Question

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

What is the average (rms) value of the electricfield?

Answer 1

1341.03 V/m

Step-by-step explanation:

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

`S = (P)/(A)` = cε₀
`E^(2) _(rms)`

⇒
`(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 = (d)/(2) = (0.002)/(2) = 0.001 m`

Solving for average (rms) value of electric field;

`E_(rms) = \sqrt{(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