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# A laser used in eye surgery emits a $3.00-\mathrm{m}$ J pulse in 1.00 ns, focused to a spot 30.0$\mu \mathrm{m}$ in diameter on the retina. (a) Find (in SI units) the power per unit area at the retina. (This quantity is called the irradiance.) (b) What energy is delivered per pulse to an area of molecular size (say, a circular area 0.600 $\mathrm{nm}$ in diameter)?

## a) $4.24 \times 10^{15} \mathrm{W} / \mathrm{m}^{2}$b) $1.20 \times 10^{-12} J$

Atomic Physics

Nuclear Physics

### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

##### Aspen F.

University of Sheffield

### Video Transcript

In this exercise, you have a laser polls that has an energy of three million jewels, unless for a time off one nanosecond. And it shines on a spot that has a diameter of 30 micro meters and in question A. We want to find the find the intensity off The pulse on this spot that has 30 has a diameter of 30 mecca meters. So the first thing you have to do is to find the power of a pulse, and the power is the energy divided by the time. So the total energy is three Milly jewels. That's three times that of the minus three healed, divided by the time that's 10 and or seconds or a tentative 10 to the minus nine seconds. So the power is three times. Stand to the six. What and the intensity is the power divided by the area. Ah, so the powers three times 10 to the sixth. What in the area is pie times the diameter, uh, times the radius squared and the radius is half the diameter so that if the diameter is is 30 my commuters that the radios this 15 my perimeters, which is 1.5 times into the minus fifth meters square and this is equal to 4.24 times 10 to the 15th. What? We're square meter and then questioned me. We have to find the energy that's delivered by the polls to a spot that has a diameter of 0.6 millimeters. Okay, so the total energy delivered is the power times the time. Okay, The time is how much the post last and the power is the intensity of the pulse times the area. So all we have is that the energy is the intensity which is which was calculated in question. A in is 4.24 times 10 to the 15th. What's we're square meters times the area, which is pi times the radius squared. And since the diameter is 0.6 millimeters, that the radius is 30.3 centimeters, which is three times 10 to the minus 10 meters times the times of the T. That's just 10 to the minus nine nine seconds. So e is equal to 1.2 time stands, the modest 12 juice. That's the energy delivered and this concludes the question