Field intensity: A 7.5 × 10¹⁴ Hz laser emits a 7.7-μs pulse, 5.0 mm in diameter, with a beam energy density of 0.51 J/m³. What is the amplitude of the electric field of the emitted waves? (c = 3.00 × 10⁸ m/s, μ₀ = 4π × 10^-7 T ∙ m/A, ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A) 340 kV/m B) 480 kV/m C) 240 kV/m D) 150 kV/m E) 120 kV/m

Question

Quizplus · Accepted Answer

First, we need to find the energy of the pulse:

E = energy density * volume
volume = area * length = π * (5 mm/2)^2 * (7.7 μs) * (3.00 × 10^8 m/s) = 2.34 × 10^-7 J

E = 0.51 J/m^3 * 2.34 × 10^-7 m^3 = 1.19 × 10^-7 J

Next, we can use the energy to find the amplitude of the electric field:

E = hc/λ * N^2/2ε0 * A^2
A = √(2ε0 * E/ hc/λ * N^2)

where h is Planck's constant, c is the speed of light, λ is the wavelength, N is the number of photons, and ε0 is the electric constant.

We can assume that all the energy from the pulse is carried by a single photon, so N = 1.

λ = c/f = 3.00 × 10^8 m/s / 7.5 × 10^14 Hz = 4.00 × 10^-7 m

A = √(2 * 8.85 × 10^-12 C^2/N·m^2 * 1.19 × 10^-7 J / (6.63 × 10^-34 J·s * 3.00 × 10^8 m/s / 4.00 × 10^-7 m * 4π × 10^-7 T·m/A)^2) = 340 kV/m

Therefore, the answer is A) 340 kV/m.

Field Intensity: a 7