A modulating wave of frequency `100 MHz` and amplitude `100V` is superimposed on a carrier wave of frequency `300 GHz` and amplitude `400V`. The value of modulating index and difference between the maximum frequency and minimum frequency of modulated wave are respectively :

Options

A. `0.25, 1xx10^(8)Hz`
B. `4.2xx10^(8)Hz`
C. `4.1xx10^(8)Hz`
D. `0.25xx10^(8)Hz`

Answer

To solve the problem, we need to find the modulating index and the difference between the maximum frequency and minimum frequency of the modulated wave.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency of the modulating wave, \( f_m = 100 \text{ MHz} = 100 \times 10^6 \text{ Hz} \)
- Amplitude of the modulating wave, \( A_m = 100 \text{ V} \)
- Frequency of the carrier wave, \( f_c = 300 \text{ GHz} = 300 \times 10^9 \text{ Hz} = 300 \times 10^3 \text{ MHz} \)
- Amplitude of the carrier wave, \( A_c = 400 \text{ V} \)

2. **Calculate the Modulating Index:**
The modulating index \( m \) is given by the formula:
\[ m = \frac{A_m}{A_c} \]
Substituting the values:
\[ m = \frac{100 \text{ V}}{400 \text{ V}} = 0.25 \]

3. **Calculate the Difference Between Maximum and Minimum Frequency:**
The maximum and minimum frequencies of the modulated wave can be calculated using the following formulas:
- Maximum frequency, \( f_{max} = f_c + f_m \)
- Minimum frequency, \( f_{min} = f_c - f_m \)

The difference between the maximum and minimum frequency is:
\[ f_{max} - f_{min} = (f_c + f_m) - (f_c - f_m) = 2f_m \]
Now substituting the value of \( f_m \):
\[ f_{max} - f_{min} = 2 \times 100 \text{ MHz} = 200 \text{ MHz} \]

4. **Convert to Hertz:**
To express the frequency difference in Hertz:
\[ 200 \text{ MHz} = 200 \times 10^6 \text{ Hz} = 2 \times 10^8 \text{ Hz} \]

### Final Answers:
- Modulating index \( m = 0.25 \)
- Difference between maximum and minimum frequency \( = 2 \times 10^8 \text{ Hz} \)