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Anton [14]
3 years ago
13

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

Physics
1 answer:
Murljashka [212]3 years ago
6 0

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} = cε₀E^{2} _{rms}

                             ⇒ \frac{P}{A} = cε₀E^{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

                             

                         

                                 

                                 

                                 

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