The path difference between two interfering waves at a point on the screen is `lambda // 6`, The ratio of intensity at this point and that at the central bright fringe will be (assume that intensity due to each slit is same)

Options

A. 0.853
B. 8.53
C. 0.75
D. 7.5

Explanation

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 bright fringe.

### Step-by-step Solution:

1. **Determine the Phase Difference:**
The phase difference \( \Delta \phi \) between two waves can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Given that \( \Delta x = \frac{\lambda}{6} \), we substitute this into the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

2. **Calculate the Resultant Intensity:**
The resultant intensity \( I \) at a point where the phase difference is \( \Delta \phi \) is given by:
\[ I = 4I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
Here, \( I_0 \) is the intensity due to each slit. We need to find \( \cos^2\left(\frac{\Delta \phi}{2}\right) \):
\[ \Delta \phi = \frac{\pi}{3} \implies \frac{\Delta \phi}{2} = \frac{\pi}{6} \]
Thus,
\[ I = 4I_0 \cos^2\left(\frac{\pi}{6}\right) \]
We know that \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), so:
\[ \cos^2\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
Therefore,
\[ I = 4I_0 \cdot \frac{3}{4} = 3I_0 \]

3. **Determine the Maximum Intensity:**
The maximum intensity \( I_{\text{max}} \) at the central bright fringe (where the path difference is zero) is:
\[ I_{\text{max}} = 4I_0 \]

4. **Calculate the Ratio of Intensities:**
The ratio of the intensity at the point with path difference \( \frac{\lambda}{6} \) to the intensity at the central bright fringe is given by:
\[ \frac{I}{I_{\text{max}}} = \frac{3I_0}{4I_0} = \frac{3}{4} \]

### Final Answer:
The ratio of intensity at the point with path difference \( \frac{\lambda}{6} \) to that at the central bright fringe is \( \frac{3}{4} \) or 0.75.

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