A 2.4 x $10^{14}$ Hz laser emits a 3.5-μs pulse that is 5.0 mm in diameter.The average energy density in the beam is $0.65 \mathrm {~J} / \mathrm { m } ^ { 3 }$ What average power is emitted by this laser? (c = 3.0 × 10⁸ m/s,μ₀ = 4π × 10^-7 T ∙ m/A,ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)3.8 kW B)7.7 kW C)12 kW D)15 kW E)19 kW

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Quizplus · Accepted Answer

The average power emitted by the laser can be calculated using the formula for power, which is energy divided by time. Given the average energy density ($0.65 \mathrm {~J} / \mathrm { m } ^ { 3 }$) and the volume of the laser pulse, we can find the total energy of the pulse. The volume of the cylinder-shaped pulse is given by $\pi r^2 h$, where $r$ is the radius of the beam and $h$ is the height (or length) of the pulse, which can be found by multiplying the speed of light ($c$) by the pulse duration.First, convert the diameter to radius: $r = 5.0 \, \text{mm} / 2 = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m}$.The pulse duration is $3.5 \, \mu\text{s} = 3.5 \times 10^{-6} \, \text{s}$.The length of the pulse ($h$) is $c \times \text{pulse duration} = 3.0 \times 10^8 \, \text{m/s} \times 3.5 \times 10^{-6} \, \text{s} = 1.05 \, \text{m}$.The volume of the pulse is $\pi r^2 h = \pi (2.5 \times 10^{-3} \, \text{m})^2 \times 1.05 \, \text{m} = 2.07 \times 10^{-5} \, \text{m}^3$.The total energy of the pulse is $0.65 \, \text{J/m}^3 \times 2.07 \times 10^{-5} \, \text{m}^3 = 1.35 \times 10^{-5} \, \text{J}$.The average power is $1.35 \times 10^{-5} \, \text{J} / 3.5 \times 10^{-6} \, \text{s} = 3.86 \, \text{W}$, which rounds to approximately 3.8 kW.

A 24 X 101410^{14}1014 Hz Laser Emits a 3 0.65 J/m30.65 \mathrm {~J} / \mathrm { m } ^ { 3 }0.65 J/m3

A 24 X $10^{14}$ Hz Laser Emits a 3 $0.65 \mathrm {~J} / \mathrm { m } ^ { 3 }$