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

Answer

To find the amplitude of the electric field (E₀) of light emitted by a red LED with a power output of 0.1 watts at a distance of 1 meter, we can follow these steps:

### Step 1: Understand the relationship between power, intensity, and electric field
The intensity (I) of the light at a distance r from the LED can be expressed as:
\[ I = \frac{P}{A} \]
where \( P \) is the power of the LED and \( A \) is the surface area of a sphere with radius r, given by:
\[ A = 4\pi r^2 \]

### Step 2: Calculate the intensity at 1 meter
Given that \( P = 0.1 \, \text{W} \) and \( r = 1 \, \text{m} \):
\[ A = 4\pi (1^2) = 4\pi \]
Thus, the intensity at 1 meter is:
\[ I = \frac{0.1}{4\pi} \]

### Step 3: Relate intensity to the electric field amplitude
The intensity can also be expressed in terms of the electric field amplitude (E₀):
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
where \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \)) and \( c \) is the speed of light in vacuum (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

### Step 4: Set the two expressions for intensity equal to each other
From the two expressions for intensity, we have:
\[ \frac{0.1}{4\pi} = \frac{1}{2} \epsilon_0 c E_0^2 \]

### Step 5: Solve for E₀
Rearranging the equation to solve for E₀ gives:
\[ E_0^2 = \frac{0.1}{4\pi} \cdot \frac{2}{\epsilon_0 c} \]
Substituting the known values:
\[ E_0^2 = \frac{0.1 \cdot 2}{4\pi \cdot (8.85 \times 10^{-12}) \cdot (3 \times 10^8)} \]

### Step 6: Calculate E₀
Calculating the right side:
1. Calculate \( 4\pi \):
\[ 4\pi \approx 12.566 \]
2. Calculate \( \epsilon_0 c \):
\[ \epsilon_0 c \approx (8.85 \times 10^{-12}) \cdot (3 \times 10^8) \approx 2.655 \times 10^{-3} \]
3. Substitute back:
\[ E_0^2 = \frac{0.1 \cdot 2}{12.566 \cdot 2.655 \times 10^{-3}} \approx \frac{0.2}{0.0333} \approx 6 \]
4. Taking the square root:
\[ E_0 \approx \sqrt{6} \approx 2.45 \, \text{V/m} \]

### Conclusion
The amplitude of the electric field at a distance of 1 meter from the LED is approximately:
\[ E_0 \approx 2.45 \, \text{V/m} \]

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