Two waves of nearly same amplitude , same frequency travelling with same velocity are superimposing to give phenomenon of interference . If `a_(1)` and `a_(2)` be their respectively amplitudes , `omega` be the frequency for both , `v` be the velocity for both and `Delta phi` is the phase difference between the two waves then ,

Options

A. the resultant intensity varies periodically with time and distance.
B. the resulting intensity with `(I_(min))/(I_(max)) = (a_(1) - a_(2))/(a_(1) + a_(2))^(2)` is obtained.
C. both the waves must have been travelling in the same direction and must be coherent.
D. `I_(R) = I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) cos ( Delta phi)`, where constructive interference is obtained for path difference that are odd multiple of `1//2 lambda` and destructive interference is obtained for path difference that are even multiple of `1//2 lambda`.

Explanation

To solve the problem regarding the interference of two waves, we will analyze the given information and derive the necessary expressions step by step.

### Step-by-Step Solution:

1. **Identify the Given Parameters**:
- Amplitudes of the two waves: \( a_1 \) and \( a_2 \)
- Frequency of both waves: \( \omega \)
- Velocity of both waves: \( v \)
- Phase difference between the two waves: \( \Delta \phi \)

2. **Understand the Resultant Amplitude**:
The resultant amplitude \( A_R \) of two superimposing waves can be expressed as:
\[ A_R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos(\Delta \phi)} \]

3. **Relate Amplitude to Intensity**:
The intensity \( I \) of a wave is proportional to the square of its amplitude:
\[ I \propto A^2 \]
Therefore, the intensities of the two waves can be expressed as:
\[ I_1 \propto a_1^2 \quad \text{and} \quad I_2 \propto a_2^2 \]

4. **Calculate Resultant Intensity**:
The resultant intensity \( I_R \) can be derived from the resultant amplitude:
\[ I_R = k A_R^2 \]
where \( k \) is a proportionality constant. Substituting for \( A_R \):
\[ I_R = k \left( a_1^2 + a_2^2 + 2a_1a_2 \cos(\Delta \phi) \right) \]

5. **Find Maximum and Minimum Intensities**:
- **Maximum Intensity \( I_{max} \)** occurs when \( \cos(\Delta \phi) = 1 \):
\[ I_{max} = k \left( a_1 + a_2 \right)^2 \]
- **Minimum Intensity \( I_{min} \)** occurs when \( \cos(\Delta \phi) = -1 \):
\[ I_{min} = k \left( a_1 - a_2 \right)^2 \]

6. **Calculate the Ratio of Intensities**:
The ratio of minimum to maximum intensity is:
\[ \frac{I_{min}}{I_{max}} = \frac{k (a_1 - a_2)^2}{k (a_1 + a_2)^2} = \frac{(a_1 - a_2)^2}{(a_1 + a_2)^2} \]

### Conclusion:
The ratio of the minimum intensity to the maximum intensity for the two waves is given by:
\[ \frac{I_{min}}{I_{max}} = \frac{(a_1 - a_2)^2}{(a_1 + a_2)^2} \]