Lasers can be used to drill or cut material. One such laser generates a series o
ID: 2012046 • Letter: L
Question
Lasers can be used to drill or cut material. One such laser generates a series of high-intensity pulses rather than a continuous beam of light. Each pulse contains 460 mJ of energy and lasts 10 ns. The laser fires 10 such pulses per second.
Part A -
What is the peak power of the laser light? The peak power is the power output during one of the 10 ns pulses.
Express your answer using two significant figures.
Ppeak = __________W
Part B -
What is the average power output of the laser? The average power is the total energy delivered per second.
Express your answer using two significant figures.
Pavg = _________W
Part C -
A lens focuses the laser beam to a 10 m diameter circle on the target. During a pulse, what is the light intensity on the target?
Express your answer using two significant figures.
Ilaser = _________W/m2
Part D -
The intensity of sunlight at the surface of the earth at midday is about 1100 W/m2 . What is the ratio of the laser intensity on the target to the intensity of the midday sun?
Express your answer using two significant figures.
Ilaser/Isun= _________
Explanation / Answer
The energy of the each pulse E = 460 mJ (a) The peak power is the power output during one of the 10 ns pulses Therefore the peak power P = 460*10^-3 J / 10*10^-9 s = 4.6*10^7 W (b) The average power is the total energy delivered per second Therefore the average power P = 10 (460*10^-3J) = 4.6 W (c) The radius of the beam r = 5*10^-6 m Then the intensity of the light I = P / r^2 = 4.6*10^7 / (5*10^-6m) ^2 = 5.85*10^17 W/m^2 (d) The intensity of sunlight at the surface of the earth at midday is about 1100 W/m2 Therefore the ratio of the laser intensity to the intensity of the sun IL /IS = 5.85*10^17 W/m^2 / 1100 W/m^2 = 5.33*10^14Related Questions
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