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Butoxors [25]
3 years ago
6

A LASIK vision-correction system uses a laser that emits 10-ns-long pulses of light, each with 3.0 mJ of energy. The laser beam

is focused to a 0.80-mm-diameter circle on the corneaWhat is the electric field amplitude of the light wave at the cornea?.
Physics
1 answer:
lyudmila [28]3 years ago
7 0
 <span>P = energy/t = 0.0025/1E-8 = 250000 W 
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2 
I(peak) = 2I(ave) = 8.8113463E11 W/m^2 
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where 
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms</span>
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Answer:

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N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

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