In young double slit experiment, the intensity at a point where path difference is `lamda/6` is l. IF `l_0` denotes the maximum intensity `l/l_0`.

Options

A. `1/sqrt2`
B. `sqrt3/2`
C. `1/2`
D. `3/4`

Answer

To solve the problem, we need to determine the ratio of the intensity \( I \) at a point where the path difference is \( \frac{\lambda}{6} \) to the maximum intensity \( I_0 \) in the Young's double-slit experiment.

### Step-by-Step Solution:

1. **Identify the Path Difference**:
The path difference \( \Delta x \) is given as \( \frac{\lambda}{6} \).

2. **Calculate the Phase Difference**:
The phase difference \( \Delta \phi \) can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \Delta x = \frac{\lambda}{6} \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

3. **Use the Intensity Formula**:
The intensity \( I \) at a point in the Young's double-slit experiment is given by:
\[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
Substituting \( \Delta \phi = \frac{\pi}{3} \):
\[ I = I_0 \cos^2\left(\frac{\pi}{6}\right) \]

4. **Calculate \( \cos\left(\frac{\pi}{6}\right) \)**:
We know that:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \cdot \frac{3}{4} \]

5. **Find the Ratio \( \frac{I}{I_0} \)**:
Now, we can find the ratio:
\[ \frac{I}{I_0} = \frac{3}{4} \]

### Final Answer:
The ratio \( \frac{I}{I_0} \) is \( \frac{3}{4} \).

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