A $2.4 \times 10 ^ { 14 } \mathrm {~Hz}$ laser emits a $3.5 - \mu$ s pulse that is $5.0 \mathrm {~mm}$ 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? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \mu _ { 0 } = 4 \pi \times \right.$ $10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } , \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 }$ )
A) $15 \mathrm {~kW}$
B) $7.7 \mathrm {~kW}$
C) $3.8 \mathrm {~kW}$
D) $12 \mathrm {~kW}$
E) $19 \mathrm {~kW}$

Question

Quizplus · Accepted Answer

The average power emitted by the laser can be calculated using the formula: Power = Energy density × Volume / Time. The volume of the cylindrical pulse is given by $ V = \pi r^2 h $, where $ r = 2.5 \times 10^{-3} $ m and $ h = c \times 3.5 \times 10^{-6} $ s. Substituting these values and the given energy density, the average power is approximately 3.8 kW.

A 2.4×1014 Hz2.4 \times 10 ^ { 14 } \mathrm {~Hz}2.4×1014 Hz Laser Emits A

A $2.4 \times 10 ^ { 14 } \mathrm {~Hz}$ Laser Emits A