Question

$2.5\mathrm{mJ}$A LASIK vision-correction system uses a laser that emits$10\mathrm{ns}$-long pulses of light, each with$2.5\mathrm{mJ}$of energy. The laser beam is focused to a $0.85-\mathrm{mm}$-diameter circle on the cornea. What is the electric field amplitude of the light wave at the cornea$?$

Short answer

The light wave has electric field amplitude is$1.8\times {10}^7\frac{\mathrm{V}}{\mathrm{m}}$.

Step-by-step solution

Calculation for intensity

Given info:

Time is $10\times {10}^{-9}\mathrm{s}$.

Energy is $2.5\times {10}^{-3}\mathrm{J}$.

Radius is $0.425\times {10}^{-3}\mathrm{m}$.

Intensity is,

${\mathrm{I}}_{\mathrm{avg}}=\frac{\text{Power}}{{\mathrm{A}}_{\text{surface}}}$

$=\frac{\mathrm{E}}{\mathrm{t}\times {\mathrm{\pi r}}^2}$

${\mathrm{I}}_{\mathrm{avg}}=\frac{2.5\times {10}^{-3}\mathrm{J}}{10\times {10}^{-9}\mathrm{s}\times \mathrm{\pi }{\left(0.425\times {10}^{-3}\right)}^2{\mathrm{m}}^2}$

${\mathrm{I}}_{\mathrm{avg}}=44\times {10}^{11}\frac{\mathrm{J}}{{\mathrm{m}}^2}$${\mathrm{I}}_{\mathrm{avg}}=\frac{\text{Power}}{{\mathrm{A}}_{\text{surface}}}$

Calculation of electric field amplitude

Amplitude is,

$\mathrm{E}={\left(\frac{2\times {\mathrm{I}}_{\mathrm{avg}}}{\mathrm{c}\times {\mathrm{\epsilon }}_{\mathrm{o}}}\right)}^{1/2}$

${\mathrm{E}}_{\mathrm{o}}={\left(\frac{2\times 44\times {10}^{11}{\displaystyle \frac{\mathrm{J}}{{\mathrm{m}}^2}}}{3\times {10}^8{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\times 8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{{\mathrm{Nm}}^2}}}\right)}^{1/2}$

${\mathrm{E}}_{\mathrm{o}}=1.8\times {10}^7\frac{\mathrm{V}}{\mathrm{m}}$