Question

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

• A. $448.8 \, \text{NC}^{-1}$
• B. $388.8 \, \text{NC}^{-1}$
• C. $380.8 \, \text{NC}^{-1}$
• D. $38.8 \, \text{NC}^{-1}$

Hint:

The electric field amplitude is related to the intensity of the electromagnetic wave, with higher intensity corresponding to higher field amplitudes.

Answer 1

The Correct Option is B

The relationship between the intensity \( I \) and the amplitude \( E_0 \) of the electric field for an electromagnetic wave is given by: \[ I = \frac{1}{2} \epsilon_0 c E_0^2 \] where \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \) is the permittivity of free space and \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light. Solving for \( E_0 \): \[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2 \times 2.0}{8.854 \times 10^{-12} \times 3 \times 10^8}} \approx 388.8 \, \text{N/C}. \]