What is the value of frequency at which electromagnetic wave must be propagated for the D - region of atmosphere to have a refractive index of `0.5`. Electron density for D - region is `400` electrons/c.c.

Options

A. 200 kHz
B. 104.2 kHz
C. 208.4 kHz
D. 312.6 kHz

Answer

To find the frequency at which an electromagnetic wave must be propagated for the D-region of the atmosphere to have a refractive index of 0.5, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Values**:
- Refractive index (n) = 0.5
- Electron density (N) = 400 electrons/cm³ = 400 × 10^6 electrons/m³ (since 1 cm³ = 10^-6 m³)

2. **Use the Refractive Index Formula**:
The relationship between the refractive index (n) and electron density (N) in the D-region can be expressed as:
\[ n^2 = 1 - \frac{81.45 \cdot N}{f^2} \]
where \( f \) is the frequency in Hz.

3. **Substitute the Values**:
Substitute \( n = 0.5 \) and \( N = 400 \times 10^6 \) into the equation:
\[ (0.5)^2 = 1 - \frac{81.45 \cdot (400 \times 10^6)}{f^2} \]

4. **Calculate \( n^2 \)**:
\[ 0.25 = 1 - \frac{81.45 \cdot (400 \times 10^6)}{f^2} \]

5. **Rearrange the Equation**:
\[ \frac{81.45 \cdot (400 \times 10^6)}{f^2} = 1 - 0.25 \]
\[ \frac{81.45 \cdot (400 \times 10^6)}{f^2} = 0.75 \]

6. **Solve for \( f^2 \)**:
\[ f^2 = \frac{81.45 \cdot (400 \times 10^6)}{0.75} \]

7. **Calculate the Right Side**:
\[ f^2 = \frac{81.45 \cdot 400 \cdot 10^6}{0.75} \]
\[ f^2 = \frac{32.58 \times 10^6}{0.75} \approx 4.344 \times 10^8 \]

8. **Take the Square Root**:
\[ f = \sqrt{4.344 \times 10^8} \approx 208.42 \times 10^3 \text{ Hz} = 208.42 \text{ kHz} \]

### Final Answer:
The frequency at which the electromagnetic wave must be propagated is approximately **208.42 kHz**.