Total power emitted =100×1005=5W
Now intensity due to electric field will be half of the total intensity. Therefore,
IE=21×areapower=21×4π×525
=40π1m2W
JEE Main 2023 — Physics Modern Physics
A point source of 100W emits light with 5 efficiency. At a distance of 5m from the source, the intensity produced by the electric field component is:
Held on 30 Jan 2023 · Verified 6 Jul 2026.
2π1m2W
40π1m2W
10π1m2W
201m2W
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