A lamp emits monochromatic green light uniformly in all directions. The lamp is `3%` efficient in converting electrical power to electromagnetic waves and consumes `100 W` of power. The amplitude of the electric field associated with the electromagnetic. radiation at a distance of `10 m` from the lamp will be

Options

A. `1.34 V//m`
B. `2.68 V//m`
C. `5.36 V//m`
D. `9.37 V//m`

Explanation

To find the amplitude of the electric field associated with the electromagnetic radiation at a distance of 10 m from the lamp, we can follow these steps:

### Step 1: Calculate the Power Output of the Lamp
The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power.

\[ \text{Power Output} (P) = \text{Efficiency} \times \text{Input Power} = 0.03 \times 100 \, \text{W} = 3 \, \text{W} \]

### Step 2: Calculate the Surface Area of the Sphere at Distance 10 m
The lamp emits light uniformly in all directions, so we can treat it as a point source. The surface area \( A \) of a sphere is given by:

\[ A = 4 \pi r^2 \]

Where \( r = 10 \, \text{m} \):

\[ A = 4 \pi (10)^2 = 4 \pi \times 100 = 400 \pi \, \text{m}^2 \]

### Step 3: Calculate the Intensity of the Radiation
Intensity \( I \) is defined as power per unit area:

\[ I = \frac{P}{A} = \frac{3 \, \text{W}}{400 \pi \, \text{m}^2} \]

### Step 4: Relate Intensity to Electric Field Amplitude
The intensity of an electromagnetic wave can also be expressed in terms of the amplitude of the electric field \( E_0 \):

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

Where:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (permittivity of free space)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

### Step 5: Solve for Electric Field Amplitude \( E_0 \)
Rearranging the equation gives:

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

### Step 6: Substitute the Values
First, calculate \( I \):

\[ I = \frac{3}{400 \pi} \approx \frac{3}{1256.64} \approx 0.00239 \, \text{W/m}^2 \]

Now substitute \( I \):

\[ E_0 = \sqrt{\frac{2 \times 0.00239}{8.85 \times 10^{-12} \times 3 \times 10^8}} \]

Calculating the denominator:

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

Now substitute this back into the equation for \( E_0 \):

\[ E_0 = \sqrt{\frac{0.00478}{2.655 \times 10^{-3}}} = \sqrt{1.80} \approx 1.34 \, \text{V/m} \]

### Final Answer
The amplitude of the electric field associated with the electromagnetic radiation at a distance of 10 m from the lamp is approximately:

\[ E_0 \approx 1.34 \, \text{V/m} \]

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