If the path difference between the interfering waves is `n lambda`, then the fringes obtained on the screen will be

Options

A. dark
B. bright
C. coloured
D. white

Answer

To solve the problem, we need to analyze the relationship between the path difference of two interfering waves and the resulting interference pattern on the screen.

### Step-by-Step Solution:

1. **Understanding Path Difference**:
The path difference between two interfering waves is given as \( n \lambda \), where \( n \) is an integer and \( \lambda \) is the wavelength of the waves.

2. **Relating Path Difference to Phase Difference**:
The phase difference \( \phi \) between the two waves can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
where \( \Delta x \) is the path difference. Substituting \( \Delta x = n \lambda \):
\[ \phi = \frac{2\pi}{\lambda} (n \lambda) = 2n\pi \]

3. **Determining the Resultant Amplitude**:
The resultant amplitude \( R \) at the point of interference can be calculated using the formula:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \phi} \]
where \( A \) and \( B \) are the amplitudes of the two waves. Substituting \( \phi = 2n\pi \):
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(2n\pi)} \]

4. **Evaluating the Cosine Term**:
The cosine of \( 2n\pi \) is always equal to 1 (since \( \cos(2n\pi) = 1 \)):
\[ R = \sqrt{A^2 + B^2 + 2AB \cdot 1} = \sqrt{A^2 + B^2 + 2AB} = \sqrt{(A + B)^2} = A + B \]

5. **Interpreting the Resultant Amplitude**:
Since the resultant amplitude \( R \) is equal to \( A + B \), this indicates that the amplitude is at its maximum.

6. **Relating Amplitude to Intensity**:
The intensity \( I \) of the interference pattern is proportional to the square of the amplitude:
\[ I \propto R^2 \]
Therefore, if the amplitude is maximum, the intensity is also maximum, resulting in a bright fringe on the screen.

### Conclusion:
Since the path difference is \( n \lambda \), the fringes obtained on the screen will be bright fringes.