In Young's double-slit experiment, the path difference between two interfering waves at a point on the screen is 13.5 times the wavelength. The point is

Options

A. Bright but not central bright
B. Neither bright not dark
C. Central bright
D. Dark

Answer

To solve the problem, we need to determine the type of fringe (bright or dark) at a point on the screen in Young's double-slit experiment where the path difference is given as 13.5 times the wavelength (λ).

### Step-by-Step Solution:

1. **Understanding Path Difference**:
In Young's double-slit experiment, the path difference (Δx) between the two waves arriving at a point on the screen determines whether the point is a bright or dark fringe. The path difference is given as Δx = 13.5λ.

2. **Identifying Conditions for Bright and Dark Fringes**:
- **Bright Fringe (Constructive Interference)**: Occurs when the path difference is an integer multiple of the wavelength, i.e., Δx = nλ, where n is an integer (0, 1, 2, ...).
- **Dark Fringe (Destructive Interference)**: Occurs when the path difference is an odd multiple of half the wavelength, i.e., Δx = (n + 0.5)λ, where n is an integer (0, 1, 2, ...).

3. **Analyzing the Given Path Difference**:
The given path difference is Δx = 13.5λ. We need to check if this value fits the criteria for bright or dark fringes.
- Since 13.5 is not an integer, it cannot represent a bright fringe (which requires Δx = nλ).
- We can express 13.5λ in terms of half-wavelengths:
\[ 13.5λ = 27 \cdot \frac{λ}{2} \quad \text{(since } 13.5 = 2 \cdot 13 + 1 \text{)} \]
This indicates that the path difference can be expressed as:
\[ \Delta x = (2n + 1) \frac{λ}{2} \quad \text{where } n = 13 \]
Thus, it satisfies the condition for destructive interference.

4. **Conclusion**:
Since the path difference of 13.5λ corresponds to destructive interference, the point on the screen is a dark fringe.

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

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