find the amplitude of the electric field in a parallel beam of lightv of intensity 8-0w/m^2?

To find the amplitude of the electric field in a parallel beam of light given its intensity, we can use a relationship from electromagnetic theory. The intensity \( I \) of an electromagnetic wave is related to the amplitude \( E_0 \) of the electric field by the formula:

Understanding the Relationship

The intensity \( I \) of a light wave is defined as the power per unit area. For an electromagnetic wave, the intensity can be expressed in terms of the amplitude of the electric field as follows:

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

Where:

• I = intensity of the light (in watts per square meter, W/m²)
• \(\epsilon_0\) = permittivity of free space (approximately \(8.85 \times 10^{-12} \, \text{F/m}\))
• c = speed of light in a vacuum (approximately \(3 \times 10^8 \, \text{m/s}\))
• E_0 = amplitude of the electric field (in volts per meter, V/m)

Calculating the Amplitude

Given that the intensity \( I \) is \( 80 \, \text{W/m}^2 \), we can rearrange the formula to solve for the amplitude \( E_0 \):

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

Now, let's plug in the values:

• I = \( 80 \, \text{W/m}^2 \)
• \(\epsilon_0\) = \( 8.85 \times 10^{-12} \, \text{F/m} \)
• c = \( 3 \times 10^8 \, \text{m/s} \)

Substituting these values into the equation:

E_0 = \sqrt{\frac{2 \times 80}{8.85 \times 10^{-12} \times 3 \times 10^8}}

Calculating the denominator:

8.85 \times 10^{-12} \times 3 \times 10^8 \approx 2.655 \times 10^{-3}

Now substituting back into the equation:

E_0 = \sqrt{\frac{160}{2.655 \times 10^{-3}}} \approx \sqrt{60253.5} \approx 245.4 \, \text{V/m}

Final Result

The amplitude of the electric field in the parallel beam of light with an intensity of \( 80 \, \text{W/m}^2 \) is approximately 245.4 V/m.

This calculation illustrates how the intensity of light is directly related to the amplitude of its electric field, showcasing the fundamental principles of electromagnetic waves.