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Step-by-Step Solution
Step 1: Recall the relationship between intensity and the peak electric field
For an electromagnetic wave in free space, the intensity I is given by:
I = \frac{1}{2}\,\varepsilon_0\,c\,E_0^2
where:
• \varepsilon_0 is the permittivity of free space,
• c is the speed of light in vacuum,
• E_0 is the peak (maximum) electric field.
Step 2: Express the peak electric field E_0 in terms of the intensity I
Rearranging the above formula for E_0 , we get:
E_0 = \sqrt{\frac{2 I}{\varepsilon_0\,c}}
Step 3: Relate the rms electric field E_{\mathrm{rms}} to E_0
The rms (root mean square) electric field is given by:
E_{\mathrm{rms}} = \frac{E_0}{\sqrt{2}}
Substituting E_0 = \sqrt{\frac{2 I}{\varepsilon_0\,c}} into the above,
E_{\mathrm{rms}} = \sqrt{\frac{I}{\varepsilon_0\,c}}
Step 4: Substitute the given values into the expression for E_{\mathrm{rms}}
We have:
• Intensity I = \frac{315}{\pi}\,\mathrm{W/m^2} ,
• \varepsilon_0 = 8.86 \times 10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}} ,
• c = 3 \times 10^8\,\mathrm{m/s} .
Thus,
E_{\mathrm{rms}}
= \sqrt{\frac{\frac{315}{\pi}}{\left(8.86 \times 10^{-12}\right)\left(3 \times 10^8\right)}}
Step 5: Calculate the numerical value
Evaluating the expression numerically gives approximately:
E_{\mathrm{rms}} \approx 194\,\mathrm{V/m}
Step 6: State the final answer
Therefore, the rms electric field associated with the given laser intensity is about
194 V/m, to the nearest integer.
