Question

A 15.0 mW laser puts out a narrow beam 2.00 mm indiameter.

Answer 1

Answer:

1341.03 V/m

Explanation:

The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.

                                 $S = \frac{P}{A}$ = cε₀$E^{2} _{rms}$

                             ⇒ $\frac{P}{A}$ = cε₀$E^{2}_{rms}$

Where;

P is the power output

A is the area of the beam

c is speed of light

ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m

$E_{rms}$ is the average (rms) value of electric field

Making electricfield $E_{rms}$ the subject of the equation

                                 $E^{2}_{rms}$ = P / Acε₀

                                 $E_{rms}$ = √(P / Acε₀)

But area A = πr²

                                 $E_{rms}$ = √(P / πr²cε₀)                    

Given:

Output power, P = 15 mW = 0. 015 W

Diameter, d = 2 mm = 0.002 m

⇒ Radius, $r = \frac{d}{2} = \frac{0.002}{2} = 0.001 m$

Solving for average (rms) value of electric field;     

$E_{rms} = \sqrt{\frac{0.015 W}{\pi * (0.001 m)^2 * (3 * 10^8 m/s) * (8.85 * 10^-12) C^2/Nm^2} }$

                                $E_{rms}$ = 1341.03 V/m