A point source of 100   W emits light with 5 % efficiency. At a distance of 5   m from the source,…

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:
  1. 12πWm2
  2. 140πWm2
  3. 110πWm2
  4. 120Wm2

Solution

Total power emitted =100×5100=5 W

Now intensity due to electric field will be half of the total intensity. Therefore,

IE=12×powerarea=12×54π×52

=140πWm2

Asked in: JEE Main 2023 (30 Jan Shift 2)

Practice more Dual Nature of Matter and Radiation questions on Aicharya