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JEE Main202330 Jan 2023Evening ShiftPhysicsDual Nature of MatterActual

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:

Options

  1. A1 2 π W m 2
  2. B1 40 π W m 2
  3. C1 10 π W m 2
  4. D1 20 W m 2

Correct answer

B. 1 40 π W m 2

Step-by-step solution

Total power emitted = 100 × 5 100 = 5   W Now intensity due to electric field will be half of the total intensity. Therefore, I E = 1 2 × power area = 1 2 × 5 4 π × 5 2 = 1 40 π W m 2

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