Question

A 27 mW laser beam has a cross-sectional area of 10 mm². The magnitude of the maximum electric field in this electromagnetic wave is given by [Take, permittivity of space, ε_o = 9 × 10^-12 SI units and speed of light, c = 3 × 10⁸ m/s]
1. 1 kV/m
2. 0.7 kV/m
3. 2 kV/m
4. 1.4 kV/m

Answer 1

Correct Answer - Option 4 : 1.4 kV/m

Concept:

An electromagnetic wave is a wave composed of both electric and magnetic fields that are oriented orthogonal to each other. The ratio of the magnitude of the electric field to that of the magnetic field is equal to the speed of light in vacuum.

\({\rm{I}} = \frac{1}{2}{\rm{nc}}{{\rm{\varepsilon }}_0}{{\rm{E}}^2}\)

Where c is the speed of light

ɛ₀ is the permittivity of free space

Calculation:

Given,

Power of laser beam (P) = 27mW = 27 × 10^-3 W

Area of cross-section (A) = 10mm² = 10 × 10^-6 m²

Permittivity of free space (ɛ₀) = 9 × 10^-12 SI unit

Speed of light c = 3 × 10⁸ m/s

Intensity of electromagnetic wave is given by the relation

\({\rm{I}} = \frac{1}{2}{\rm{nc}}{{\rm{\varepsilon }}_0}{{\rm{E}}^2}\)

Where, n is refractive index, for air n = 1.

\(\therefore {\rm{\;I}} = \frac{1}{2}{\rm{c}}\cdot{{\rm{\varepsilon }}_0}{{\rm{E}}^2}\;\)      ----(1)

Also \({\rm{I}} = \frac{{\rm{P}}}{{\rm{A}}}\) ----(2)

From Eq. (1) and (2), we get

\(\frac{1}{2}{\rm{c}}{{\rm{\varepsilon }}_0}{{\rm{E}}^2} = \frac{{\rm{P}}}{{\rm{A}}}{\rm{\;or\;}}{{\rm{E}}^2} = \frac{{2{\rm{P}}}}{{{\rm{Ac}}{{\rm{\varepsilon }}_0}}}\)

or \({\rm{E}} = \sqrt {\frac{{2 \times 27 \times {{10}^{ - 3}}}}{{10 \times {{10}^{ - 6}} \times 3 \times {{10}^8} \times 9 \times {{10}^{ - 12}}}}} \)

= 1.4 × 10³ V/m = 1.4 kV/m