Lasers can be used to drill or cut material. One such lasergenerates a series of
ID: 1665309 • Letter: L
Question
Lasers can be used to drill or cut material. One such lasergenerates a series of high-intensity pulses rather than acontinuous beam of light. Each pulse contains 490 mJ of energy and lasts 10 ns. The laser fires 10such pulses per second. (a) What is the peak power of the laser light?The peak power is the power output during one of the 10 ns pulses.W(b) What is the average power output of the laser? The averagepower is the total energy delivered per second. W
(c) A lens focuses the laser beam to a 10µm diameter circle on the target. During a pulse, what is thelight intensity on the target? W/m2
(d) The intensity of sunlight at midday is about 1100W/m2. What is the ratio of the laser intensity on thetarget to the intensity of the midday sun?
Ilaser / Isun
(a) What is the peak power of the laser light?The peak power is the power output during one of the 10 ns pulses.W
(b) What is the average power output of the laser? The averagepower is the total energy delivered per second. W
(c) A lens focuses the laser beam to a 10µm diameter circle on the target. During a pulse, what is thelight intensity on the target? W/m2
(d) The intensity of sunlight at midday is about 1100W/m2. What is the ratio of the laser intensity on thetarget to the intensity of the midday sun?
Ilaser / Isun
Explanation / Answer
a. Peakpower = energy per pulse / pulseduration Ppeak = 490* 10-3 J / 10 *10-9 s = 4.9* 107 W b. Averagepower = total energy / totaltime Pavg = 490* 10-3 J * 10 pulses / 1 s = 4.90 W c. Intensity = Powerin one pulse / area Ilaser = 4.90* 107 / * r2 = 4.90* 107 / 3.14 * (10 * 10-6 /2)2 = 4.90* 107 / 7.85 * 10-11 = 6.24*1017 W/m2 d. Ilaser /Isun = 6.24 *1017 / 1100 = 5.67* 1014 d. Ilaser /Isun = 6.24 *1017 / 1100 = 5.67* 1014Related Questions
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