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? , ,

Question

Quizplus · Accepted Answer

First, we can find the energy per pulse (E) using the given beam energy density and the area of the pulse:

E = energy density x area
E = (0.51 J/m) x π(2.5 x 10^-3 m)^2
E = 1.99 x 10^-6 J

Next, we can find the peak electric field (E0) using the formula:
E0 = sqrt(2E/ε0πr^2t)

where ε0 is the permittivity of free space, r is the radius of the pulse (2.5 x 10^-3 m), and t is the pulse duration (7.7 µs or 7.7 x 10^-6 s).

Plugging in the values and solving gives us:
E0 = sqrt(2(1.99 x 10^-6 J)/(8.85 x 10^-12 C^2/Nm^2 π(2.5 x 10^-3 m)^2(7.7 x 10^-6 s)))
E0 = 340 kV/m

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

A 75 × 1014 Hz Laser Emits a 7

A 75 × 10¹⁴ Hz Laser Emits a 7