Question:

Find the amplitude of the electric field in a parallel beam of light of intensity 2.0W/m22.0 \, \text{W/m}^2:

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The electric field amplitude is related to the intensity of the electromagnetic wave, with higher intensity corresponding to higher field amplitudes.
Updated On: Mar 17, 2025
  • 448.8NC1448.8 \, \text{NC}^{-1}
  • 388.8NC1388.8 \, \text{NC}^{-1}
  • 380.8NC1380.8 \, \text{NC}^{-1}
  • 38.8NC138.8 \, \text{NC}^{-1}
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The Correct Option is B

Solution and Explanation

The relationship between the intensity I I and the amplitude E0 E_0 of the electric field for an electromagnetic wave is given by: I=12ϵ0cE02 I = \frac{1}{2} \epsilon_0 c E_0^2 where ϵ0=8.854×1012C2/Nm2 \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 is the permittivity of free space and c=3×108m/s c = 3 \times 10^8 \, \text{m/s} is the speed of light. Solving for E0 E_0 : E0=2Iϵ0c=2×2.08.854×1012×3×108388.8N/C. E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2 \times 2.0}{8.854 \times 10^{-12} \times 3 \times 10^8}} \approx 388.8 \, \text{N/C}.

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