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JEE Main201911 Jan 2019Evening ShiftPhysicsElectromagnetic WavesActual

A 27 ~mW laser beam has a cross-sectional area of 10 ~mm ² . The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space ₀=9 10⁻¹² SI units, Speed of light . c =3 10⁸ ~m / s ]

Options

  1. A2 kV / m
  2. B0.7 kV / m
  3. C1 kV / m
  4. D1.4 kV / m

Correct answer

D. 1.4 kV / m

Step-by-step solution

EM wave intensity I = Power Area = 1 2 ₀ E ₀² c [where E ₀= maximum electric field] ] 27 10⁻³ 10 10⁻⁶ = 1 2 9 10⁻¹² E ₀² 3 10⁸ E ₀= 2 10³ kV / m =1.4 kV / m

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