Question

A 30 mW laser beam has a cross-sectional area of 15 mm\(^2\). The magnitude of the maximum electric field in this electromagnetic wave is given by: \[ \text{Permittivity of space, } \varepsilon_0 = 9 \times 10^{-12}, \quad \text{Speed of light, } c = 3 \times 10^8 \text{ m/s} \]

• A. \( 1.22 \text{ kV/m} \)
• B. \( 12 \text{ kV/m} \)
• C. \( 10 \text{ kV/m} \)
• D. \( 201 \text{ kV/m} \)

Hint:

When calculating the electric field from the intensity, use the relationship \( I = \frac{1}{2} \varepsilon_0 c E_{\text{max}}^2 \) and make sure the units are consistent.

Answer 1

The Correct Option is A

The intensity \( I \) of the wave is related to the maximum electric field \( E_{\text{max}} \) by the formula: \[ I = \frac{1}{2} \varepsilon_0 c E_{\text{max}}^2 \] Given the power of the laser beam, the intensity \( I \) is: \[ I = \frac{P}{A} = \frac{30 \times 10^{-3} \text{ W}}{15 \times 10^{-6} \text{ m}^2} = 2 \text{ W/m}^2 \] Substituting the intensity into the formula for \( E_{\text{max}} \): \[ 2 = \frac{1}{2} \times (9 \times 10^{-12}) \times (3 \times 10^8) \times E_{\text{max}}^2 \] Solving for \( E_{\text{max}} \), we get: \[ E_{\text{max}} = 1.22 \times 10^3 \text{ V/m} = 1.22 \text{ kV/m} \] Thus, the correct answer is (A).