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;
[tex]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[/tex]
Therefore, the peak value of the electric field is 489.64 V/m.
