The path difference between two interfering waves at a point on the screen is `lambda"/"6`from central maximum. The ratio of intensity at this point and that at the central fringe will be

Options

A. `0.75`
B. `7.5`
C. `85.3`
D. `853`

Answer

To solve the problem, we need to find the ratio of the intensity at a point on the screen where the path difference is \( \frac{\lambda}{6} \) to the intensity at the central maximum.

### Step-by-Step Solution:

1. **Understanding the Intensity at Central Maximum:**
The intensity at the central maximum is denoted as \( I_{\text{max}} \).

2. **Finding the Phase Difference:**
The phase difference \( \phi \) can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \times \text{(path difference)} \]
Given the path difference is \( \frac{\lambda}{6} \):
\[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

3. **Calculating Intensity at the Point:**
The intensity at any point on the screen can be expressed as:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
Substituting the phase difference we found:
\[ I = I_{\text{max}} \cos^2\left(\frac{\pi}{3}/2\right) = I_{\text{max}} \cos^2\left(\frac{\pi}{6}\right) \]

4. **Finding \( \cos\left(\frac{\pi}{6}\right) \):**
We know that:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ I = I_{\text{max}} \left(\frac{\sqrt{3}}{2}\right)^2 = I_{\text{max}} \cdot \frac{3}{4} \]

5. **Calculating the Ratio of Intensities:**
The ratio of the intensity at the point to the intensity at the central maximum is:
\[ \frac{I}{I_{\text{max}}} = \frac{I_{\text{max}} \cdot \frac{3}{4}}{I_{\text{max}}} = \frac{3}{4} \]

6. **Final Answer:**
The ratio of intensity at the point where the path difference is \( \frac{\lambda}{6} \) to the intensity at the central maximum is:
\[ \frac{3}{4} \quad \text{or} \quad 0.75 \]