A laser with a power of 1.0 mW has a beam radius of 1.0 mm.What is the peak value of the electric field in that beam? (c = 3.0 × 10⁸ m/s,μ₀ = 4π × 10^-7 T ∙ m/A, ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)490 V/m B)840 V/m C)65 V/m D)120 V/m E)22 V/m

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Quizplus · Accepted Answer

The peak value of the electric field in a laser beam is given by:$$E_0 = \sqrt{\frac{2P}{\varepsilon_0 c \pi r^2}}$$where:* $E_0$ is the peak electric field (V/m)* $P$ is the power of the laser (W)* $\varepsilon_0$ is the permittivity of free space (8.85 × 10^-12 C^2/N m^2)* $c$ is the speed of light (3.0 × 10^8 m/s)* $r$ is the radius of the beam (m)Plugging in the given values, we get:$$E_0 = \sqrt{\frac{2(1.0 	imes 10^{-3} 	ext{ W})}{(8.85 	imes 10^{-12} 	ext{ C}^2/	ext{N m}^2)(3.0 	imes 10^8 	ext{ m/s})(\pi (1.0 	imes 10^{-3} 	ext{ m})^2)}}$$$$E_0 \approx 490 	ext{ V/m}$$Therefore, the peak value of the electric field in the laser beam is approximately 490 V/m.

A Laser with a Power of 1