Question

The amplitude of electric field in a parallel light beam of intensity 4 Wm^-2 is?

Answer 1

54.9 V/m

Answer 2

The intensity ( $I$ ) of a parallel light beam (an electromagnetic wave) is related to the amplitude of the electric field ( $E_0$ ) by the formula:

$I = \frac{1}{2} c \epsilon_0 E_0^2$

Where:

• $I$ is the intensity of the light beam.
• $c$ is the speed of light in vacuum ($3 \times 10^8 \text{ m/s}$).
• $\epsilon_0$ is the permittivity of free space ($8.854 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2}$ or $\text{F/m}$).
• $E_0$ is the amplitude of the electric field.

We are given: $I = 4 \text{ Wm}^{-2}$

We need to find $E_0$. Rearranging the formula to solve for $E_0$:

$E_0^2 = \frac{2I}{c \epsilon_0}$ $E_0 = \sqrt{\frac{2I}{c \epsilon_0}}$

Now, substitute the given values and constants into the formula:

$E_0 = \sqrt{\frac{2 \times 4}{(3 \times 10^8 \text{ m/s}) \times (8.854 \times 10^{-12} \text{ F/m})}}$ $E_0 = \sqrt{\frac{8}{26.562 \times 10^{-4}}}$ $E_0 = \sqrt{\frac{8}{0.0026562}}$ $E_0 = \sqrt{3011.064}$ $E_0 \approx 54.87 \text{ V/m}$

The amplitude of the electric field is approximately 54.9 V/m.