The path difference between two interfering waves at a point on screen is `70.5` tomes the wave length. The point is

Options

A. Dark
B. Bright
C. Not possible
D. Green in colour

Answer

To determine whether the point with a path difference of 70.5 times the wavelength corresponds to a bright or dark fringe, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Path Difference**:
The path difference between two interfering waves is given as \( \Delta x = 70.5 \lambda \), where \( \lambda \) is the wavelength of the waves.

2. **Identifying Conditions for Bright and Dark Fringes**:
- Bright fringes (maxima) occur when the path difference is an integer multiple of the wavelength:
\[ \Delta x = n \lambda \quad (n = 0, 1, 2, \ldots) \]
- Dark fringes (minima) occur when the path difference is an odd multiple of half the wavelength:
\[ \Delta x = \left(n + \frac{1}{2}\right) \lambda \quad (n = 0, 1, 2, \ldots) \]

3. **Analyzing the Given Path Difference**:
We can express the given path difference in terms of half-wavelengths:
\[ \Delta x = 70.5 \lambda = 141 \left(\frac{1}{2}\lambda\right) + \frac{1}{2}\lambda \]
This means that the path difference can be seen as:
\[ \Delta x = 141 \left(\frac{1}{2}\lambda\right) + \frac{1}{2}\lambda \]
Here, \( 141 \) is an integer, and the additional \( \frac{1}{2}\lambda \) indicates that the total path difference is an odd multiple of \( \frac{1}{2}\lambda \).

4. **Conclusion**:
Since the path difference \( 70.5 \lambda \) can be expressed as an odd multiple of \( \frac{1}{2}\lambda \), this means that at this point, destructive interference occurs, resulting in a dark fringe.

### Final Answer:
The point is a **dark fringe**.

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