A laser uniformly illuminates an area with green light that has an average intensity of 550 `W//m^(2)` . What is the rms value of the electric field of this light ?

Options

A. 322 N/C
B. 455 N/C
C. 405 N/C
D. 643 N/C

Answer

To find the RMS value of the electric field (E_rms) of the green light given its average intensity (I), we can use the relationship between intensity and the electric field in electromagnetic waves. The formula we will use is:

\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]

Where:
- \(I\) is the intensity,
- \(\epsilon_0\) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{F/m}\)),
- \(c\) is the speed of light in vacuum (\(3 \times 10^8 \, \text{m/s}\)),
- \(E_0\) is the peak electric field.

The RMS value of the electric field is related to the peak electric field by:

\[ E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \]

From the intensity formula, we can express \(E_0\) in terms of intensity:

\[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \]

Now, substituting this into the equation for \(E_{\text{rms}}\):

\[ E_{\text{rms}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{I}{\epsilon_0 c}} \]

Now we can plug in the values:

1. Given \(I = 550 \, \text{W/m}^2\)
2. \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\)
3. \(c = 3 \times 10^8 \, \text{m/s}\)

Now, substituting these values into the equation:

\[ E_{\text{rms}} = \sqrt{\frac{550}{(8.85 \times 10^{-12})(3 \times 10^8)}} \]

Calculating the denominator:

\[ \epsilon_0 c = (8.85 \times 10^{-12})(3 \times 10^8) \approx 2.655 \times 10^{-3} \]

Now substituting this back into the equation for \(E_{\text{rms}}\):

\[ E_{\text{rms}} = \sqrt{\frac{550}{2.655 \times 10^{-3}}} \]

Calculating the value:

\[ E_{\text{rms}} \approx \sqrt{207,000} \approx 455.84 \, \text{N/C} \]

Thus, the RMS value of the electric field of the green light is approximately:

\[ E_{\text{rms}} \approx 455 \, \text{N/C} \]