) W m⁻². In an electromagnetic wave, electric and magnetic fields share energy equally, so electric-field contribution is half: 1/(40π) W m⁻².