A sky wave with a frequency 55 MHz is incident on D-region of the earth.s atmosphere at `45^(@)`. The angle of refraction is (electron density for D-region is 400 electron/c.c)

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `15^(@)`

Answer

To solve the problem of finding the angle of refraction for a sky wave with a frequency of 55 MHz incident on the D-region of the Earth's atmosphere at an angle of 45 degrees, we can follow these steps:

### Step 1: Convert Electron Density
The electron density given is 400 electrons/cm³. We need to convert this to electrons/m³.

\[ \rho = 400 \text{ electrons/cm}^3 = 400 \times 10^6 \text{ electrons/m}^3 = 4 \times 10^8 \text{ electrons/m}^3 \]

### Step 2: Identify the Incident Angle
The angle of incidence \( i \) is given as 45 degrees.

### Step 3: Determine the Refractive Index of Air
The refractive index of air \( N_1 \) is approximately 1.

### Step 4: Calculate the Refractive Index of the D-region
Using the formula for the refractive index \( N_2 \) of the D-region:

\[ N_2 = N_1 \sqrt{1 - \frac{81.45 \times \rho}{f^2}} \]

Where:
- \( \rho = 4 \times 10^8 \text{ electrons/m}^3 \)
- \( f = 55 \text{ MHz} = 55 \times 10^6 \text{ Hz} \)

Substituting the values:

\[ N_2 = 1 \sqrt{1 - \frac{81.45 \times 4 \times 10^8}{(55 \times 10^6)^2}} \]

Calculating \( (55 \times 10^6)^2 \):

\[ (55 \times 10^6)^2 = 3025 \times 10^{12} = 3.025 \times 10^{15} \]

Now substituting back into the equation for \( N_2 \):

\[ N_2 = 1 \sqrt{1 - \frac{81.45 \times 4 \times 10^8}{3.025 \times 10^{15}}} \]

Calculating the fraction:

\[ \frac{81.45 \times 4 \times 10^8}{3.025 \times 10^{15}} = \frac{325.8 \times 10^8}{3.025 \times 10^{15}} = \frac{325.8}{3.025} \times 10^{-7} \approx 1.075 \times 10^{-7} \]

Thus,

\[ N_2 \approx 1 \sqrt{1 - 1.075 \times 10^{-7}} \approx 1 \text{ (since the term is very small)} \]

### Step 5: Apply Snell's Law
According to Snell's Law:

\[ N_1 \sin(i) = N_2 \sin(r) \]

Substituting \( N_1 = 1 \) and \( N_2 \approx 1 \):

\[ \sin(45^\circ) = \sin(r) \]

Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), we have:

\[ \sin(r) = \frac{\sqrt{2}}{2} \]

### Step 6: Conclusion
This means that the angle of refraction \( r \) is also 45 degrees.

\[ r = 45^\circ \]

### Final Answer
The angle of refraction is \( 45^\circ \).

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