The wavelength of the light used in Young's double slit experiment is `lambda`. The intensity at a point on the screen is I, where the path difference is `(lambda)/(6)`. If `I_(0)` denotes the maximum intensity, then the ratio of I and `I_(0)` is

Options

A. 0.866
B. 0.5
C. 0.707
D. 0.75

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understanding the relationship between intensity and phase difference
In Young's double slit experiment, the intensity \( I \) at a point on the screen is related to the maximum intensity \( I_0 \) and the phase difference \( \phi \) by the formula:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]

### Step 2: Calculate the phase difference
The phase difference \( \phi \) can be calculated using the path difference. The path difference \( \Delta x \) is given as \( \frac{\lambda}{6} \). The phase difference is given by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting the value of \( \Delta x \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Step 3: Substitute the phase difference into the intensity formula
Now we substitute \( \phi \) into the intensity formula:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Substituting \( \phi = \frac{\pi}{3} \):
\[ I = I_0 \cos^2\left(\frac{\pi}{3 \cdot 2}\right) = I_0 \cos^2\left(\frac{\pi}{6}\right) \]

### Step 4: Calculate \( \cos\left(\frac{\pi}{6}\right) \)
We know that:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \cos^2\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

### Step 5: Substitute back to find the ratio \( \frac{I}{I_0} \)
Now substituting this value back into the intensity equation:
\[ I = I_0 \cdot \frac{3}{4} \]
To find the ratio \( \frac{I}{I_0} \):
\[ \frac{I}{I_0} = \frac{3}{4} \]

### Conclusion
The ratio of the intensity \( I \) at the point on the screen to the maximum intensity \( I_0 \) is:
\[ \frac{I}{I_0} = \frac{3}{4} \]