If the path difference between the two light waves at a point is equal to intergral multiple of wavelength of then points appears as

Options

A. bright
B. dark
C. may be bright or dark
D. none of these

Answer

To solve the question, we need to understand the relationship between path difference, wavelength, and the resulting intensity of light at a point due to interference.

### Step-by-Step Solution:

1. **Understanding Path Difference**:
- The path difference (\( \Delta x \)) between two light waves at a point is defined as the difference in the distance traveled by the two waves to reach that point.

2. **Condition for Bright Fringes**:
- For constructive interference (which results in bright spots), the path difference must be an integral multiple of the wavelength (\( \lambda \)). This can be mathematically expressed as:
\[ \Delta x = n\lambda \quad (n = 0, 1, 2, 3, \ldots) \]

3. **Phase Difference**:
- The phase difference (\( \phi \)) corresponding to the path difference is given by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
- Substituting the path difference, we get:
\[ \phi = \frac{2\pi}{\lambda} (n\lambda) = 2n\pi \]

4. **Evaluating Cosine of Phase Difference**:
- The cosine of the phase difference is:
\[ \cos(2n\pi) = 1 \]
- This indicates that the resultant amplitude of the waves will be at its maximum.

5. **Resultant Amplitude and Intensity**:
- The intensity (\( I \)) of the resultant wave is proportional to the square of the amplitude. Since the amplitude is maximum when the phase difference is an integral multiple of \( 2\pi \), the intensity is also maximum at these points.
- Therefore, points where the path difference is an integral multiple of the wavelength will appear bright.

### Final Answer:
If the path difference between the two light waves at a point is equal to an integral multiple of the wavelength, then the points appear as **bright spots**.