In a Young's double slit experiment the intensity at a point where tha path difference is `(lamda)/(6)` (`lamda` being the wavelength of light used) is I. If `I_0` denotes the maximum intensity, `(I)/(I_0)` is equal to

Options

A. `K`
B. `K//4`
C. `K//2`
D. `zero`

Answer

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

### Step-by-Step Solution:

1. **Understanding Path Difference and Phase Difference**:
The path difference \( \Delta d \) is given as \( \frac{\lambda}{6} \). The phase difference \( \phi \) corresponding to this path difference can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta d \]
Substituting the value of \( \Delta d \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

2. **Using the Intensity Formula**:
The intensity \( I \) at a point in a Young's double slit experiment is related to the maximum intensity \( I_0 \) and the phase difference \( \phi \) by the formula:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Substituting \( \phi = \frac{\pi}{3} \):
\[ I = I_0 \cos^2\left(\frac{\pi}{3 \cdot 2}\right) = I_0 \cos^2\left(\frac{\pi}{6}\right) \]

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

4. **Finding the Intensity Ratio**:
Now substituting back into the intensity equation:
\[ I = I_0 \cdot \frac{3}{4} \]
To find the ratio \( \frac{I}{I_0} \):
\[ \frac{I}{I_0} = \frac{3}{4} \]

### Final Answer:
Thus, the ratio \( \frac{I}{I_0} \) is equal to \( \frac{3}{4} \).