The amplitude of electric field in a parallel light beam of intensity 4 W m⁻² is:
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Question

The amplitude of electric field in a parallel light beam of intensity 4 W m⁻² is:

G1 Physics · Accepted Answer

For a sinusoidal electromagnetic wave, average intensity is I=(1/2)cε₀E₀². Therefore E₀=√[2I/(cε₀)]. With I=4 W m⁻², c=3×10⁸ m s⁻¹ and ε₀=8.85×10⁻¹² F m⁻¹, E₀≈54.9 N C⁻¹. Hence option D.