Question

A laser with a power of 1.0 mW has a beam radius of 1.0 mm. What is the peak value of the electric field in that beam

Answer 1

Answer:

The peak value of the electric field is 489.64 V/m

Explanation:

Given;

power of the laser, P = 1.0 mW = 1 x 10⁻³ W

Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m

Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²

The average intensity of the light = P / A

The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)

The average intensity of the light = 318.27 W/m²

The peak value of the electric field is given by;

$E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m$

Therefore, the peak value of the electric field is 489.64 V/m.