A red LED emits light of 0.1 watt uniformaly around it. The amplitude of the electric field of the light at a distance of 1m from the diode is

Options

A. `1.73 V//m`
B. `2.45 V//m`
C. `5.48 V//m`
D. `7.75 V//m`

Explanation

To find the amplitude of the electric field (E₀) of the light emitted by a red LED at a distance of 1 meter, we can follow these steps:

### Step-by-Step Solution

1. **Identify Given Values:**
- Power (P) = 0.1 W
- Distance (r) = 1 m

2. **Calculate the Intensity (I):**
The intensity of the light is given by the formula:
\[ I = \frac{P}{A} \]
For a point source emitting uniformly in all directions, the area (A) is the surface area of a sphere:
\[ A = 4\pi r^2 \]
Substituting the value of r:
\[ A = 4\pi (1)^2 = 4\pi \, \text{m}^2 \]
Now substituting the values into the intensity formula:
\[ I = \frac{0.1}{4\pi} \]

3. **Relate Intensity to Electric Field Amplitude:**
The intensity (I) is also related to the amplitude of the electric field (E₀) by the formula:
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
Where:
- \(\epsilon_0\) (Permittivity of free space) = \(8.85 \times 10^{-12} \, \text{F/m}\)
- \(c\) (Speed of light) = \(3 \times 10^8 \, \text{m/s}\)

4. **Rearranging the Formula:**
Rearranging the intensity formula to solve for E₀:
\[ E_0^2 = \frac{2I}{\epsilon_0 c} \]
Therefore,
\[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \]

5. **Substituting the Values:**
First, calculate the intensity:
\[ I = \frac{0.1}{4\pi} \approx \frac{0.1}{12.566} \approx 0.00796 \, \text{W/m}^2 \]
Now substituting I, \(\epsilon_0\), and \(c\) into the equation for E₀:
\[ E_0 = \sqrt{\frac{2 \times 0.00796}{8.85 \times 10^{-12} \times 3 \times 10^8}} \]

6. **Calculating the Denominator:**
\[ \epsilon_0 c \approx 8.85 \times 10^{-12} \times 3 \times 10^8 \approx 2.655 \times 10^{-3} \]

7. **Final Calculation for E₀:**
\[ E_0 = \sqrt{\frac{0.01592}{2.655 \times 10^{-3}}} \approx \sqrt{6.00 \times 10^{3}} \approx 77.46 \, \text{V/m} \]

### Final Answer:
The amplitude of the electric field (E₀) at a distance of 1 meter from the LED is approximately **77.46 V/m**.