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 analyze the given information about the path difference in Young's double-slit experiment and determine the nature of the interference at that point on the screen.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The path difference (Δ) between the two waves at a point on the screen is given as 13.5 times the wavelength (λ).
- Mathematically, this can be expressed as:
\[ \Delta = 13.5 \lambda \]

2. **Convert Path Difference to a Fraction of Wavelength:**
- We can express this path difference in terms of half-wavelengths:
\[ \Delta = 13.5 \lambda = \frac{27}{2} \lambda \]

3. **Calculate the Phase Difference:**
- The phase difference (φ) corresponding to the path difference can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta \]
- Substituting the value of Δ:
\[ \phi = \frac{2\pi}{\lambda} \left(\frac{27}{2} \lambda\right) \]
- The λ cancels out:
\[ \phi = 27\pi \]

4. **Determine the Nature of the Fringe:**
- The phase difference of \(27\pi\) can be analyzed to determine if it corresponds to a bright or dark fringe.
- Since \(27\pi\) is an odd multiple of \(\pi\) (i.e., \(27 = 2n + 1\) for \(n = 13\)), it indicates that the point will be a dark fringe.

5. **Conclusion:**
- In Young's double-slit experiment, a phase difference of an odd multiple of \(\pi\) results in destructive interference, leading to a dark fringe.
- Therefore, the point on the screen is a dark fringe.

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

---